Lottery / Wheeling (long!)
Mat Newman
>
>>Barry.J...@brunel.ac.uk (Barry John Good) wrote:
>>>
>>>Does anybody understand what is meant by wheeling? I think it's some
>>>way of choosing sets of numbers on multiple boards to try to maximise
>>>one's chances of winning, but does anyone have any details as to the
>>>method?
>
>When you pick 7 numbers, you can form 7 different tickets with those
>numbers like you mention above. Then, if you purchase those 7 tickets
>in a Lotto 6/49 there will be exactly 432,824 tickets out of the
>possible 13,983,816 that will make you win something. This represents
>3.10%
>
>Now, the question that the first wheel inventor may have asked was
>probably something like: "Can I buy less than the 7 tickets and still
>have a 3.10% chance of winning?" It turns out that you can indeed.
>Actually, in this particular case you can select any 4 of the 7
>tickets and you will still win 3.10% of the time! This is a 3 on 3
>system since even if only 3 of the 7 numbers turn out to be on the
>ticket drawn you are guaranteed that at least one of your tickets
>will win a prize (at least a 3 on 6). But, since your chances of
>winning something are so small, why waste the extra 3 dollars?
>
[snip]
>
>I don't know if this is the minimum number of tickets needed to still
>guarantee a win when only 3 of the 12 numbers appear on the winning
>ticket since I did not "invent" it. I just checked that the claim
>was true.
[snip]
>There are all types of wheels available: 3 on 3, 3 on 4, 3 on 5,
>3 on 6, 4 on 5, 4 on 6, 5 on 6 and others that don't use this
>minimum win guarantee approach. Personally, I am interested in
>finding minimal 3 on 6 wheels for all numbers from 6 to 49. I
>have found them up to 15 and someone gave me a 3 on 6 for 18
>numbers. Anyone else have contributions?
>
I wrote a program to generate "good" sets of tickets, which is included
at the end of this post. In this sense I've answered a slightly different
question to that posed above. In the example you've got four tickets
and you've covered all triples in the range 1-7. However, given that
the lottery is on the range 1-49 you'd be better off buying the tickets
1,2,3,4,5,6 7,8,9,10,11,12 13,14,15,16,17,18 19,20,21,22,23,24
if you just want to increase your chances of winning *any* prize.
You can also use this program to generate the wheels you're after (but
see the disclaimer later!) by setting the range to be whatever you
want. The program outputs a set of tickets and a "cost", the cost
being defined as the number of triples of numbers in the specified
range that aren't covered by the ticket set.
If you run the program with the number of tickets set to 15 and the
range set to 12, it produces a set of tickets with zero cost, i.e. all
triples are covered. This is in keeping with the previous posters set
of 15 tickets in his 1-12 3 on 3 wheel. Setting the number of tickets
to 14, and the program only finds sets of tickets with a cost of 1, i.e.
there is some triple which isn't covered by the tickets. This suggests that
the number 15 is optimal for the given wheeling problem.
Notes:
Remove this header stuff and save the program as filename.c
Compile the program as gcc -o out_file_name filename.c -lm
The usage can be found by typing out_file_name -help
The algorithm used is a simulated annealing one (use for solving
all sorts of combinatorial optimization problems - I took this
one from a chip layout program I wrote last year) which is a
randomized algorithm (i.e. it gives different results each time you
run it!). However, with suitable choices of parameters (see later) it
will give good answers. So if the program does find a set of tickets, you
know it exists, but if it doesn't find a set with a certain cost, its
either because it can't be done, or else it just didn't find it!
The two command line parameters are #tickets and range, so running:
out_file_name 47 16
will (with enough luck and a good random number generator!) yield
a set of 47 tickets which covers all triples in the range [1..16].
Running
out_file_name 46 16
will produce a set of tickets with a cost of slightly more than zero. This
indicates that the program didn't find an "optimal" set, but changing the
run-time parameters might change this, who knows?
The parameters are (with current values) :
inner = 80;
control = 40.0;
threshold = 1.0;
inner is the base for the number of iterations per control value.
Control is the initial control parameter value, and threshold is the
stopping value for the control value. Making inner bigger will make the
program take longer, but the it's more likely to find the best
solution - no guarantees though (that was the disclaimer).
Things to do.
At the moment it tests how many triples aren't covered. For a true "I've
got $x to spend on tickets, what's the best set of tickets to buy" answer
it needs to be changed to test how many choices of 6 numbers don't win
you anything. Problem with this is there are lots of choices of 6 numbers!
Depending on how large the jackpot gets here, I may or may not implement
this change - if I do, would people like the new code?
Here is the program:
---------------------------Cut this and above----------------------------
/************************************************************************\
* Wheel generator (well, aproximately anyway). Mat Newman 5/12/94. *
* mne...@comlab.ox.ac.uk *
* *
* This program tries to generate the best set of 'no_tickets' tickets *
* to cover the range of numbers 1..range. Here 'best' is defined as *
* covering most triples. *
* *
* Feel free to tweak to your heart's content - but keep this message! *
* *
\************************************************************************/
/************************************************************************\
* P.S. To compile use "gcc -o out_file_name this_file.c -lm" since *
* without the `-lm' flag it won't link the maths library properly! *
\************************************************************************/
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#define max_range 50
#define max_tickets 100
#define TRUE 1
#define FALSE 0
int tck_num[max_tickets][6];
int best_tck[max_tickets][6];
int ticket[max_tickets][max_range];
int range=12;
int no_tickets=14;
int last_ticket, last_number, last_val;
double drand48();
double exp();
void print_tickets();
/*
A ticket has six numbers, tck_num[i][0]..tck[i][5], and we have
ticket[i][tck_num[i][j]] = 1, with all other ticket[i] entries 0.
*/
/* The 'cost' is the number of missing triples */
int cost()
{
int a,b,c,i;
int cost = 0;
for (a=0;a<range-2;a++)
for (b=a+1;b<range-1;b++)
for (c=b+1;c<=range-1;c++){
i = 0;
while(ticket[i][a]==0 || ticket[i][b]==0 || ticket[i][c]==0){
i++;
if (i==no_tickets){
cost++;
break;
}
}
}
return(cost);
}
/* Quickly (!) compute the change in cost due to the last move */
int delta_cost()
{
int a,b,c,i,j,k;
int delta = 0;
/* First count all triples no longer covered */
a = last_val;
for (j=0;j<5;j++)
for (k=j+1;k<6;k++){
if (j!=last_number && k!=last_number){
b = tck_num[last_ticket][j];
c = tck_num[last_ticket][k];
/* The triple abc was covered by the old ticket */
i = 0;
while(ticket[i][a]==0 || ticket[i][b]==0 || ticket[i][c]==0){
i++;
if (i==no_tickets){
delta++;
break;
}
}
}
}
/* Now count all newly covered triples */
a = tck_num[last_ticket][last_number];
for (j=0;j<5;j++)
for (k=j+1;k<6;k++){
if (j!=last_number && k!=last_number){
b = tck_num[last_ticket][j];
c = tck_num[last_ticket][k];
/* The triple abc is covered by the new ticket */
i = 0;
while(i==last_ticket || ticket[i][a]==0
|| ticket[i][b]==0 || ticket[i][c]==0){
i++;
if (i==no_tickets){
delta--;
break;
}
}
}
}
return(delta);
}
/* A move changes a number on one of the tickets */
void move()
{
/* Make a random move */
int tkt, num, rep, count=0;
tkt = (int) (drand48() * no_tickets);
num = (int) (drand48() * 6);
rep = (int) (drand48() * range);
while (ticket[tkt][rep]!=0){
rep = (int) (drand48() * range);
if (++count==20){
printf("Maybe stuck making a move! tkt %d num %d rep %d\n", tkt,num,rep);
print_tickets();
}
}
last_val = tck_num[tkt][num];
ticket[tkt][tck_num[tkt][num]] = 0;
tck_num[tkt][num] = rep;
ticket[tkt][rep] = 1;
last_ticket = tkt;
last_number = num;
}
void unmove()
{
/* Undo last move */
ticket[last_ticket][last_val] = 1;
ticket[last_ticket][tck_num[last_ticket][last_number]] = 0;
tck_num[last_ticket][last_number] = last_val;
}
/* Generate initial tickets */
void set_tickets()
{
int i,n,v;
for (i=0;i<no_tickets;i++)
for (n=0;n<range;n++)
ticket[i][n] = 0;
for (i=0;i<no_tickets;i++)
for (n=0;n<6;n++){
v = (int) (drand48() * range);
v=(n+i*6)%range;
while (ticket[i][v]==1)
v = (int) (drand48() * range);
tck_num[i][n] = v;
ticket[i][v] = 1;
}
}
void save_best()
{
int i,n;
for (i=0;i<no_tickets;i++)
for (n=0;n<6;n++)
best_tck[i][n] = tck_num[i][n];
}
void get_best()
{
int i,n;
for (i=0;i<no_tickets;i++){
for (n=0;n<range;n++)
ticket[i][n] = 0;
for (n=0;n<6;n++){
tck_num[i][n] = best_tck[i][n];
ticket[i][tck_num[i][n]] = 1;
}
}
}
/* Decide whether or not to accept a given move */
int accept (int delta, double control)
{
double change, dr;
if (delta <= 0) /* Always accept good moves */
return (TRUE);
if (control < 0.1) /* We'll never accept a bad move now */
return (FALSE);
change = ((double) delta)*5; /* !? */
dr = drand48();
if ( dr < exp(-change/control) )
return (TRUE);
else
return (FALSE);
}
double update_control (double control, int loop)
{
double alpha;
if (control > 30)
alpha = 0.88;
else
if (control > 5)
alpha = 0.92;
else
if (control > 3)
alpha = 0.95;
else
if (control > 1)
alpha = 0.98;
else
alpha = 0.9;
return (control * alpha);
}
/* Compute the inner loop iteration value */
int calc_inner (int inner, double control)
{
int variable;
if (control > 20)
variable = inner;
else
if (control > 10)
variable = inner*10;
else
if (control > 2)
variable = inner*(inner-1)/2;
else
if (control > 1.5)
variable = inner*20;
else
variable = inner*5;
return (variable);
}
void print_tickets()
{
int i,n;
for (i=0;i<no_tickets;i++){
printf("Ticket %2d : ",i+1);
for (n=0;n<6;n++)
printf("%2d ",tck_num[i][n]+1);
printf("\n");
}
}
/* Comparison function for qsort */
int comp(int *i, int *j)
{
if ((*i)<(*j))
return(-1);
if ((*i)==(*j))
return(0);
return(1);
}
void sort_tickets()
{
int i,n;
for (i=0;i<no_tickets;i++){
qsort(tck_num[i], 6, sizeof(int), (int (*)(void*,void*)) comp);
}
}
/* Use simulated annealing to find good ticket sets. */
int main(int argc, char *argv[])
{
int inner, loop_count, iteration, no_improvement, print=1;
int inner_mod, inner_count;
double control, threshold;
int best_cost, curr_cost, delta;
long time_secs;
if (argc>1)
no_tickets = atoi(argv[1]);
if (argc>2)
range = atoi(argv[2]);
if (no_tickets<1 || no_tickets>max_tickets || range<6 || range > max_range){
fprintf(stderr,"Usage %s [no_tickets [range]]\n\n", argv[0]);
fprintf(stderr,"Where 1 <= no_tickets <= %d and 6 <= range <= %d\n",
max_tickets, max_range);
exit(1);
}
time_secs = (long) time(NULL);
srandom (time_secs);
srand48 (time_secs);
inner = 80;
control = 40.0;
threshold = 1.0;
printf("Trying to find a wheel for the range [1..%d] using %d tickets.\n",
range, no_tickets);
if (print)
printf("Using inner %d control %7.6g threshold %5.4g\n",
inner, control, threshold);
set_tickets();
save_best();
loop_count = 0;
iteration = 0;
no_improvement = 0;
best_cost = curr_cost = cost();
while ((control > threshold) || (no_improvement < 5)){
inner_mod = calc_inner (inner, control);
if (print && (loop_count>0) && (loop_count%5==0))
{
printf ("%3d Inner %4d Control %7.6g Current %4d Best %4d\n",
loop_count, inner_mod, control, curr_cost, best_cost);
}
inner_count = 0;
while (inner_count <= inner_mod){
move();
delta = delta_cost();
if (accept(delta, control)){
curr_cost += delta;
if (curr_cost < best_cost){
best_cost = curr_cost;
save_best();
no_improvement = 0;
if (best_cost==0){
control = threshold;
inner_count = inner_mod;
}
}
}
else{
unmove();
}
inner_count++;
iteration++;
}
control = update_control(control, loop_count);
loop_count++; no_improvement++;
}
get_best();
sort_tickets();
print_tickets();
printf("Final cost is %d.\n", cost());
}
Angus Walker
> range set to 12, it produces a set of tickets with zero cost, i.e. all
> triples are covered. This is in keeping with the previous posters set
> of 15 tickets in his 1-12 3 on 3 wheel. Setting the number of tickets
> to 14, and the program only finds sets of tickets with a cost of 1, i.e.
