Probability Question

[qcan]

I would just like some confirmation on the following....

- Wager $ 90.00 ( 9 combinations @ $ 10.00 each )
- Probability of winning = 71 % ( in other words, there is a 71 %
chance that one of these 9 combinations will win).
- If win, return will be $ 105.00 + $ 10.00 on inital wager on the
one
winning combination.
- 39 % chance of loosing $ 90.00.


I am not a probability expert, but my thought is that over 100 events
if I win exactly 71 times the retrun will be:


71 x $ 115.00 = $ 8165.00 (total retrun on win)


39 x $ 90.00 = $ 3510.00 (complete loss)
61 x $ 80.00 = $ 4880.00 (8 other combinations that loss)
loss = $ 8390.00


Net loss = - $ 225.00


Is my math correct ?


Thanks.

[Ray Vickson]

No, it is incorrect. Look at one game. Let W = 1 if you win the game
and W = 0 if you lose. The probability P{W = 1} = 0.71.
The 1-game net return (i.e., profit) is NR = 115*W - 90*(1-W) = 205 W
- 90; this equals 115 if W = 1 and equals -90 if W = 0. The expected
net return is ENR = 205*(0.71) - 90 = 55.55.
In other words, in an average game your net profit will be $55.55. The
100-game expected profit is $100*55.55 = $5,550.
Of course, that actual profit in 100 games could range from an extreme
of -9000 (a $9000 loss) to a high of $115,000, but it is likely that
the actual net will come out "near" the average value $5,550. In fact,
one can use the binomial distribution to look at the probability
distribution of the actual net. The probability that you lose money
(giving a net < 0 in 100 games) is about 0.5x10^(-8) = 0.000000005.
The probability you make less than $4000 net is 0.0515, or about one
chance in 20.

RGV

[Ray Vickson]

Sorry: that should say "the actual profit in 100 games could range
from an extreme
of -$9000 to a high of $11,500".

RGV

[qcan]

> RGV- Hide quoted text -
>
> - Show quoted text -

Thanks for the explanation Ray. Much appreciated !

[John Morriss]

On Wednesday, 1 February 2012 13:11:34 UTC-5, qcan wrote:
> I would just like some confirmation on the following....
>
> - Wager $ 90.00 ( 9 combinations @ $ 10.00 each )
> - Probability of winning = 71 % ( in other words, there is a 71 %
> chance that one of these 9 combinations will win).
> - If win, return will be $ 105.00 + $ 10.00 on inital wager on the
> one
> winning combination.
> - 39 % chance of loosing $ 90.00

Uhhh... If the 39% probability of losing $90.00 comes from subtracting 71% from 100%, then your problem doesn't lie in the area of probability theory...

The game has no memory, so just look at one game. You pay out $90.00 <<every time>>, and expect to win $115, 71% of the time, or an average of $81.65 per game.

That's it. No more calculations. You lose $8.35 on average every time you play. Period

Nice game for the house...

[Ray Vickson]

Actually, you win a net of $55.55 per game, for an expected net of
$5,555 over the 100 games. As indicated in my first post, the
probability that you net out at < 0 (i.e., lose money) is about
0.5*10^(-8), or about 1 in 200,000,000. The chance you net out at <
$4000 is about 1 in 20.

RGV

[qcan]

> RGV- Hide quoted text -
>
> - Show quoted text -

Interesting discussion.... of course I am just a wee bit confused now
with what the answer truly is. Just to add to this confusion - I
neglected to mention that each the 9 combinations of 2 numbers has an
equal chance (1 in 13) of winning. Further, it is possible that more
than one combination can win. Not sure at all how this effects the
odds or if this was factored into the calculation - but it does make a
difference.

[John Morriss]

On Wednesday, 1 February 2012 15:34:59 UTC-5, Ray Vickson wrote:
> On Feb 1, 10:11 am, qcan <comeonove...@gmail.com> wrote:
> > I would just like some confirmation on the following....
> >
> > - Wager $ 90.00 ( 9 combinations @ $ 10.00 each )
> > - Probability of winning = 71 % ( in other words, there is a 71 %
> > chance that one of these 9 combinations will win).
> > - If win, return will be $ 105.00 + $ 10.00 on inital wager on the
> > one
> > winning combination.
> > - 39 % chance of loosing $ 90.00.
> >
> > I am not a probability expert, but my thought is that over 100 events
> > if I win exactly 71 times the retrun will be:
> >
> > 71 x  $ 115.00 = $ 8165.00  (total retrun on win)
> >
> > 39 x  $   90.00 = $ 3510.00 (complete loss)
> > 61 x  $   80.00 = $ 4880.00 (8 other combinations that loss)
> > loss = $ 8390.00
> >
> > Net loss = - $ 225.00
> >
> > Is my math correct ?
> >
> > Thanks.
>
>
> No, it is incorrect. Look at one game. Let W = 1 if you win the game
> and W = 0 if you lose. The probability P{W = 1} = 0.71.
> The 1-game net return (i.e., profit) is NR = 115*W - 90*(1-W) = 205 W
> - 90; this equals 115 if W = 1 and equals -90 if W = 0. The expected
> net return is ENR = 205*(0.71) - 90 = 55.55.

