6/49 lottery
Roddie McLure
1/13983816
if you buy 163 tickets that would be 163/13983816 or 1 in 85790.28
approx
If 163 tickets guarantee at least one of the tickets has 3 numbers
correct for a 6/49 draw could you not say that your chances of winning
are now
1 in 46!/(3!)(43!) = 1/15180 since you know you have a ticket with 3
numbers right, you just need 3 numbers from the remaining 46. Does the
fact that I don't know which of my 163 tickets has the 3 numbers
right, factor in?
How can choosing 163 tickets at random give you different chances
than selecting 163 tickets from a pre-selected list?
D & A Klinkenberg
If you already have an answer, please compare it with my answer at the
bottom and post your reaction.
I don't play the lottery. You talk of a guarantee. Who makes the
guarantee? Doesn't that matter?
You discuss having a list of 163 tickets, and being guaranteed that you have
three numbers out of 6 that are 'correct'. Then you do some calculations
based on "at least one of the tickets has 3" correct numbers. But, of
course, those aren't the same thing.
Then you ask, "How can choosing 163 tickets at random give you different
chances than selecting 163 tickets from a pre-selected list?" My suggestion
as an answer to this: It is unlikely that 163 tickets at random can have
any guarantee, even if a those in a pre-selected list do. Doesn't the
guarantee have to come from the 'person' who pre-selects the list?
Dan in NY
&&& ----- Original Message -----
"Roddie McLure" <rdmc...@pei.sympatico.ca> wrote in message
news:cccfa469.02043...@posting.google.com...
Bob Harris
> The chances of winning when buying 1 ticket in a 6/49 lottery is
> 1/13983816
Correct-- that's the chance of matching all six numbers chosen, to win the
grand prize.
> if you buy 163 tickets that would be 163/13983816 or 1 in 85790.28
> approx
Yep; that's your chance of winning the grand prize with 163 tickets.
> If 163 tickets guarantee at least one of the tickets has 3 numbers correct for
> a 6/49 draw ...
Provided you select the correct set of 163 tickets, of course.
> ... could you not say that your chances of winning are now 1 in
> 46!/(3!)(43!) = 1/15180 since you know you have a ticket with 3 numbers right,
> you just need 3 numbers from the remaining 46.
After they've drawn the first three numbers, are you guaranteed to have a
ticket with those three numbers on it? If so, then you could apply your
logic, and you'd have a 46-choose-3 chance to win (or better, if you have
more than one ticket with those three numbers).
But you *don't* have those three numbers; at least not but rarely.
--
-- Bob Harris =======================================================+
| Birds of a feather flock together; so be certain to park your car |
| in the garage. |
+====================================================================+
Peter Webb
> After they've drawn the first three numbers, are you guaranteed to have a
> ticket with those three numbers on it? If so, then you could apply your
> logic, and you'd have a 46-choose-3 chance to win (or better, if you have
> more than one ticket with those three numbers).
>
> But you *don't* have those three numbers; at least not but rarely.
>
Why should it matter which numbers are pulled out first? There is nothing
intrinsic to the rules of the game that has anything to do with the order.
When we know the six winning numbers, why can't we just pull the entry (out
of the 163) that has 3 correct. Cross these three off the entry, and out of
the list of 6 winning numbers. Aren't we left with 3 numbers selected at
random (on our winning card) and 3 random winning numbers, out of 46?
If your argument is correct, you would think it would make C(6,3) or P(6,3)
difference to the probability (the chances that the three numbers you
selected are the first three numbers in the winning set, in no particular
order or correct order respectively). Instead, the ratio between the chance
based upon 163 cards at random and 163 cards with a guaranteed match of 3 is
5.651515 ... (85790.28 / 15180) unless you can think of a good explanation
for such an unusual number appearing, I remain unconvinced.
By the way, nor am Iconvinced by the thread which stated that 163 entries
formed a covering set which guaranteed at list 3 matches in a 6/49 lottery.
None of the posted arguments appeared correct (to me), and nobody posted the
actual 163 groups of 6 numbers or a construction thereof.