> there is some triple which isn't covered by the tickets. This suggests that
> the number 15 is optimal for the given wheeling problem.
I seem to be missing some intervening posts here, but this seems to have
been generated by my original lottery question. I have recently noticed
that if you find a set of tickets with zero cost (as defined above) for
the range 1-24, then copy this set into the range 25-48 by adding 24 to
each number, you will then have a set of tickets which will guarantee
winning in the range 1-49, since any six numbers chosen between 1 and 49
must either have at least three in the range 1-24 and/or at least three in
the range 25-48.
If someone could run the program (I don't have a C compiler) and find the
least number of tickets with zero cost in the range 1-24 I would be very
grateful.
--
Angus Walker
Andrew Halket
>I wrote a program to generate "good" sets of tickets, which is included
>at the end of this post. In this sense I've answered a slightly different
>question to that posed above. In the example you've got four tickets
>and you've covered all triples in the range 1-7. However, given that
>the lottery is on the range 1-49 you'd be better off buying the tickets
>1,2,3,4,5,6 7,8,9,10,11,12 13,14,15,16,17,18 19,20,21,22,23,24
>if you just want to increase your chances of winning *any* prize.
No, I'm afraid it is all a load of bollocks - I suppose you are one of those
who believes in the non-existent "law of averages". Every number has an equal
chance of coming up and every group of six has an equal chance of 3 to 6
numbers coming out of it. Just show me the statistics to prove otherwise.
The only good set of numbers is a set that others are less likely to pick, so
maximising your winnings if those come up. However, those numbers (in whole
or part) are no more likely to come up. And frankly I couldn't be buggered
whether I win 1 million or 6 million.
Andrew
========================================================================
Mail: Andrew Halket #####
The Technology Partnership Ltd #####
Melbourn, Royston #####
Herts. SG8 6EE #####
Tel: +44 763 262626 #####
Fax: +44 763 261582 THE TECHNOLOGY PARTNERSHIP
EMail: ar...@techprt.co.uk #####
========================================================================
robert and stimets
Andrew Halket <ar...@techprt.co.uk> wrote:
>
>>I wrote a program to generate "good" sets of tickets, which is included
>>at the end of this post. In this sense I've answered a slightly different
>>question to that posed above. In the example you've got four tickets
>>and you've covered all triples in the range 1-7. However, given that
>>the lottery is on the range 1-49 you'd be better off buying the tickets
>>1,2,3,4,5,6 7,8,9,10,11,12 13,14,15,16,17,18 19,20,21,22,23,24
>>if you just want to increase your chances of winning *any* prize.
>
>No, I'm afraid it is all a load of bollocks - I suppose you are one of those
>who believes in the non-existent "law of averages". Every number has an equal
>chance of coming up and every group of six has an equal chance of 3 to 6
>numbers coming out of it. Just show me the statistics to prove otherwise.
Sorry. You have all the "bollocks" today.
The fact of the matter is that one does maximize one's chances of
winning by buying *disjoint* sets of tickets. That is, if one purchases
(say) 4 tickets, then the way to maximize the chances of winning _something_
is to make sure there is no overlap of numbers between tickets.
If you do, indeed, purchase 4 such tickets, then your chances of winning
are as good as anyone else who purchased 4 disjoint tickets, but better
than anyone who purchased 4 tickets with any amount of overlap.
RStimets
--
/----/ /----/ /----\ /----/ /----\ /----/
/ / / / / /____ / / /____
/ -/ / / /----< / / / /
/----/ /----/ / / /----/ /-----/ /----/
Mat Newman
>
>I seem to be missing some intervening posts here, but this seems to have
>been generated by my original lottery question. I have recently noticed
>that if you find a set of tickets with zero cost (as defined above) for
>the range 1-24, then copy this set into the range 25-48 by adding 24 to
>each number, you will then have a set of tickets which will guarantee
>winning in the range 1-49, since any six numbers chosen between 1 and 49
>must either have at least three in the range 1-24 and/or at least three in
>the range 25-48.
>
>If someone could run the program (I don't have a C compiler) and find the
>least number of tickets with zero cost in the range 1-24 I would be very
>grateful.
>
>--
>Angus Walker
Neat idea! I left my "wheeling" program running whilst I was out this
evening, and when I got back it had found a set of 173 tickets which form
a 3 from 3 wheel for the range 1-24, so as you point out this implies a set
of 346 tickets would guarantee a win in the British lottery. This number
(346) is almost certainly just an upper bound and not optimal, but it's
better than any other specific sets I've seen (as opposed to non-constructive
estimates).
It took the program 147 runs with 173 tickets before it found a zero-cost
set, so it may well be that there are such sets with less than 173 tickets.
I'll leave the program running tonight to see if it gets any further.
There are two things I could do with these numbers :
1) Post them if there's enough interest
2) Start a lottery ticket service along the lines of people email me how
many tickets they want, and I send them numbers from the set. This way
(if enough people play) at least someone will win!
Mat.
PS I tried [and failed:-(] modifying my "cost" function to count how many
choices of 6 numbers a given set of tickets wins with (well, I can do
it the "look through all 14m combinations" way, but that takes far too
long), so this doesn't look to be a viable way of finding the true
minimum number of tickets needed (unless anyone can donate a few hours
of Cray time!).
Andrew Halket
>>>the lottery is on the range 1-49 you'd be better off buying the tickets
>>>1,2,3,4,5,6 7,8,9,10,11,12 13,14,15,16,17,18 19,20,21,22,23,24
>>>if you just want to increase your chances of winning *any* prize.
>>
>>No, I'm afraid it is all a load of bollocks - I suppose you are one of those
>>who believes in the non-existent "law of averages". Every number has an equal
>>chance of coming up and every group of six has an equal chance of 3 to 6
>>numbers coming out of it. Just show me the statistics to prove otherwise.
>Sorry. You have all the "bollocks" today.
>The fact of the matter is that one does maximize one's chances of
>winning by buying *disjoint* sets of tickets. That is, if one purchases
>(say) 4 tickets, then the way to maximize the chances of winning _something_
>is to make sure there is no overlap of numbers between tickets.
>If you do, indeed, purchase 4 such tickets, then your chances of winning
>are as good as anyone else who purchased 4 disjoint tickets, but better
>than anyone who purchased 4 tickets with any amount of overlap.
Mmmmm. I'll have to think about this. What you say is in fact what most
people (including me) will do out of instinct and it will certainly increase
the chances of picking some of the right numbers (if you had nine tickets you
would get all 49 numbers somewhere). However, these numbers are just as
likely to be one on each of six tickets as 3 on one ticket (or is that more
likely?!)
I think that it is a case of instinct above pure statistics here but I'll have
to think about it (immortal words from HHGTTG?)
Mat Newman
I wrote a program to generate "good" sets of tickets, which is included
at the end of this post. In this sense I've answered a slightly different
question to that posed above. In the example you've got four tickets
and you've covered all triples in the range 1-7. However, given that
the lottery is on the range 1-49 you'd be better off buying the tickets
1,2,3,4,5,6 7,8,9,10,11,12 13,14,15,16,17,18 19,20,21,22,23,24
if you just want to increase your chances of winning *any* prize.
You can also use this program to generate the wheels you're after (but
see the disclaimer later!) by setting the range to be whatever you
want. The program outputs a set of tickets and a "cost", the cost
being defined as the number of triples of numbers in the specified
range that aren't covered by the ticket set.
If you run the program with the number of tickets set to 15 and the
range set to 12, it produces a set of tickets with zero cost, i.e. all
triples are covered. This is in keeping with the previous posters set
of 15 tickets in his 1-12 3 on 3 wheel. Setting the number of tickets
to 14, and the program only finds sets of tickets with a cost of 1, i.e.
there is some triple which isn't covered by the tickets. This suggests that
the number 15 is optimal for the given wheeling problem.
Notes:
out_file_name 47 16
out_file_name 46 16
Things to do.
Here is the program:
---------------------------Cut this and above----------------------------
return(delta);
}
void save_best()
{
int i,n;
void get_best()
{
int i,n;
return (control * alpha);
}
return (variable);
}
void print_tickets()
{
int i,n;
void sort_tickets()
{
int i,n;
inner_count = 0;
while (inner_count <= inner_mod){
move();
delta = delta_cost();
Newsgroups: rec.puzzles,sci.math,uk.misc
Subject: Re: Lottery / Wheeling (long!)