If you win, your NET profit is only $25. Put down $90, get back $115, according yo the OP... If you lose your net lose is $90. And if you take 71% of $25 and subtract 29% of $90, you get ..-$8.35

[hagman]

To calculate the expected win, it doesn't even matter whether or not
the
events of winning different games are independent

[John Morriss]

On Thursday, 2 February 2012 18:56:27 UTC-5, qcan wrote:
>
> Interesting discussion.... of course I am just a wee bit confused now
> with what the answer truly is. Just to add to this confusion - I
> neglected to mention that each the 9 combinations of 2 numbers has an
> equal chance (1 in 13) of winning. Further, it is possible that more
> than one combination can win. Not sure at all how this effects the
> odds or if this was factored into the calculation - but it does make a
> difference.

Well, it DOES make a difference. This post and the original post cannot both be true. If you have 1 chance in 13 of winning a single combo, then you have 12 chances in 13 of losing and then you have 48.6% chance of losing all nine, not 29% A much more expensive game.

It doesn't matter if you chose to divide the sequence of $10 bets into blocks of 9. Look at one bet: Cost $10 to get in the game and returns $115 when you win 1 out of 13 times. Expected profit = $115/13-$10 = -$1.15.

So your loss for a 9-game block is $10.38.



It doesn

[Ray Vickson]

The original question was maybe a bit ambiguous; I now realize that I
likely chose the wrong interpretation. Assuming he means: net win =
105 - 80 = 25 if there is a winning combination (get back 10 on the
winning one, and lose 80 on the others), or net win = -90 otherwise,
then the net in a game = N = 25W - 90(1-W) = 115W - 90, where W = 1 if
get a winner, and W=0 otherwise. The expected net per game is
115(0.71) - 90 = -8.35. In my original interpretation, I assumed the
winning amount 115 applied to the total, which I now agree the OP
probably did not mean.

RGV

[qcan]

> RGV- Hide quoted text -
>
> - Show quoted text -

Correct.... Net win on 1 combo of 9 = 105.00 - 80 = 25. This is
clear. What is not so clear is how John arrives at the 48.9 % chance
of loosing all nine. To be crystal clear the chance of ONE combo
winning is 7.8674948 % ( 1 in 12.71 ). In my naivety, I incorrectly
assumed that finiding out the winning probability of 1 set winning was
as simple as 9 x .078674948 or 70.8074534 %. With this logic, that
would mean that I would have a 100 % of winning if I chose 13 sets.
Silly me.

This is actually a Keno type of problem. There would be 70 (yes 70)
balls. 20 would be drawn. 2 would be chosen by the player. The
question.... what is the probability of one set of nine sets would
match the 20 balls chosen. If we want to get a bit deeper.... there is
of course a possibilty of multiple wins on the 9 sets.

Yes, my original question was ambigious. I did it that way thinking
that I would simpify it somewhat. My appologies.

[Ray Vickson]

The analysis is doable when all the pairs chosen by the player are non-
overlapping--that is, if one chosen pair is (i,j), none of the other
pairs can contain i or j. If this is not the case (so, for example, a
player can choose (1,2) on one ticket and (2,3) on another ticket),
things are more complicated. So, let's look at the non-overlapping
case.

In one game the outcome is a string B1 B2 B3 ... B20, where the order
does not matter; the B1, B2, ..., B20
are just twenty distinct numbers drawn without replacement from 1,
2, ..., 70. Assuming an absolutely fair, unbiased, choice method,
each possible string of 20 has the same probability, namely, 1/N,
where N = number of ways of choosing 20 numbers from 70
= C(70,20), the binomial coefficient. In one choice of two numbers by
a player, what is the probability the player wins (i.e, has his two
numbers among the B1, B2, ..., B 20)? Say the player chooses i and j.
In how many outcomes do i and j occur? Well, remove i and j; now we
are left with 68 numbers from which we choose 18 (then add back i and
j later). The number of relevant outcomes is C(68,18), so P(win) = p1
= C(68,18)/C(70,20) = 38/483 =~= 0.0786749, as you have stated.

Next, suppose a player chooses 2 non-overlapping pairs, say (i,j) and
(k,l). Obviously, by symmetry, P{(i,j) wins} = P{(k,l) wins} = p1. Now
P{(i,j) wins & (k,l) wins} = P{(i,j) wins}*P{(k,l) wins | (i,j) wins};
this conditional probability is obtained by removing i and j and
looking at the chance of getting (k,l) when choosing 18 numbers from
68: it = C(66,16)/C(68,18). Thus, P{both win} = C(68,18)/C(70,20) *
C(66,16)/C(68,18) = C(66,16)/C(70,20). Thus, P{at least 1 win} =
P{(i,j) wins} + P{(k,l) wins} - P{both win) = 2*p1 - p2, where p2 =
P{both win} = C(66,16)/(C(70,20) = 57/10787 =~= .52841e-2. We can also
get P{exactly 1 win} = P{at least 1 win} - P{2 wins} = 2p1 - 2p2, and
P{0 wins} = 1 - P{at least one win}.