Mr McClure's argument makes me suspect that the 163 is actually wrong, as
the probability should match up for the reasons he gave.
Bob Harris
> ... nobody posted the actual 163 groups of 6 numbers or a construction
> thereof.
I posted a link to them in a previous thread, back on apr/24.
: The best know answer to date seems to be 163, discovered by Dragan
: Stojiljkovic and Rade Belic in Nov/2000. You can find a set of 163
: specific tickets at
: http://lottery.merseyworld.com/Wheel/wheel163.html
: though of course you could apply any permutation of 1..49 to get another
: set of tickets that'd do the same thing.
> Why should it matter which numbers are pulled out first? There is nothing
> intrinsic to the rules of the game that has anything to do with the order.
> When we know the six winning numbers, why can't we just pull the entry (out
> of the 163) that has 3 correct. Cross these three off the entry, and out of
> the list of 6 winning numbers. Aren't we left with 3 numbers selected at
> random (on our winning card) and 3 random winning numbers, out of 46?
Well, offhand I don't see any flaw in your logic. Perhaps my proposal is
wrong.
But, since I believe in 163 (thought I have not verified it myself), I'm
pretty sure there's got to be some way to explain the paradox that Roddie
posted. I'll think about it a little more. Maybe an example with a smaller
covering would be instructive.
Bob Harris
> Maybe an example with a smaller covering would be instructive.
Consider a lottery in which you choose four numbers from 1..10. Buying the
three tickets 1-2-3-4, 4-5-6-7, and 7-8-9-10 guarantee that I'll match at
least one pair, regardless of the winning numbers. Two tickets would not be
enough to make this guarantee (e.g. 1-2-3-4, 7-8-9-10 comes up short if the
winning numbers are 5-6 with one low number and one high).
Take a look at Roddie's paradox in that context and see if it helps clear
anything up.
Bob H
Jonathan Miller
>Why should it matter which numbers are pulled out first? There is nothing
>intrinsic to the rules of the game that has anything to do with the order.
>
It doesn't. What matters is if you already know that you've matched 3
numbers: the conditional probablility.
>When we know the six winning numbers, why can't we just pull the entry (out
>of the 163) that has 3 correct. Cross these three off the entry, and out of
>the list of 6 winning numbers. Aren't we left with 3 numbers selected at
>random (on our winning card) and 3 random winning numbers, out of 46?
>
Not really.
>If your argument is correct, you would think it would make C(6,3) or P(6,3)
>difference to the probability (the chances that the three numbers you
>selected are the first three numbers in the winning set, in no particular
>order or correct order respectively). Instead, the ratio between the chance
>based upon 163 cards at random and 163 cards with a guaranteed match of 3 is
>5.651515 ... (85790.28 / 15180) unless you can think of a good explanation
>for such an unusual number appearing, I remain unconvinced.
>
What unusual number? approx. 5.65 instead of 6? Or 6?
>By the way, nor am Iconvinced by the thread which stated that 163 entries
>formed a covering set which guaranteed at list 3 matches in a 6/49 lottery.
>None of the posted arguments appeared correct (to me), and nobody posted the
>actual 163 groups of 6 numbers or a construction thereof.
>
Well, a link was posted. Try
http://lottery.merseyworld.com/Wheel/wheel163.html .
For verication, this page includes c source code:
http://lottery.merseyworld.com/Wheel/
In addition, this page gives results for the past 662 drawings (or more,
depending when you read this and how often the page is updated
[apparently manually, no less]): http://lottery.merseyworld.com/Wheel/
. Note that this doesn't really answer the question you asked, since
what you really want to know is winnings by drawing and not by number
chosen. Since I don't have the data, I can't update it for you.
>Mr McClure's argument makes me suspect that the 163 is actually wrong, as
>the probability should match up for the reasons he gave.
>
Well, if you have a random number generator (like in your spreadsheet
program), you can do a real quick test. Just compare these to about
10,000 random drawings. If you get even one non-winner, you know the
list is incorrect. If all 10,000 have a winner, then you can have a lot
more confidence in the result, even though it still isn't proven.