Summary:
Expires:
References: <1994Dec7.1...@client44.comlab.ox.ac.uk> <arbh.141...@techprt.co.uk> <3c7o5m$q...@usenet.ucs.indiana.edu>
Sender:
Followup-To:
Distribution:
Organization: Oxford University Computing Laboratory
Keywords:
In article <3c7o5m$q...@usenet.ucs.indiana.edu> rsti...@silver.ucs.indiana.edu (robert and stimets) writes:
>In article <arbh.141...@techprt.co.uk>,
>Andrew Halket <ar...@techprt.co.uk> wrote:
>>
>>>I wrote a program to generate "good" sets of tickets, which is included
>>>at the end of this post. In this sense I've answered a slightly different
>>>question to that posed above. In the example you've got four tickets
>>>and you've covered all triples in the range 1-7. However, given that
>>>the lottery is on the range 1-49 you'd be better off buying the tickets
>>>1,2,3,4,5,6 7,8,9,10,11,12 13,14,15,16,17,18 19,20,21,22,23,24
>>>if you just want to increase your chances of winning *any* prize.
>>
>>No, I'm afraid it is all a load of bollocks - I suppose you are one of those
>>who believes in the non-existent "law of averages". Every number has an equal
>>chance of coming up and every group of six has an equal chance of 3 to 6
>>numbers coming out of it. Just show me the statistics to prove otherwise.
What you've said is true, but how you apply it the the problem in hand
is totally wrong.
>
>Sorry. You have all the "bollocks" today.
Big hairy ones by the sounds of things...
>The fact of the matter is that one does maximize one's chances of
>winning by buying *disjoint* sets of tickets. That is, if one purchases
>(say) 4 tickets, then the way to maximize the chances of winning _something_
>is to make sure there is no overlap of numbers between tickets.
>
>If you do, indeed, purchase 4 such tickets, then your chances of winning
>are as good as anyone else who purchased 4 disjoint tickets, but better
>than anyone who purchased 4 tickets with any amount of overlap.
>
>RStimets
(Thanks for the support)
Yes, one shouldn't douse oneself in petrol before flaming :-)
Suppose you bought two tickets, both of which have the numbers
1,2,3,4,5,6. You've got a 1 in 54 chance of winning a prize (i.e.
exactly the same as if you'd just bought one ticket). True, if you
do win a prize, you win more than if you'd just bought one ticket, but
the question asked for *any* prize. Now suppose you buy two disjoint
tickets, then the chance of winning any prize is 1 in 26.85. This seems
better than 1 in 54 to me, but I could be talking bollocks...
Mat.
Mat Newman
newsserver went down and appended both the articles into a dead.article
and posted it later :-(
For those of you who (sensibly) bailed out before the end...
Two tickets, same numbers on each => 1 in 54 chance of winning
Two tickets, disjoint numbers => 1 in 27 chance of winning
Conclusion : Andrew Halket <ar...@techprt.co.uk> was talking gonads.
Mat.
P.S. 312 tickets are enough for a guaranteed win on the lottery.
Dave Dodson
Mat Newman <mne...@comlab.ox.ac.uk> wrote:
>
>P.S. 312 tickets are enough for a guaranteed win on the lottery.
It seems to me that 312 tickets are enough for a guaranteed loss on almost
all of them. Buying multiple tickets may increase your chances of winning,
but it decreases your expected return per dollar "invested."
----------------------------------------------------------------------
Dave Dodson dod...@convex.com
Convex Computer Corporation Richardson, Texas (214) 497-4234
Seth Breidbart
Dave Dodson <dod...@convex.COM> wrote:
>In article <1994Dec9.1...@client44.comlab.ox.ac.uk>,
>Mat Newman <mne...@comlab.ox.ac.uk> wrote:
>>
>>P.S. 312 tickets are enough for a guaranteed win on the lottery.
>
>It seems to me that 312 tickets are enough for a guaranteed loss on almost
>all of them.
Nope; in the common state lotteries, you'd have to buy millions of
tickets to have a guaranteed loss. If you buy only a few, you might
win.
(Yes, I know the words "win" and "loss" are being used in very
different senses here. If you can't deal with intentional ambiguity,
I hope you're not reading this thread in rec.puzzles.)
Seth
Dave Dodson
But you will agree that if you buy 312 state lottery tickets, you will
expect to lose on almost all of them, won't you?!? That is what I said.
Mat Newman
>In article <1994Dec9.1...@client44.comlab.ox.ac.uk>,
>Mat Newman <mne...@comlab.ox.ac.uk> wrote:
>>
>>P.S. 312 tickets are enough for a guaranteed win on the lottery.
>
>It seems to me that 312 tickets are enough for a guaranteed loss on almost
>all of them. Buying multiple tickets may increase your chances of winning,
>but it decreases your expected return per dollar "invested."
>
Absolutely - I was just trying to answer the question "what's the least
number of tickets you need to buy to guarantee winning a prize?". Once
you factor in the cost of the ticket you are onto a loosing proposition!
I've now got the number down to 308 tickets - I'll post on Monday how much they
would have won if I'd bought them in this weeks lottery - that should put
any hopeful punters off actually doing it in practise!
I think my poor PC is at it's limit in trying to solve this problem, what we
need is someone with access to a massively parallel machine...
Mat.
David Karr
>all of them.
True, but entirely beside the point. The question is merely whether the
payoff on the *winning* tickets can be expected to exceed the cost of
all the tickets. (I'm fairly sure the answer to this question is "no.")
>Buying multiple tickets may increase your chances of winning,
>but it decreases your expected return per dollar "invested."
I don't believe this. There are two kinds of prizes in lotteries that
I know of:
(a) Prizes of fixed amount, e.g. you get $1 back if your ticket
matches on three numbers. The contribution of such prizes to the
expected value of each ticket is fixed in the lottery rules and is
completely independent of what other tickets you or anyone
else may have bought.
(b) "Jackpot" type prizes whereby some amount may be split among all
the tickets that match the required pattern. The expected value
of such a ticket *does* depend on other tickets that may have been
bought. You have no control over whether other people buy tickets
that might take a share of your jaackpot, but if you buy two
tickets yourself that could both match for a jackpot (e.g., two
identical tickets), the jackpot contributes only once rather than
twice to the expected value of your purchase.
So as long as you avoid tickets that could simultaneously hit a single
jackpot on two or more tickets (and I suspect the "wheeling" schemes
do), the expected value of N tickets is exactly N times the expected
value of a ticket bought singly. Which is of course less than the
cost of the tickets in most cases; that's why lotteries are profitable
for those who run them.
-- David A. Karr (ka...@cs.cornell.edu)
-- http://www.cs.cornell.edu/Info/People/karr/home.html
Seth Breidbart
Dave Dodson <dod...@convex.COM> wrote:
>In article <3casgp$9...@panix3.panix.com>,
>Seth Breidbart <se...@panix.com> wrote:
>>In article <3ca8ir$5...@wagner.convex.com>,
>>Dave Dodson <dod...@convex.COM> wrote:
>>>In article <1994Dec9.1...@client44.comlab.ox.ac.uk>,
>>>Mat Newman <mne...@comlab.ox.ac.uk> wrote:
>>>>
>>>>P.S. 312 tickets are enough for a guaranteed win on the lottery.
>>>It seems to me that 312 tickets are enough for a guaranteed loss on almost
>>>all of them.
>>Nope; in the common state lotteries, you'd have to buy millions of
>>tickets to have a guaranteed loss. If you buy only a few, you might
>>win.
>>(Yes, I know the words "win" and "loss" are being used in very
>>different senses here. If you can't deal with intentional ambiguity,
>>I hope you're not reading this thread in rec.puzzles.)
>But you will agree that if you buy 312 state lottery tickets, you will
>expect to lose on almost all of them, won't you?!? That is what I said.
Well, in that case, I can buy 3 tickets and be guaranteed to lose on
at least one of them. With 5 tickets, I can be guaranteed to lose on
a majority of them.
Seth
Dave Dodson
I was talking about what you call "jackpot" type prizes. Suppose your
lottery returns 50 cents on the dollar. If the jackpot is $1 million
after you have bought one ticket, it will be $1.0000005 million after you
have bought two tickets. If there are N different lottery tickets,
assuming your two tickets were different:
With only one ticket: you have 1/N chance of winning ($1 million - $1),
so your expected return is $999,999/N and your expected return per dollar
"invested" is $999,999/N.
With two tickets: you have 2/N chances of winning ($1000000.50 - $2),
so your expected return is $999,998.50 * 2/N and your expected return per
dollar "invested" is $999,998.50/N.
Which has the better expected return per dollar "invested?"
Andrew Halket
>. However, given that
>>>>the lottery is on the range 1-49 you'd be better off buying the tickets
>>>>1,2,3,4,5,6 7,8,9,10,11,12 13,14,15,16,17,18 19,20,21,22,23,24
>>>>if you just want to increase your chances of winning *any* prize.
>>>
>>>No, I'm afraid it is all a load of bollocks - I suppose you are one of those
>>>who believes in the non-existent "law of averages". Every number has an equal
>>>chance of coming up and every group of six has an equal chance of 3 to 6
>>>numbers coming out of it. Just show me the statistics to prove otherwise.
>>Sorry. You have all the "bollocks" today.
>>The fact of the matter is that one does maximize one's chances of
>>winning by buying *disjoint* sets of tickets. That is, if one purchases
>>(say) 4 tickets, then the way to maximize the chances of winning _something_
>>is to make sure there is no overlap of numbers between tickets.
>>If you do, indeed, purchase 4 such tickets, then your chances of winning
>>are as good as anyone else who purchased 4 disjoint tickets, but better
>>than anyone who purchased 4 tickets with any amount of overlap.
>Mmmmm. I'll have to think about this. What you say is in fact what most
>people (including me) will do out of instinct and it will certainly increase
>the chances of picking some of the right numbers (if you had nine tickets you
>would get all 49 numbers somewhere). However, these numbers are just as
>likely to be one on each of six tickets as 3 on one ticket (or is that more
>likely?!)
>I think that it is a case of instinct above pure statistics here but I'll have
>to think about it (immortal words from HHGTTG?)
Monday morning and I've thought about it. You are right on the basis that you
stand slightly more chance of winning a prize, but the average value of your
prize will be correspondingly slightly lower. You don't stand any more chance
of winning the jackpot though and an infintessimally bigger chance of winning
some of the middle prizes. You will win the Ł10 very slightly more often with
non-overlapping numbers but very occasionally I will win multiple Ł10 wins
with my overlapping numbers (and this is how the two approaches compensate).
There is no way to improve your overall chances of winning (except for picking
non-common numbers which some lucky bastard has just done).
An illustration - suppose the lottery is pick two numbers from 5. I choose
1,2, and 1,3. You choose 1,2 and 3,4. our winnings are
I win You win
1,2
1,3
1,4
1,5
2,3
2,4
2,5
3,4
3,5
4,5
Andrew Halket
>Sorry for the unexpectedly long message I posted last time - our
>newsserver went down and appended both the articles into a dead.article
>and posted it later :-(
>For those of you who (sensibly) bailed out before the end...