Next, if the player selects 3 non-overlapping pairs, say (i1,j1),
(i2,j2) and (i3,j3), let W1={(i1,j1) wins}, etc. We have P{W1 or W2 or
W3} = P{W1} + P{W2} + P{W3} - P{W1 & W2} - P{W1 & W3} - P{W2 & W3} +
P{W1 & W2 & W3} = 3P{W1} - 3P{W1 & W2} + P{W1 & E2 & W3} = 3p1 - 3p2 +
p3, by obvious symmetry. Here, p3 = P{all three win} can be obtained
as p3 = C(64,14)/C(70,20) = 456/1542541 =~= .295616e-3. We can also
get P{exactly 1 win} and P{exactly 2 wins} using generalizations of
the inclusion-exclusion formulas, as developed in detail in Feller,
Introduction to Probability Theory, Vol. I, Wiley 1968.

It is more-or-less obvious how to do the computations now for 4, 5, 6,
7, ... , 9 *non-overlapping* choices by the player. The case with
overlapping choices is a lot messier.

RGV

[qcan]

Thanks for the explanation Ray !

[Ray Vickson]

On Feb 3, 7:08 am, qcan <comeonove...@gmail.com> wrote:

To follow up on the my previous posting above: suppose the player
picks r non-overlapping pairs (i1,j1),...,(ir,jr),
corresponding to lotto tickets t1, t2, ...,tr. Non-overlapping means
that all the i's and j's are distinct numbers (that is,
our plyer choose 2r distinct numbers). Let Wu = {ticket u wins}, u =
1,..,r, and note that P{Wu} = p[1] = P{t1 wins} for all u,
because of symmetry and non-overlap. Similarly, P{Wu & Wv} = p[2] =
P{W1 & W2} for all pairs u < v, P{Wu & Wv & Ww} = p[3] =
P{W1 & W2 & W3}, etc. In general, let p[k] = P{W1 & W2 & ... & Wk}. We
have, for k = 1, 2, ...,r, that p[k] = C(70-2k,20-2k)/C(70,20),
because for W1,...,Wk to all occur we need 2k numbers to come up, and
the number of outcomes in which this occurs is the number
of ways of choosing 20-2k other numbers from the remaining 70-2k
numbers after removing the relevant 2k. Finally, let s[1] =
sum{ P{Wi}, i=1..r}, s[2] = sum{ P{Wi & Wj}: 1 <= i < j <= r}, s[3] =
sum{P{Wi & Wj & Wk : 1 <= i < j < k <= r}, etc. The quantity s[u] is
the sum over all u-tuples of the probability of that u-tuple. For the
non-overlapping case we have s[1] = r*p[1], s[2] = c(r,2)*p[2], ...,
s[k] = C(r,k)*p[k], ..., s[r] = p[r].

Inclusion-exclusion gives P{at least one of the r tickets wins} = s[1]
- s[2] + s[3] - s[4] + ... +- s[r]. Furthermore, from Feller, Vol. I,
we have: P{exactly k tickets win} = s[k] - C(k+1,k)*s[k+1] + C(k
+2,k)*s[k+2] - ... +- C(r,k)*s[r].

Here are the results for r = 9 tickets:
P{at least 1 win} =~= 0.54105 (cf. your estimate 0.71), and the
distribution of the number of winning tickets is:
j P{exactly j winners}
1 0.39567
2 0.12525
3 0.018674
4 0.0014026
5 5.2865e-05
6 9.4455e-07
7 7.0800e-09
8 1.7508e-11
9 8.1910e-15

All the computations were done in Maple, which allows exact rational
calculations throughout, later converted to floats for
purposes of output and comparison. For example, the exact value of
P{>= 1 winner} is 10667185690951/19715577111516.

What about problems with overlaps? They are a bit trickier because
there are more things to keep track of. Suppose, for example, t1 =
(1,2), t2 = (2,3) and all other tk do not overlap t1, t2 or each
other. Then P{W1 & W2} and all other P{Wi & Wj} are different. For W1
& W2 the three numbers 1,2,3 must occur, so the probability is p12 =
C(70-3,20-3)/C(70,20), but for all other P{Wu & Wv} the probabilities
are equal to p13 = C(70-4,20-4)/C(70,20). Now s[2] = p12 + (r-1)*p13,
etc. In any k-tuple we need to keep track of whether or not both (1,2)
and (2,3) are present, or whether only one (or none) of these occurs.
Aside from that, the computations are not overwhelmingly more
difficult. The above results for P{k winners} in terms of the s[m]
hold unchanged, but computations of the s[m] need to be modified.

RGV