Depending on your memory size, you should be able to do them in batches
of 100 (that's a 163 x 600 table, plus a little extra, the way I do it).
Jon Miller
Bob Harris
> Why should it matter which numbers are pulled out first? There is nothing
> intrinsic to the rules of the game that has anything to do with the order.
> When we know the six winning numbers, why can't we just pull the entry (out
> of the 163) that has 3 correct. Cross these three off the entry, and out of
> the list of 6 winning numbers. Aren't we left with 3 numbers selected at
> random (on our winning card) and 3 random winning numbers, out of 46?
to which I answered:
> Well, offhand I don't see any flaw in your logic. Perhaps my proposal is
> wrong.
Having gicen it more thought, it seems like this is all just a variation of
the Monty Hall paradox.
Bob H
Peter Webb
This is also consistent with Jonathon Miller's contention that this is a
conditional probability question. Still doesn't explain the 5.65 factor, but
on the other hand 163 is a kind of strange number itself.
This spanning set stuff came as somewhat of a surprise when I first saw it
in this ng a few days ago. Such a simple question to be unsolved.
Jonathan Miller
>Yes ... nice connection.
>
>This is also consistent with Jonathon Miller's contention that this is a
>conditional probability question. Still doesn't explain the 5.65 factor, but
>on the other hand 163 is a kind of strange number itself.
>
>This spanning set stuff came as somewhat of a surprise when I first saw it
>in this ng a few days ago. Such a simple question to be unsolved.
>
Lots of simple questions are still unsolved. I guess it's a flaw of our
educational system that we don't let students know that.
Jon Miller
Bob Harris
Peter Webb wrote:
> This spanning set stuff came as somewhat of a surprise when I first saw it in
> this ng a few days ago. Such a simple question to be unsolved.
to which Jonathan Miller replied:
> Lots of simple questions are still unsolved. I guess it's a flaw of our
> educational system that we don't let students know that.
Ironically, any teacher that brought up this subject in his/her class,
describing it in terms of a lottery, would likely get into trouble for
teaching his/her students how to gamble. At least, that's ironic in the
state I live in, since lottery dollars are used to fund education.
--
-- Bob Harris =======================================================+
| If stupidity got us into this mess, how come it can't get us out |
| of it? |
+====================================================================+
Jonathan Miller
>Peter Webb wrote:
>
>>This spanning set stuff came as somewhat of a surprise when I first saw it in
>>this ng a few days ago. Such a simple question to be unsolved.
>>
>
>to which Jonathan Miller replied:
>
>>Lots of simple questions are still unsolved. I guess it's a flaw of our
>>educational system that we don't let students know that.
>>
>
>Ironically, any teacher that brought up this subject in his/her class,
>describing it in terms of a lottery, would likely get into trouble for
>teaching his/her students how to gamble. At least, that's ironic in the
>state I live in, since lottery dollars are used to fund education.
>
Maybe. I have used lotteries in a discrete math class (college level --
covered some basic probability and statistics and an introduction to
some of the math used in operations research). But a simple analysis
tells you that you can't expect to make money playing the lottery. The
fact is that 50% of the lottery revenue goes to education and the other
50% to prizes (this is the US lotteries that I'm familiar with). I
forget which 50% the expenses fall into, I think the education part.
But anyway, if the odds are stacked so heavily against you, how can a
subtle strategy win? Any winning strategy would have to be blindingly
obvious (well, at least with hindsight) because it would have to be an
incredibly strong strategy. Are you so much smarter than the other
people looking to make easy money?
Not that I advocate not playing the lottery. Assuming you don't mind
supporting what the lottery supports (or don't overly want to support
it), the basic equation is that your dollar gets you, on average, 50
cents of return and some amount of pleasure. If the value of the
pleasure is more than 50 cents, you should play the lottery and if it's
less, you shouldn't. Just like any other consumer item.
You could also argue that the prizes are skewed so that you won't, in
fact, win a prize (well, not in a million years). In that case, the
value of the entertainment must be more than $1. It's still a consumer
choice argument.