>Two tickets, same numbers on each => 1 in 54 chance of winning
>Two tickets, disjoint numbers => 1 in 27 chance of winning
>Conclusion : Andrew Halket <ar...@techprt.co.uk> was talking gonads.
..but if I do win, I win twice as much as you!
Why is everyone so fussed about slightly increasing their chances of winning
£10. No one plays to win £10, or £100 for that matter. £17,808,003 will do
me very nicely and the only way to improve your chances of winning that is to
buy more tickets - selcction of numbers on those tickets by some fancy bit of
software won't help one iota!
Andrew
David Karr
>In article <3ca8ir$5...@wagner.convex.com> dod...@convex.COM (Dave Dodson) writes:
>>Buying multiple tickets may increase your chances of winning,
>>but it decreases your expected return per dollar "invested."
>
>value of a ticket bought singly.
OK, Mr. Dodson has shown me my error. Two distinct tickets bought in
the same lottery cannot both win the jackpot (at least in a typical
case in a typical lottery). So in my simplistic analysis (I assumed
the jackpot was fixed before the tickets for this particular lottery
were sold, but I'm not that familiar with the rules of any real
lottery), these tickets have slightly less expected value than the
same two tickets bought one at a time in two different lotteries
(where a double jackpot hit *is* possible)---and it is the latter
expected value that is twice the value of one ticket.
If a part of the tickets you buy is plowed back into the jackpot of
the same running of the lottery (as opposed to, say, the *next*
running of the lottery as I'd assumed), then this will offset the
effect of "no double hits" slightly, but as Mr. Dodson pointed out,
it will offset it completely only if 100% of your tickets' cost
is added to the jackpot value.
In fact, my impression was that the amount added to the jackpot was a
lot less than 30 cents on the dollar, and moreover, in a realistic
analysis you ought to take into account also that the real value of
the jackpot is a lot less than the advertised value, due to the tax
laws and the obnoxious payout schemes that are typically employed.
All of this is pretty much irrelevant to "wheeling" schemes or most
other multiple-distinct-ticket schemes, however. Unless you buy an
enormous number of tickets, the effect of the lack of multiple hits is
so slight that only in rare borderline cases would it distinguish
between the tickets being a "good investment" or a "bad investment."
The deciding question, it seems to me, is whether the expectation of
winning the jackpot can justify even one ticket purchase; if it can,
then it seems it would almost certainly justify a rather large number
of distinct ticket purchases in the same running of the lottery.
I can even see a possible benefit of a "wheeling" scheme. The
strategy adds absolutely nothing to your expected value vs. any other
set of N distinct tickets. In fact it lowers your chance of making
any real amount of money off the non-jackpot prizes. But for someone
who is playing for the jackpot and not for the lesser prizes, this
is good---we have lowered the risk associated with the lesser prizes,
and increased the odds of having enough cash left over to buy a lot
of tickets in next week's lottery (when, who knows, the jackpot may
be even larger).
Note that all the above is analyzed in the context of "*if* it ever
makes any sense to play the lottery at all." I'm not advocating any
real course of action.
Normand Veilleux
>From: mne...@comlab.ox.ac.uk (Mat Newman)
>Ok, here's how my set of 308 tickets faired in this weeks lottery :
[stats, etc. deleted]
I've been searching for ways of finding the minimal set of tickets
required to guarantee a win (3 on 6 or above) in a Lotto 6/49. I
saw a very interesting post here a few months ago about L(n,m,t,k).
WHERE: n is the total amount of numbers,
m is the amount of numbers drawn,
t is the amount of numbers on a purchased ticket, and
k is the amount of numbers to match
I'm not a mathematician, so I had a hard time figuring what it was
all about. Nonetheless, I managed to find a book that contained
several wheeling systems. After some time I realized I could
combine some of those wheels to generate a cover for a Lotto 6/49.
I realized that I could combine L(26, 3, 6, 3) with L(23, 3, 6, 3)
to be guaranteed a win for L(49, 6, 6, 3) if you use the numbers
1 to 26 in the first set and the numbers 27 to 49 in the second
set. Well, this was quite a shock since this generated exactly
234 tickets that were indeed a cover.
I later realized that both sets won when the 6 draw numbers are
split 3 - 3. Since this didn't look like "normal behaviour" for
a minimal set, I looked into it a bit further and found that I could
combine L(26, 3, 6, 3) with L(22, 3, 6, 3) to form another cover
for L(49, 6, 6, 3). The worse possible case here happens when
the 49th number is drawn. This means you are left with 5 numbers
to guarantee a win. But, no matter how you divide these 5 numbers
among the two sets, there will always be at least 3 in one of the
sets. Hence, this is also a cover. And it only takes 207 tickets!
This sounds rather similar to what you have been doing.
Normand Veilleux
>From: Dave Dodson <dod...@convex.COM>
>I was talking about what you call "jackpot" type prizes. Suppose your
>lottery returns 50 cents on the dollar. If the jackpot is $1 million
>after you have bought one ticket, it will be $1.0000005 million after you
>have bought two tickets. If there are N different lottery tickets,
>assuming your two tickets were different:
>
>With only one ticket: you have 1/N chance of winning ($1 million - $1),
>so your expected return is $999,999/N and your expected return per dollar
>"invested" is $999,999/N.
>
>With two tickets: you have 2/N chances of winning ($1000000.50 - $2),
>so your expected return is $999,998.50 * 2/N and your expected return per
>dollar "invested" is $999,998.50/N.
>
>Which has the better expected return per dollar "invested?"
Aren't you forgetting something ? What if my neighbour also buys a
ticket. Wouldn't that add to my expected return, but not to my cost!?
What if 20,000,000 other people bought tickets too ?
Normand Veilleux
>From: ar...@techprt.co.uk (Andrew Halket)
>Why is everyone so fussed about slightly increasing their chances of winning
>$10. No one plays to win $10, or $100 for that matter. $17,808,003 will do
>me very nicely and the only way to improve your chances of winning that is to
>buy more tickets - selection of numbers on those tickets by some fancy bit of
>software won't help one iota!
Elementary my dear Andrew! :-)
Do you realize that millions of people will buy Lotto tickets all their
lives and never win the jackpot! I agree with you that buying more tickets
will give you a better chance at winning the jackpot. But, for those of us
not fortunate enough to win the jackpot, it makes sense to try and select
tickets that increase our odds of winning smaller prizes.
The debate still rages whether this is possible or not. Some people claim
wheeling is the best thing since hot cakes, while others claim that
theoretically, two sets of tickets that contain the same amount of tickets
will have the same expected return.
Who's right, who's wrong !? I don't know, but wheeling does seem to have
potential, in my opinion. At least from what I've seen so far.
Dave Dodson
Yes, other people buying tickets will increase your expected return (after
all, that's where that $1 million prize in my example came from) assuming
there aren't so many tickets sold that the winner will have to share the
jackpot with other bettors using the same number, but the expected return
of your second and subsequent tickets is still less than the expected
return of your first ticket.
The Famous Druid
In article <arbh.149...@techprt.co.uk>, Andrew Halket (ar...@techprt.co.uk) writes:
>
>Why is everyone so fussed about slightly increasing their chances of winning
>£10. No one plays to win £10, or £100 for that matter. £17,808,003 will do
>me very nicely and the only way to improve your chances of winning that is to
>buy more tickets - selcction of numbers on those tickets by some fancy bit of
>software won't help one iota!
>
I have a way to substantially increase your chances of winning a
few million. Just take one pound with you to one of those fancy
casinos in Monte Carlo, and bet it on the black, keep doubling
your stake until you've won 4 and a bit million (the most likely
first prize in the National VAT supplement). This requires 22
consecutive blacks to come up, much more likely than picking 6
from 49, and the 'house percentage' at a casino is a lot lower.
----------------------------------------------------------------------------
dr...@stonehenge.win-uk.net Just bang the rocks together guys
"It's OK to do the right thing... as long as you don't get caught."
-- The Lone Contractor
----------------------------------------------------------------------------
Mat Newman
>
>The debate still rages whether this is possible or not. Some people claim
>wheeling is the best thing since hot cakes, while others claim that
>theoretically, two sets of tickets that contain the same amount of tickets
>will have the same expected return.
>
>Who's right, who's wrong !? I don't know, but wheeling does seem to have
>potential, in my opinion. At least from what I've seen so far.
>
Consider the simple example of buying two tickets :
Case A: Ticket 1 = Ticket 2 = 1,2,3,4,5,6
Case B: Ticket 1 = 1,2,3,4,5,6 Ticket 2 = 7,8,9,10,11,12
What are the expected returns? (Where expected return = sum over
all prizes of prize value * probability of winning that prize)
Case A
------
P(Match 3) = 6C3*43C3/49C6 = 246,820/13,983,816 = 1.765%
Prize(Match 3) = 2*10 = 20
Expected return from Match 3 = 0.35
P(Match 4) = 6C4*43C2/49C6 = 13,545/13,983,816 = 0.0969%
Prize(Match 4) = 2*M4
P(Match 5) = 252/13,983,816
Prize(Match 5) = 2*M5
P(5+Bonus) = 6/13,983,816
Prize(5+Bonus) = 2*M5B
P(Match 6) = 1/13,983,816
Prize(Match 6) = 2*M6
Where M6 is one ticket's worth of the share of the Match 6 prize (so
if one other ticket matched 6, you'd get 2/3 of the Jackpot).
Case B
------
P(Match 3 on one ticket) = 2*6C3*[6C2*37C1+6C1*37C2+6C0*37C3]/49C6
= 492,840/13,983,816 = 3.524%
Prize(Match 3 on one ticket) = 10
P(Match 3 on both tickets) = 6C3*6C3/6C49 = 400/13,983,816
Prize(Match 3 on both tickets) = 2*10 = 20
Expected return from Match 3 = 0.35
(This is *exactly* the same as the case A for 10 pound prizes)
P(Match 4) = 2*13,545/13,983,816 = 2*0.0969%
Prize(Match 4) = M4
P(Match 5) = 2*252/13,983,816
Prize(Match 5) = M5
P(5+Bonus) = 2*6/13,983,816
Prize(5+Bonus) = M5B
P(Match 6) = 2/13,983,816
Prize(Match 6) = M6
So it would seem that the expected return is identical. However, we have
yet to account for variable prizes. If all the prizes were of fixed value
then the expected returns for Cases A and B would indeed be identical.
If there are no shared prizes then the expected return for Case A would be
less than that for Case B. Why? Well, look at the Jackpot. In Case A you
have a one in 13,983,816 chanve of winning the lot, whereas in Case B you
have a two in 13,983,816 of winning the lot, so your expected return
from the Jackpot prize is twice as much in Case B as in Case A. A similar
reasoning applies to the lesser prizes too.
If you share the Match 6 prize with k other tickets, then in Case A
the Match 6 expected return is :
[Jackpot Value/(k+2)]*2*1/13,983,816
whereas the Case B expected return is :
[Jackpot Value/(k+1)]*2/13,983,816
and since 1/(k+2) < 1/(k+1) we see that again the expected return in
Case B is higher than that in case A.
The argument that you should overlap tickets because "when you win you
win more" as published e.g. in the Daily Mail is plainly flawed, because
it doesn't account for the decrease in probability that you do actually
win. When you combine both probability of winning and value of winnings
you see the true picture.
Mat.
Angus Walker
> split 3 - 3. Since this didn't look like "normal behaviour" for
> a minimal set, I looked into it a bit further and found that I could
> combine L(26, 3, 6, 3) with L(22, 3, 6, 3) to form another cover
> for L(49, 6, 6, 3). The worse possible case here happens when
> the 49th number is drawn. This means you are left with 5 numbers
> to guarantee a win. But, no matter how you divide these 5 numbers
> among the two sets, there will always be at least 3 in one of the
> sets. Hence, this is also a cover. And it only takes 207 tickets!
This is very useful. I would like to know how your magical L function works!
More immediately, how do you know that L(26,3,6,3),L(22,3,6,3) is the
best pair, and not, for example L(24,3,6,3),L(24,3,6,3), which
would intuitively seem better, or even L(42,3,6,3),L(6,3,6,3), which is
the other limiting case. What are the results for these?
--
Angus Walker
Andrew Halket
>Elementary my dear Andrew! :-)
>Do you realize that millions of people will buy Lotto tickets all their
>lives and never win the jackpot! I agree with you that buying more tickets
>will give you a better chance at winning the jackpot. But, for those of us
>not fortunate enough to win the jackpot, it makes sense to try and select
>tickets that increase our odds of winning smaller prizes.
>The debate still rages whether this is possible or not. Some people claim
>wheeling is the best thing since hot cakes, while others claim that
>theoretically, two sets of tickets that contain the same amount of tickets
>will have the same expected return.
>Who's right, who's wrong !? I don't know, but wheeling does seem to have
>potential, in my opinion. At least from what I've seen so far.
I now believe that it is in fact elementary now :-)
Buying tickets without overlap will increase your chances slightly of winning
a small prize. Buying tickets with overlap will decrease your chances
slightly of winning but on a few occasions you would win a couple of small
prizes. The product of (number of wins X amount of win) will (and has to)be
the same.
For example I could pick 1,2,7,8,9,10 and 1,2,11,12,13,14 where someone else
would pick 1,2,3,4,5,6 and 7,8,9,10,11,12. if the numbers 1,2,7,11,15,16 came
up I would win £20 whereas someone else would win £0. It is the few cases
like this that would even out the two approaches.
With my (admittedly limited and rusty) statistics knowledge I have to conclude
that the expected overall return from any sets of tickets must be equal in the
long term. Of course there is money to be made from convincing dim old Joe
Public that there is something in wheeling. Perhaps whoever it was on the net
who advocates wheeling with that program could run a few weeks of his 308 or
312 selections from it against a random 308/312 selections and see which
combination wins more?
Normand Veilleux
>From: An...@angus.demon.co.uk (Angus Walker)
>> From nvei...@emr1.emr.ca (Normand Veilleux)
>[...]
>> I later realized that both sets won when the 6 draw numbers are
>> split 3 - 3. Since this didn't look like "normal behaviour" for
>> a minimal set, I looked into it a bit further and found that I could
>> combine L(26, 3, 6, 3) with L(22, 3, 6, 3) to form another cover
>> for L(49, 6, 6, 3). The worse possible case here happens when
>> the 49th number is drawn. This means you are left with 5 numbers
>> to guarantee a win. But, no matter how you divide these 5 numbers
>> among the two sets, there will always be at least 3 in one of the
>> sets. Hence, this is also a cover. And it only takes 207 tickets!
>
>This is very useful. I would like to know how your magical L function works!
OK, let me find the post that explained it and re-post it. I did not
invent the L function, I just pushed it to new uncharted limits :-)
(at least, as far as I know!)
>More immediately, how do you know that L(26,3,6,3),L(22,3,6,3) is the
>best pair, and not, for example L(24,3,6,3),L(24,3,6,3), which
>would intuitively seem better, or even L(42,3,6,3),L(6,3,6,3), which is
>the other limiting case. What are the results for these?
You're quite right in that:
L(49, 6, 6, 3) <= L(n, 3, 6, 3) + L(48-n, 3, 6, 3) for 6 <= n <= 24
I had originaly started with:
L(49, 6, 6, 3) <= L(n, 3, 6, 3) + L(49-n, 3, 6, 3) for 6 <= n <= 24
but, like I previously explained I saw that the first equation was
sufficient to guarantee a cover.
So the value of n that generates the lowest sum will be the best
upper bound. I did not have access to all the wheels needed to
check all 19 possibilites, but I did find a book that allowed me
to check some values of n at the top of the range.
I did not generate any of the wheels myself. I just used the
wheels from the wheeling book and combined them to produce the cover.
The book I saw contained 3 on 3 wheels for 22, 23, 24 and 26 numbers:
Number | No. of Tickets
| in wheel
---------|-----------------
22 | 77
23 | 104
24 | 119
25 | not in book
26 | 130
I don't know if all of these are minimal, but 22 and 26 certainly
are. Those two contain every 3-tuple exactly once! From this you
can see that using 24 and 24 would generate a cover with 238 tickets,
yet using 23 and 26, the cover only requires 234 tickets and using 22
and 26 the cover can be brought down to 207 tickets!
Quite an improvement over the 308 cover without any computing on
my part, except for verifying the wheels and the cover by brute force.
Now the real question: Is 207 a minimal cover ? I have no proof either
way yet, but I strongly suspect it is not. Mind boggling! :->
___ _______
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88' `Y8b,__,gd8P""' Normand Veilleux `""Y8bg,_
d8P `Y88888P"' nvei...@emr1.emr.ca `"Y88ba
Normand Veilleux
Here is the post I was taking about. I shrunk it a bit to include
only the stuff that applied directly to L(n,m,t,k) and to generating
a cover for L(49,6,6,3) and L(50,6,6,3).
-----------------------------------------------------------
From: Timothy Y. Chow <tyc...@math.mit.edu>
Date: Wed, 5 Oct 94 17:35:55 EDT
Subject: Re: Lotto 6/49 Questions
Following is some material from old articles on the subject of Lotto.
:In article <42...@bbn.COM> la...@alexander.bbn.com (Larry Denenberg) writes:
:>If there are N balls in all, and you as a player can choose m numbers
:>on each card, and the lottery authorities then choose n balls, define
:>L(N,n,m,k) as the minimum number of cards you must purchase to ensure that
:>at least one of your cards will have at least k numbers in common with the
:>balls chosen in the lottery. So, for example, Matthew Fletcher asks for
:>L(20,2,4,2) [and L(20,2,3,2)].
:
:L(20,2,3,2) = 67.
:L(20,2,4,2) = 35 or 36.
:
:The lower bounds come from the standard theory of Steiner coverings:
:In the first case, restrict attention to the cards in the covering
:and the sets balls that contain the number 1. The sets of balls
:are pairs of the form (1,n), 2<=n<=20, so there are 19 of them.
:Each card contains 2 other numbers besides one, and we see that
:we need at least 10 cards that contain 1 to cover all ball sets that
:contain 1. Therefore the density of the covering (the average number
:of times that each ball set is covered) is at least 20/19. Since
:there are 190 ball sets total, there must be at least 200/3 cards
:total. But this is not an integer, so we need at least 67 cards.
:
:From this argument we may extract a general principle:
:
:L(N,k,n,k) >= Ceiling((N-choose-k)/(n-choose-k)*
: (n-1-choose-k-1)/(N-1-choose-k-1)*L(N-1,k-1,n-1,k-1))
: = Ceiling(N/n*L(N-1,k-1,n-1,k-1))
:
:Therefore L(20,2,4,2) >= 35 also.
:
:For the upper bounds, enumerate the balls 0 through 19 in the case of
:L(20,2,3,2), and write out all cards with three distinct numbers which
:add to zero modulo 20. I leave it as an exercise that there are 57
:such cards, and they cover all pairs of balls except those of the form
:(i,j) such that 2*i+j = 0 mod 20. j must be even, and for each choice
:of j, there are two choices of i which differ by 10. So for the
:remaining 10 cards, choose (j,18-j/2,28-j/2 mod 20) for all even j.
:
:For the case of L(20,2,4,2), enumerate the balls 1 through 16 and
:a through d and view them as a group of 16 and a group of 4. There
:exists a perfect covering of pairs of balls in the first group, that is
:L(16,2,4,2) = (16-choose-2)/(4-choose-2) = 20. For the moment we don't
:have to know what it is. For all pairs of balls with at least one
:member not in the first group, we see that the 16 cards:
:
: 1, 2,a,b; 1, 2,c,d
: 3, 4,a,c; 3, 4,b,d
: 5, 6,a,d; 5, 6,b,c
: 7, 8,a,b; 7, 8,c,d
: 9,10,a,b; 9,10,c,d
:11,12,a,b;11,12,c,d
:13,14,a,b;13,14,c,d
:15,16,a,b;15,16,c,d
:
:cover all of them. These 16 cards, plus the 20 that cover the
:first 16 balls, make a covering with a total of 36 cards.
:
:Where does the perfect cover in the case of L(16,2,4,2) come from?
:Consider the set of four elements {0,1,w,w^2}. Define addition
:by 1+1=0, 1+w = w^2, w+w = 0; and multiplication by w^3 = 1, 1*x = x,
:0*x = 0, etc. Addition and multiplication are associative and commutative
:and multiplication is distributive. Non-zero elements have a reciprocal.
:Enumerate the 16 balls as pairs of elements in this set. Now define a
:line to be the set of balls (x,y) that satisfies a linear
:equation a*x+b*y = c. It is not too hard to show that:
:
:1) Lines have four points.
:2) Every pair of points is contained in a unique line.
:
:Thus the set of lines is the desired set of cards.
:
:Remark: A projective plane of order n is a set of n*(n+1) cards that
:form a cover in the case of L(n^2+n+1,2,n+1,2). It was recently shown
:with a lot of hard math and a Cray that there is no projective plane of
:order 10.
:
:
:From: gr...@manifold.berkeley.edu (Greg Kuperberg)
:Message-ID: <1162c6...@agate.berkeley.edu>
:Date: 10 Jun 92 23:14:14 GMT
:
:The question is: The German lottery (suspiciously like all
:other lotteries) has 49 numbers and the winning ticket has 6
:numbers and your tickets also have 6 numbers. How many tickets do
:you need to *guarantee* that you match 3 numbers? According
:to Stephan Neuhaus, some German company claims you can do it in 150.