Jon Miller
Roddie McLure
incorrect,I
am not able to show exactly why, It does remind me of the Monty Hall
question.If you take into account,as someone suggested,the fact that
the 163 tickets "guarantee" a 3 number match by partitioning the 49
numbers into 2 groups of 22 numbers and 27 numbers giving you 77
tickets for the first group and 86 for the second group then I don't
think my statement "your chances of winning are now
1 in 46!/(3!)(43!) = 1/15180 since you know you have a ticket with 3
numbers right, you just need 3 numbers from the remaining 46. "
1st group
of 22 numbers or the second group of 27 numbers. The odds of all 6
numbers coming from the same group are not accounted for, I think its
1 in 187.42 for
the first group or 1 in 47.27 for the second : once this condition is
met then
the I could say I only need the other 3 numbers to match tbe ones on
my ticket.I
haven't been able to do the math but am convinced.
I also looked at '19 tickets to guarantee 2' numbers by partitioning
the 49 numbers into 5 groups of 10,10,10,10,9 since 6 numbers are
drawn 1 group must have 2 numbers. The same thing happens here odds of
19 tickets winning 1 in 735990.26, but if you know you have 2 nubers
right C(47/4) is only 178365
...at least I'm consistent.
rdmc...@pei.sympatico.ca (Roddie McLure) wrote in message news:<cccfa469.02043...@posting.google.com>...
Michael Mendelsohn
> I have given this some more thought and am fairly certain my math is
> incorrect,I
> am not able to show exactly why, It does remind me of the Monty Hall
> question.
Your assumption was "I have 3 right, so let's look at the other three
and find that probability".
Well, the problem is, the three right you have could be any 3 of the 6
on either your card or the drawing. What I mean is, if they aren't the
first three on your card or the numbers drawn, you know that you can't
win because you already skipped a number that didn't fit. The chance of
that happening is, I estimate, about 1 in (3 out of 6), because the only
place for the 3 "guaranteed" numbers to be is the first 3 - ggg... ; all
others (g.gg.. , .ggg.. etc.) mean that you can't have a winning ticket.
Once you have beaten these odds and actually have the first three
numbers on your ticket and the drawing coincide we can put the random
drawing of the other three numbers in play.
I leave the computation of the actual probabilities as an exercise to
the reader ;-)
Have fun with lotto mathematics
Michael
--
When the tongue or the pen is let loose in a phrenzy of passion,
it is the man, and not the subject, that becomes exhausted.
-- Thomas Paine, "On Usenet"
Roddie McLure
possible 3 number
combinations 18424 and divide by 20 since that is how many you have
when you draw 6 numbers from 49, you get 921.2
If you were to buy 921.2 (theoretically) tickets with one set of 3 on
each ticket your chances of winning are 1 in 46!/(3!)(43!) = 1/15180,
since (theoretically) you should have 1 of the 20 correct 3 number
combos ...this agrees with 921.2 tickets /13983816 = 1 in 15180
when we apply this to the situation where 163 tickets has guaranteed 3
correct numbers 163 / 13983816 = 1 in 85790.28 but if we know we have
3 numbers right
1 in 46!/(3!)(43!) = 1/15180 what's interesting is we need to multiply
this by 921.2/163 or 5.65 to get 1 in 85790.28 I still don't know what
it means but it is interesting
rdmc...@pei.sympatico.ca (Roddie McLure) wrote in message news:<cccfa469.02050...@posting.google.com>...
Royce Penny
Roddie McLure wrote:
>
> ok I think I see where the 5.65 comes from ... if you take all
> possible 3 number
> combinations 18424 and divide by 20 since that is how many you have
> when you draw 6 numbers from 49, you get 921.2
(...snip for brevity..)
The chance of getting three correct out of 6 numbers drawn
in a lotto649 is 1 in C(49,6)/(C(43,3)*C(6,3)), or 1 in
56.6559
--
Royce Penny
Royce Penny's Money Machine
http://www.geocities.com/lottoking.geo