:
:Chris Long and Arthur Rubin claim that you need at least 54
:tickets, presumably on the grounds that each ticket meets
:x winning tickets in at least 3 numbers, there are
:y winning tickets total, and y/x > 53.
:
:Bill Mayne, using a SPARCstation, found a set of 456 tickets which are
:sufficient to match 3. Paul Zimmermann apparently also used a computer
:program and he found a solution with 323 tickets.
:
:This problem is the same idea as one proposed in this forum by Larry
:Denenberg a few years back, except that he used the Massachusetts
:lottery, which at the time had 36 numbers and not 49. But the tickets
:still have 6 numbers and he wanted to match 3 numbers.
:
:Recycling my ideas for that problem, I can show that you need at least
:84 tickets and 224 tickets suffice. For the lower bound, we note that
:each winning ticket represents 20 triples, so you can take a bound for
:the question "How many triples do you need to choose to guarantee that
:each winning ticket contains at least one of them?" and divide by 20.
:Reversing the question in quotes, we want to find a family of 46-uples
:in a set of size 49 that contain every 43-uple. This is the classical
:Steiner covering problem: A set of k-tuples that contain every t-tuple
:in a set of size v is a (v,k,t) covering system. So-and-so's bound
:says that a (v,k,t) covering system is at least v/k times as big as the
:smallest (v-1,k-1,t-1) covering system (why?). In this fashion you get
:an inductive bound, and you get to round up to the next integer at each
:step. My computer says:
:
:Bound for 7, 4, 1: 2
:Bound for 8, 5, 2: 4
:Bound for 9, 6, 3: 6
:Bound for 10, 7, 4: 9
:...
:Bound for 47, 44, 41: 1466
:Bound for 48, 45, 42: 1564
:Bound for 49, 46, 43: 1666
:
:Thus you need at least 1666/20 = 83.3 tickets, or 84 really.
:
:For the upper bound, it's simpler if we assume 50 numbers instead of 49
:(I will call them 0 through 49), and for this problem I have a system
:with 235 tickets (I can save 11 tickets by nuking one of the numbers
:and rearranging about 40 of the tickets.) First, I need a (10,4,2)
:covering with 9 4-tuples (if I had one with 8 4-tuples I could save
:another 25 tickets). Apparently 0 1 2 3, 0 4 5 6, 0 7 8 9, 1 4 7 8,
:1 5 6 9, 2 5 6 7, 2 4 8 9, 3 5 6 8, and 3 4 7 9 do the job.
:
:So for each a,b,c,d in the (10,4,2) system, and for each i and j
:from 0 to 4, make the ticket
:
:a+10i, b+10i, c+10i, d+10i, 2j+10i+10 mod 50, 2j+1+10i+10 mod 50
:
:In addition, you need the tickets 10i, 10i+1, 10i+2, 10i+3, 10i+4, 10i+5
:and 10i+4, 10i+5, 10i+6, 10i+7, 10i+8, 10i+9 for each i from 0 to 4.
:That makes 9*5*5+10 = 235 tickets. Why it works is left as an exercise
:for the reader.
[...]
-----------------------------------------------------------
BTW, this last part is false. The 235 ticket "cover" is not a cover.
I modified my brute force analyser to accept 50 numbers instead 49
and it discovered lots and lots of tickets that were not covered.
The set covers exactly 15,425,055 tickets out of the total of
15,890,700 possibilities. So, there's about 465,000 tickets that
are missing.
Here are the first few tickets that are not covered by the set:
00 01 06 07 20 21
00 01 06 07 20 .
00 01 06 07 20 .
00 01 06 07 20 .
00 01 06 07 20 39
00 01 06 07 21 22
00 01 06 07 21 .
00 01 06 07 21 .
00 01 06 07 21 .
00 01 06 07 21 39
I did modify the set a little bit here and there and managed to
increase the amount of tickets covered, but was unable to find
a true cover by this method.
It didn't really matter much since by using
L(50,6,6,3) <= L(23,3,6,3) + L(26,3,6,3) (see previous
post for logic)
I was able to find a cover for a Lotto 6/50 that required
only 234 tickets. So, basically the conclusion was right
(235 is enough to guarantee a cover) but the set was wrong.
And this could be something as simple as a couple of typos,
who knows!?
Normand Veilleux
>If you share the Match 6 prize with k other tickets, then in Case A
>the Match 6 expected return is :
>
>[Jackpot Value/(k+2)]*2*1/13,983,816
>
>whereas the Case B expected return is :
>[Jackpot Value/(k+1)]*2/13,983,816
>
>and since 1/(k+2) < 1/(k+1) we see that again the expected return in
>Case B is higher than that in case A.
Sounded convincing, even looked convinving, but somehow I was convinced
there was a mistake in there somewhere! I stared at the equations for
a good 10 minutes before I had a flash: you're comparing apples and
oranges like the saying goes. You're comparing k+1 winning tickets
and k+2 winning tickets. That's not the same amount of winners!
Instead, assume a fixed amount of winners, say W, then in Case A
the Match 6 expected return is:
[Jackpot Value/W]*2 * (1/13,983,816)
and in Case B expected return is:
[Jackpot Value/W]*1 * (2/13,983,816)
They look the same to me now! :-)
Even then, who says both tickets in Case B would have shared with
W-1 other winners ? Perhaps the other ticket would not have shared
at all, or perhaps it shares with more winners! Are you confused
yet ? I am! :-( Can someone clear this up ?
David Karr
>
>>If you share the Match 6 prize with k other tickets, [...]
>>and since 1/(k+2) < 1/(k+1) we see that again the expected return in
>>Case B is higher than that in case A.
>
>Sounded convincing, even looked convinving, but somehow I was convinced
>there was a mistake in there somewhere! I stared at the equations for
>a good 10 minutes before I had a flash: you're comparing apples and
>oranges like the saying goes. You're comparing k+1 winning tickets
>and k+2 winning tickets. That's not the same amount of winners!
>Are you confused yet ? I am! :-( Can someone clear this up ?
Well, this discussion certainly has gotten confusing, but Mat Newman
isn't comparing apples and oranges at all; he's using very simple and
fundamental assumptions of decision theory.
Suppose I buy lottery ticket number 1,2,3,11,12,13. I have no way
of knowing how many other people bought this ticket. Maybe none did.
Maybe ten such tickets were bought by others. Likewise, the vast
majority of people who had the opportunity to purchase 1,2,3,11,12,13
had no idea what tickets I bought, so my purchasing decisions have
no influence on theirs.
What this means is that if I look at the results of last week's
lottery and there was a jackpot split among 3 winning tickets, it is
reasonable for me to say, "If I'd bought that combination too, the
prize would have been split 4 ways." It's reasonable to say, "If I'd
bought *two* tickets with that combination, the prize would have been
split 5 ways." It is *not* reasonable to say, "If I'd bought two
tickets with that combination, the prize would still have been split 3
ways." That's saying that my purchase would have caused two other
total strangers to change their minds. (This is assuming, of course,
that you don't know any of the winners. If you do, why are you
wasting your time here instead of asking to borrow money from them?)
A similar situation holds for the *next* drawing of the lottery,
except this time there are some variables. First, I don't know what
the winning combination will be. Second, I don't know how many other
people will buy tickets with the winning combination. But I do know
that number is out of my control. (Well, not entirely. I could take
out an advertisement saying, "I'm buying ticket 1,2,3,11,12,13 so
don't take that combination if you don't want to have to split the
jackpot with me." But I've never heard of anyone doing this, so let's
assume I don't do it.)
OK, so whatever combination I pick, other people will buy k tickets
identical to mine. I don't know what k is; it might be 0; it might be
99; but it's out of my control. That means that if I decide to buy 1
ticket, there will be k+1 tickets in the lottery with that combination
(some of them possibly bought after I bought mine, it doesn't matter
---all that matters is how many are sold before the deadline); if I
buy 2 tickets, there will then be k+2 identical tickets (including
mine). That means that *if* I happen to win the jackpot on that ticket
(which has odd of 1-in-whatever), I will get either 1/(k+1) of the
jackpot or 2/(k+2) of the jackpot, depending on whether I bought
1 or 2 tickets with that combination.
Now, clearly a 1-in-whatever shot at 2/(k+2) is not worth twice as
much as a 1-in-whatever shot at 1/(k+1), even though it would cost me
twice as much. Especially if you consider that 2/(k+2) = 1/(k+1) for
k = 0 (which, assuming I wisely pick unpopular combinations, is a
reasonably likely occurrence). My extra dollar (or pound) would be
much better spent on another ticket with a different unpopular
combination. The chances given by these two distinct tickets *would*
be worth twice as much (or near enough for all practical intents) as
the chance given by one ticket.
Now this says you should not buy any duplicate tickets. But it
doesn't say whether you should "wheel." I wrote before that I see a
possible advantage in "wheeling" to someone who wanted to control the
risk associated with the small fixed prizes. In other words, you buy,
say, 230 tickets that pretty much ensure that you'll make a loss
unless you hit a really big prize, but the loss is of a more
predictable magnitude than if you just bought 230 random distinct
tickets. IMHO, however, I still don't see that this gets you anything
worth the effort. After all, suppose you mess up and forget to buy
one of the tickets, and it happens that the winning combination
matches 3 on the ticket you forgot. So you lose 229 dollars from
buying 229 losing tickets, instead of 230 tickets of which just one
wins you back a dollar. (Or maybe it's two dollars. Big deal.)
Angi Long
>If a part of the tickets you buy is plowed back into the jackpot of
>the same running of the lottery (as opposed to, say, the *next*
>running of the lottery as I'd assumed), then this will offset the
>effect of "no double hits" slightly, but as Mr. Dodson pointed out,
>it will offset it completely only if 100% of your tickets' cost
>is added to the jackpot value.
I don't know about other states, but in WA the jackpot for the week
is fixed before any tickets for that week are sold, and (some portion
of) the ticket sale revenues are added to the next week's jackpot.
>Note that all the above is analyzed in the context of "*if* it ever
>makes any sense to play the lottery at all." I'm not advocating any
It makes sense if you see the lottery as a voluntary donation to the
state with the added bonus that there's a slight chance you might win.
It doesn't make sense to see it as a pure "investment."
Andrew Halket
>>
>I have a way to substantially increase your chances of winning a
>few million. Just take one pound with you to one of those fancy
>casinos in Monte Carlo, and bet it on the black, keep doubling
>your stake until you've won 4 and a bit million (the most likely
>first prize in the National VAT supplement). This requires 22
>consecutive blacks to come up, much more likely than picking 6
>from 49, and the 'house percentage' at a casino is a lot lower.
a lot , lot lower. they only win on the 0 on roulette. If it weren't there
then it would all be evens.
John S. McGowan
(RE roulette)
>
> a lot , lot lower. they only win on the 0 on roulette. If it weren't there
They don't win on 0 (or double 00 or triple 000 depending on the wheel,
he originals having but a single 0). (you can bet on 0) What they do is
figure the payback based upon fewer possibilities that there are. (like
having a six sided die... play a dollar to play, if your umber comes up I
pay you your dollar back and four more... not five more!) They simply
lower the count of the number of possibles by the number of 0 or 00 or
000 spaces... but they do not win on those... they win on everything (on
the average) (of course, if no one ever bet on 0 or 00 or 000 THEN it
would be fair except for when they come up... but you can bet on them)
--
John S. McGowan | jmcg...@bigcat.missouri.edu [COIN] (preferred)
| j.mcg...@genie.geis.com [GEnie]
----------------------------------------------------------------------
David Karr
[On the topic of "wheeling" schemes for selecting multiple lottery tickets:]
>I have a way to substantially increase your chances of winning a
>few million. Just take one pound with you to one of those fancy
>casinos in Monte Carlo, and bet it on the black, keep doubling
>your stake until you've won 4 and a bit million [...]
OK, let's consider ways to make big bucks at a casino instead of
at Lotto.
Part of the problem is I believe any real casino will cut you off long
before you have a chance to win 4 million in one night. But suppose
you find a casino that will let you win up to, let's say, $100,000 at
the roulette wheel before they show you the door. And let's say you
walk in there with $100 in your pocket.
So you sit down at that roulette wheel with the motto, "100,000 or
bust." That is, you will play until either you reach your goal of
winning $100,000 or you lose all the money you're carrying.
What's the chance that you will reach your goal?
If the roulette wheel were perfectly fair with no house percentage (no
0 or 00) and you just bet $100 on black every time until you're done,
it's fairly easy to show that you have exactly 1/1000 chance of
reaching your goal. The reason is that the expected value of each
play is 0, and (1/1000) * 100,000 = 100.
But if there's a house percentage, your chance is obviously less.
Suppose the wheel has 0 and 00, so your chance to win on black is only
18/38, not 1/2. According to simple random walk theory, if you bet
$100 on black every time your chance to reach your goal is
(1 - 20/18)/(1 - (20/18)^1000) < 10^-46
so you should just forget about that strategy. On the other hand, if
you put down your $100 on black and just let it ride, your chance of
reaching $100,000 (actually $102,400) is
(18/38)^10 = 0.57 * 10^-3
so the house percentage only reduces the chance of reaching your
goal by 43% in comparison to a wheel that gives even odds.
Now the puzzle: What is the best strategy for reaching $100,000, and
what is your chance to reach it?
Obviously you can slightly improve your odds by taking some of the
extra chips out of the last bet so you can start over in case you
lose. But how much does this improve the odds, and is there another
betting protocol that does even better?
Normand Veilleux
>From: ka...@cs.cornell.edu (David Karr)
>Well, this discussion certainly has gotten confusing, but Mat Newman
>isn't comparing apples and oranges at all; he's using very simple and
>fundamental assumptions of decision theory.
Thanks for the extra explanations. I'm not a mathematician and I
know nothing about decision theory. Although I think I understand
what you're saying, I still have my doubts.
>A similar situation holds for the *next* drawing of the lottery,
>except this time there are some variables. First, I don't know what
>the winning combination will be. Second, I don't know how many other
>people will buy tickets with the winning combination. But I do know
>that number is out of my control. (Well, not entirely. I could take
>out an advertisement saying, "I'm buying ticket 1,2,3,11,12,13 so
>don't take that combination if you don't want to have to split the
>jackpot with me." But I've never heard of anyone doing this, so let's
>assume I don't do it.)
>
>OK, so whatever combination I pick, other people will buy k tickets
>identical to mine. I don't know what k is; it might be 0; it might be
>99; but it's out of my control. That means that if I decide to buy 1
>ticket, there will be k+1 tickets in the lottery with that combination
>(some of them possibly bought after I bought mine, it doesn't matter
>---all that matters is how many are sold before the deadline); if I
>buy 2 tickets, there will then be k+2 identical tickets (including
>mine). That means that *if* I happen to win the jackpot on that ticket
>(which has odd of 1-in-whatever), I will get either 1/(k+1) of the
>jackpot or 2/(k+2) of the jackpot, depending on whether I bought
>1 or 2 tickets with that combination.
My view seems just as valid as this, but I can't seem to pin-point
why it differs from yours. In my case, the number of winning tickets
remains the same, say W, and either I win 1 part of the jackpot or I
win 2 parts, depending if I bought 2 different or 2 identical tickets.
What I find really strange about the above equations is that for a
fixed value of k you are comparing two different amounts of winning
tickets. For a specific draw, k will have a certain value for some
players, but a different value for other players. On the other hand,
W will be the same for all players.
On a similar note, someone posted a table in sci.math, not too long
ago, that showed a Poisson distribution of the expected number of
winners for certain amounts of tickets sold. I don't have the table,
but it showed that the probability of having exactly 2 winners is not
the same as the probability of having exactly 3 winners in a draw.
BTW, how does the decision theory explain the case where I buy W+1
identical tickets ? From my point of view, even if I had lost the
tickets and forgotten which 6 numbers I had bought, I would still
know that I had not won the jackpot because if I had, there would
have been at least W+1 winners! What is the value of k in such a
case?
The closest I can come to understand the differences is by asking
a question. In your case you seem to ask, "What can I expect if
I purchase the same winning ticket as the declared winners of the
draw ?" Whereas, I'm asking, "What can I expect if I am among the
declared winners with the tickets I purchase ?"
Does this make any sense to you ? I'm not even sure the questions
are right!
Tapani Lindgren
[You play roulette starting with $100. Numbers 0 and 00 are included.]
...
>Now the puzzle: What is the best strategy for reaching $100,000, and
>what is your chance to reach it?
The house has at least 2/38 edge on every round, so the fewer rounds
it takes to reach your goal, the better are your chances. Unfortunately,
the best-paying bets have a larger edge, e.g. if you play $1 on one number
only and win, you don't have $36 as you should, but only $30-$32, depending
on the casino.
If the casino pays 31 to 1 or better, you need only win two jackpots to
reach your goal. Your chances are (1/38)^2 = .06925% . I believe most
casinos pay less; e.g., if they pay 30 to 1, you will end up with only
31*31*$100 = $96100, and you must keep playing to win that extra $3900.
Your chances to win $3900 when you have ¤96100 are about 90% (at least
1-(26/38)^6)), so your overall chances are at least .06215% .
The highest-paying bets that are almost fair are those that pay 2 to 1:
any of the columns or any of the consecutive dozens (1-12, 13-24 or 25-36).
Winning six times in a row on one of these bets gives you $100 * 3^6
= $72900 with the probability (12/38)^6 = .09917% . If you are
allowed to play two columns at a time, paying 1 to 2, play ¤54200 and you
have a 24/38 chance of having exactly $100000. If you don't make it,
you still have $18700 and you may triple and double it with probability
(12/38)*(18/38) = 14.96% . Your overall chances are therefore at least
(12/38)^6 * (1 - (1-24/38)*(1-(12/38)*(18/38))) = .06810% . If you are not
allowed to play two columns, try to triple $13550 and, if you fail, $20325
and finally ¤30488. You will reach $100000 unless you lose three times in
a row (p = (26/38)^3). Overall chances are therefore at least
(12/38)^6 * (1 - (26/38)^3) = .06741% .
Chance Strategy
.06925% Bet all on one number twice, if it pays 31-to-1 or better.
.06810% Bet all on a column six times, then $54200 on two columns.
.06741% Bet all on a column six times, then smaller amounts until you win.
.06215% Bet all on one number twice, if it pays 30-to-1. Then play
untill you win $3900 more or run out of money.
.05687% Bet all on black and win ten times in a row.
Tapani
MARALYN sMITH
AS THE NEXT WINNER WITH A BIT OF LUCK SO ALL THE BEST TO YOU !!!
MARALYN SMITH
Maralyn Smith,
"Rimast Promotions for Business and Personal Startup"
mjs...@ibm-win.uk.net
My Disclaimer.. For and on Behalf of myself.. I can spell, it's just
that my keyboard STICKS!! if you believe that you'll believe anything
A STRESSED OUT FEMALE !!!!
The Famous Druid
In article <3csu9s$h...@eis.calstate.edu>, John S. McGowan (jom...@eis.calstate.edu) writes:
>ar...@techprt.co.uk (Andrew Halket) writes:
>
>(RE roulette)
>>
>> a lot , lot lower. they only win on the 0 on roulette. If it weren't there
>
>They don't win on 0 (or double 00 or triple 000 depending on the wheel,
>he originals having but a single 0). (you can bet on 0) What they do is
>figure the payback based upon fewer possibilities that there are
In the context of my original posting, when you bet
red/black/odds/evens in roulette, it pays even money, which would
be entirely fair if there was no 0 or 00 on the wheel. These
numbers are neither odd nor even, red nor black, so, on average, a
player betting one of the even money bets will win 18/37th of the
time, and lose 19/37ths (18/38 and 20/38 on a US wheel).
The odds of winning 22 spins in a row is therefor
(18/37)**22 = 1 / 7,663,776
which is substantially better than 1 / 13,983,816 for a 6 from 49
lottery.
----------------------------------------------------------------------------
dr...@stonehenge.win-uk.net Just bang the rocks together guys
Any two philosophers can tell each other all they know in two hours.
-- Oliver Wendell Holmes, Jr.
----------------------------------------------------------------------------
Seth Breidbart
David Karr <ka...@cs.cornell.edu> wrote:
>OK, let's consider ways to make big bucks at a casino instead of
>at Lotto.
> On the other hand, if
>you put down your $100 on black and just let it ride, your chance of
>reaching $100,000 (actually $102,400) is
>
> (18/38)^10 = 0.57 * 10^-3
:
>Now the puzzle: What is the best strategy for reaching $100,000, and
>what is your chance to reach it?
The best method (as discussed on rec.gambling.math.weenies some time
back) is to bet on any number the lesser of (1) the amount needed such
that a win reaches your goal, or (2) your entire bankroll.
That's assuming you play roulette.
If you play blackjack, and count, you can reach your goal with a
probability arbitrarily close to 1.
Seth
Matthew Russotto
Seth Breidbart <se...@panix.com> wrote:
}The best method (as discussed on rec.gambling.math.weenies some time
}back) is to bet on any number the lesser of (1) the amount needed such
}that a win reaches your goal, or (2) your entire bankroll.
}
}That's assuming you play roulette.
}
}If you play blackjack, and count, you can reach your goal with a
}probability arbitrarily close to 1.
If you play blackjack right and count, your payoff:odds ratio will be better
than 1:1. Which is why they try to make it hard to count.
Mat Newman
>
>>If you share the Match 6 prize with k other tickets, then in Case A
>>the Match 6 expected return is :
>>
>>[Jackpot Value/(k+2)]*2*1/13,983,816
>>
>>whereas the Case B expected return is :
>>[Jackpot Value/(k+1)]*2/13,983,816
>>
>>and since 1/(k+2) < 1/(k+1) we see that again the expected return in
>>Case B is higher than that in case A.
>
>
>Sounded convincing, even looked convinving, but somehow I was convinced
>there was a mistake in there somewhere! I stared at the equations for
>a good 10 minutes before I had a flash: you're comparing apples and
>oranges like the saying goes. You're comparing k+1 winning tickets
>and k+2 winning tickets. That's not the same amount of winners!
>
win with both tickets. In case B, if you win, you win with one ticket.
Hence the k+2 vs the k+1 - the assumption is that there are k other
winners of the jackpot apart from you (i.e. if you didn't play then
how many jackpot winners would there be). The only fair way to compare
the two cases is the assume they'll both face the same number of competing
jackpot winners.
>Even then, who says both tickets in Case B would have shared with
>W-1 other winners ? Perhaps the other ticket would not have shared
>at all, or perhaps it shares with more winners! Are you confused
>yet ? I am! :-( Can someone clear this up ?
>
Of course the two tickets might be shared with different numbers of other
ticket holders, but a priori we don't know the numbers so we have to
assume an equal distribution for this calculation.
Mat.
Seth Breidbart
>>A similar situation holds for the *next* drawing of the lottery,
>>except this time there are some variables. First, I don't know what
>>the winning combination will be. Second, I don't know how many other
>>people will buy tickets with the winning combination.
The assumption is that the number of tickets bought by others, and the
sets of numbers on them, is independent of your action.
>>OK, so whatever combination I pick, other people will buy k tickets
>>identical to mine. I don't know what k is; it might be 0; it might be
>>99; but it's out of my control. That means that if I decide to buy 1
>>ticket, there will be k+1 tickets in the lottery with that combination
>>(some of them possibly bought after I bought mine, it doesn't matter
>>---all that matters is how many are sold before the deadline); if I
>>buy 2 tickets, there will then be k+2 identical tickets (including
>>mine). That means that *if* I happen to win the jackpot on that ticket
>>(which has odd of 1-in-whatever), I will get either 1/(k+1) of the
>>jackpot or 2/(k+2) of the jackpot, depending on whether I bought
>>1 or 2 tickets with that combination.
>
>My view seems just as valid as this, but I can't seem to pin-point
>why it differs from yours. In my case, the number of winning tickets
>remains the same, say W, and either I win 1 part of the jackpot or I
>win 2 parts, depending if I bought 2 different or 2 identical tickets.
If I buy 1 ticket, I have a 1 in 6 million (roughly, for some lottery)
chance of winning. Suppose I buy 10 identical tickets. I should
still have the same chance of winning; but the likelihood of there
being 10 or more winners is extremely small, thus your reasoning would
have my chances of winning be much lower.
>What I find really strange about the above equations is that for a
>fixed value of k you are comparing two different amounts of winning
>tickets.
Precisely. If I am going to win, then if you buy one winning ticket,
there will be two; if you buy ten winning tickets, there will be 11.
>BTW, how does the decision theory explain the case where I buy W+1
>identical tickets ? From my point of view, even if I had lost the
>tickets and forgotten which 6 numbers I had bought, I would still
>know that I had not won the jackpot because if I had, there would
>have been at least W+1 winners! What is the value of k in such a
>case?
Are you assuming that they posted the number of winners?
>The closest I can come to understand the differences is by asking
>a question. In your case you seem to ask, "What can I expect if
>I purchase the same winning ticket as the declared winners of the
>draw ?" Whereas, I'm asking, "What can I expect if I am among the
>declared winners with the tickets I purchase ?"
>
>Does this make any sense to you ? I'm not even sure the questions
>are right!
The problem is that your results are not independent of your actions,
but other people's actions are (assumed to be).
Seth
Douglas Zare (NC)
> [...]
> I have a way to substantially increase your chances of winning a
> few million. Just take one pound with you to one of those fancy
> casinos in Monte Carlo, and bet it on the black, keep doubling
> your stake until you've won 4 and a bit million (the most likely
> first prize in the National VAT supplement). This requires 22
> consecutive blacks to come up, much more likely than picking 6
> from 49, and the 'house percentage' at a casino is a lot lower.
With this method, you run into the house percentage in series instead of
all at once. So you actually do not get nearly as good a deal as you
might think; nominally it looks *worse* than the lottery. (18/37)^22 is
about 1 in 7.66 million, whereas (1/2)^22 is about 1 in 4.2 million.
Although that is better odds than the lottery, you get no chance at
smaller prizes. I am unfamiliar with the lottery under discussion, but in
the Florida 6/49, half of the prize money is distributed in the smaller
prizes. So if 4+ million were the value of each, the lottery would be better.
Of course, the 4+ million is probably just the sum of the nominal
payments, so it is not the true value or the amount of money going into
the jackpot is really 4+ million, as was noted earlier. So the casino is
probably better.
> ----------------------------------------------------------------------------
> dr...@stonehenge.win-uk.net Just bang the rocks together guys
> "It's OK to do the right thing... as long as you don't get caught."
> -- The Lone Contractor
> ----------------------------------------------------------------------------
Douglas Zare
za...@cco.caltech.edu
Stephen H. Landrum
Tapani Lindgren <ni...@vipunen.hut.fi> wrote:
>David Karr <ka...@cs.cornell.edu>:
>[You play roulette starting with $100. Numbers 0 and 00 are included.]
>...
>>Now the puzzle: What is the best strategy for reaching $100,000, and
>>what is your chance to reach it?
>...
>
>The house has at least 2/38 edge on every round, so the fewer rounds
>it takes to reach your goal, the better are your chances. Unfortunately,
>the best-paying bets have a larger edge, e.g. if you play $1 on one number
>only and win, you don't have $36 as you should, but only $30-$32, depending
>on the casino.
Not true anywhere I've been.
All bets on a U.S. wheel pay off with exactly the same edge, with the
exception of the one 5 number bet which has a worse payoff. $1 on a
single number pays 36 for one, or 35:1. The house edge is therefore
2/38 or 5.26%. The 5 number bet pays 7 for one (6:1), with a house
edge of 3/38 or 7.89%.
European wheels have no double zero, so the house edge is 1/37 or
2.70%. In addition, in Europe you may find places that have a rule
called "en prison", where if you place an even money bet and the
0 is what comes up, the bet is placed "en prison". On the next spin
if the bet loses, it is lost, but if the bet wins, it comes back
onto the board. The "en prison" rule cuts the house edge almost in
half for the bets where it applies. I may not have the exact details
of the "en prison" rule correct, but the gist is there.
There may be casinos in some countries that really rape the players
for roulette by offering the payoffs you describe, but the bets on
roulette already suck so badly that it is difficult to imagine.
--
Stephen H. Landrum voice: (415) 261-2626 email: slan...@3do.com
The 3DO Company 600 Galveston Drive, Redwood City, CA 94063
3DO Customer service: cl...@3do.com Developer support: sup...@3do.com
The Famous Druid
In article <Pine.SUN.3.91.941220210300.15871A-100000@virtu>, NC (za...@virtu.sar.usf.edu) writes:
>On Mon, 12 Dec 1994, The Famous Druid wrote:
>
>> [...]
>> I have a way to substantially increase your chances of winning a
>> few million. Just take one pound with you to one of those fancy
>> casinos in Monte Carlo, and bet it on the black, keep doubling
>> your stake until you've won 4 and a bit million
>Although that is better odds than the lottery, you get no chance at
>smaller prizes. I am unfamiliar with the lottery under discussion, but in
>the Florida 6/49, half of the prize money is distributed in the smaller
>prizes. So if 4+ million were the value of each, the lottery would be better.
>
>Of course, the 4+ million is probably just the sum of the nominal
>payments, so it is not the true value or the amount of money going into
>the jackpot is really 4+ million, as was noted earlier. So the casino is
>probably better.
I've just done a few more sums. On an American wheel (with 0 and 00)
the odds are (18/38)^22 = 1 in 13,779,834, which is very close to
the 1 in 13,983,816 for a 6 from 49 lottery.
I don't have the information on what % of the prize pool goes to
each prize category, but the total payout is 45% of takings, so
the statistics are as follows
Game | Total prize pool | 'expected' return
------------------------------------------------
Lottery | 6,292,717 | 0.45
US Casino | 4,194,304 | 0.3044
European | 4,194,304 | 0.5473
In the case of the casinos, the expected return is the prize (2^22)
divided by the probability of winning ((18/37)^22 or (18/38)^22).
Of course, it's pretty pointless calculating the statistics for
casinos, because there is *no*way* that I could put GBP 2,097,152 on
a roulette table ! I'd be chickening out at about GBP 4096.
----------------------------------------------------------------------------
dr...@stonehenge.win-uk.net Just bang the rocks together guys
Mrs Asquith remarked indiscreetly that if Kitchener was not a
great man, he was, at least, a great poster.
-- Margot Asquith
----------------------------------------------------------------------------
Jens Joergen Nielsen
slan...@3do.com (Stephen H. Landrum) wrote:
>European wheels have no double zero, so the house edge is 1/37 or
>2.70%. In addition, in Europe you may find places that have a rule
>called "en prison", where if you place an even money bet and the
>0 is what comes up, the bet is placed "en prison". On the next spin
>if the bet loses, it is lost, but if the bet wins, it comes back
>onto the board. The "en prison" rule cuts the house edge almost in
>half for the bets where it applies. I may not have the exact details
>of the "en prison" rule correct, but the gist is there.
Yes, that is correct. In Denmark (Casino Copenhagen) the Roulette wheel
has only one zero, and you lose only half the stake on all 1:1 bets in
the event Zero wins.
The casino's house edge is very favorable compared to the national
lottery. If you really want to increase the variance of your fortune
without reducing the expectation value significantly, this is the
place to go.
*---------------------*---------------------------------------------------*
: Jens Jorgen Nielsen : Ever wondered how Lagrange points really work ? :
: j...@login.dknet.dk : Try : garbo.uwasa.fi//pc/astronomy/3d-g12.zip :
*---------------------*---------------------------------------------------*
Seth Breidbart
Jens Joergen Nielsen <j...@login.dknet.dk> wrote:
>Yes, that is correct. In Denmark (Casino Copenhagen) the Roulette wheel
>has only one zero, and you lose only half the stake on all 1:1 bets in
>the event Zero wins.
>
>The casino's house edge is very favorable compared to the national
>lottery. If you really want to increase the variance of your fortune
>without reducing the expectation value significantly, this is the
>place to go.
What happens if I bet a a single number? (What is the payoff, any
special rules, etc.)
If I'm trying to turn $1 into $1 million, given the same house
percentage on all bets, I should favor the ones with the largest odds.
Seth