blackjack
 

Has anyone worked out the formula for the probability of winning x hands before losing 2x hands in any number of deals (assuming an equal chance of winning or losing)? For example, what is the probability one would win 4 hands before losing 8 hands in 12 deals, 24 deals, x deals?

Re: blackjack
 

I haven't.

Re: blackjack
 

To determine the chances of at least winning 1/3rd of the
total hands, it would be just 1-P(winning less than 1/3rd).
There is an approximation to the binomial distribution
using the standard normal curve which is better as the
trials become very large. For the exact answer, the chances
of not winning at least 1/3rd of the hands for 3m trials
(m an integer>=1) would be determined by:

Summing of the combinatorial numbers
C(3m,k) where k runs from 0 to m-1.
Then, dividing this sum by 2**(3m).

For example, with m=4 or 12 deals, you simply sum

C(12,0) = 1
C(12,1) = 12
C(12,2) = 66
C(12,3) = 220

giving the sum of 299.

2**12 = 4096 so the chances of NOT winning 1/3rd in this
case is 299/4096 or there is a 3797/4096 chance of winning
at least 1/3rd of the hands.

Re: blackjack
 

I'm a math idiot (business major [img]/images/graemlins/smirk.gif[/img]). Is the 1/3 because of possibility of pushes?

Re: blackjack
 

Pooch, you're answering a different question. He wants to know the probability of winning 4 before losing 8. That's not the same as winning at least 4 out of 12 because it depends on which games you win. For example, there are not C(12,n) ways to win n games, since once you lose 8 or win 4, it's over, and it doesn't matter what happens the remaining games, so there are more ways to do this out of 2^12.

Here is the correct solution.

I'm ignoring pushes and taking P(win) = P(lose) = 1/2.

Count the successful cases out of 11 games. Note that if we get the 4th win on the nth game, the nth game must be a win, so we just have to count the number of ways to choose the 3 games out of n-1 to win, and multiply by 2^(11-n) possible results for the remaining 11-n games which don't matter.

4 of 4: C(3,3)*2^7
4 of 5: C(4,3)*2^6
4 of 6: C(5,3)*2^5
4 of 7: C(6,3)*2^4
4 of 8: C(7,3)*2^3
4 of 9: C(8,3)*2^2
4 of 10: C(9,3)*2^1
4 of 11: C(10,3)*1
--------------------------
total = 1816.

1816/2^11 = 88.7%. Note that this is the same regardless of the number of deals if more than 11, since we either win 4 or lose 8 out of the first 11.

It would have been easier to compute the probability of losing 8 before winning 4 and subtract from 1, since this would only have 4 terms to sum. We can do that as a check to make sure these account for all the remaining cases out of 2^11 possibilities. Note that this is the same as the last 4 terms above with 3 replaced by 7:

8 of 8: C(7,7)*2^3
8 of 9: C(8,7)*2^2
8 of 10: C(9,7)*2^1
8 of 11: C(10,7)*1
--------------------------
total = 232.

232 + 1816 = 2048 = 2^11, so this accounts for all the cases.

In general, the probability of winning x games before losing 2x games is:

sum{n = x to 3x-1} C(n-1,x-1)*2^(3x-1-n) / 2^(3x-1).

Re: blackjack
 

Serious question. How is it helpful to know the answer to this probability question. Are there any practical consequences, i.e., how much money you should wager, how large your bankroll will be, how many hands you should play, to clear a bonus?

I assume that without a bonus, you should never play even a single hand of blackjack, which I have heard is negative EV.

Another interpretation - classic problem
 

I think the poster might be trying to ask "what is the probability that if a person plays this game forever, that his number of losses never exceeds twice his number of wins?" If this was the question, the example wasn't worded properly. This is a classic problem, and it is related to random walks, and the gambler's ruin problem.

Here is the solution to that problem.

Suppose he starts with 1 dollar. Suppose that when he wins, he wins 2 dollars, and when he loses, he loses 1 dollar. Then if he ever has at least twice as many losses as wins, he will be broke. Call this probability p. Then we can write:

p = 1/2 + 1/2 * p^3

This is the standard trick for gambler's ruin problems. It says he can go broke on the first play with probability 1/2, OR he can win on the first play with probability 1/2 and go broke later. If he wins on the first play he will have 3 dollars, and the probability of losing 3 dollars is p^3, since he has to lose 1 dollar 3 times. The solution to the above equation is

p = [sqrt(5)-1] / 2 = 61.8%.

You can verify that this is the solution. This is the probability that he will eventually have twice as many losses as wins, so the probabilty that he never has twice as many losses as wins is:

1 - [sqrt(5) - 1] / 2 =

[3 - sqrt(5)] / 2 = 38.2%

Re: Another interpretation - classic problem
 

Right. It is interesting that the golden ratio (or its
inverse) appears here! I see that for a player to have at
some time not won at least 1/k of the hands where k is an
integer >=2, the probability is the solution to

p = 1/2 + (1/2)*p**k

but is that also true for k that is any real number >2?

Nevetheless, good answer to the post as I agree that you
answered the question that was probably asked.

Re: Another interpretation - classic problem
 

fwiw - BJ is not quite a 50-50 game....i believe these numbers are appropriate to a basic strategy player....

42% - win
49% - loss
9% - push

the player closes the gap towards a 50-50 game with 3:2 BJ payout and double-downs.

i know this isn't really relevant to the question at hand....but i just thought i would nit-pick here and point out that BJ is not exactly a coin-flip game.

oherwise, i can only barely understand what in the world you guys are talking about.

corrections
 

Suppose he starts with 1 dollar. Suppose that when he wins, he wins 2 dollars, and when he loses, he loses 1 dollar. Then if he ever has at least twice as many losses as wins, he will be broke.

He will be broke if he has more than twice as many losses as wins, which is what we want. This is the same as saying that he always wins x times before he loses 2x times as stated in the original problem. He can have x wins and 2x losses as long as he got the x wins first.


p = [sqrt(5)-1] / 2 = 61.8%.

You can verify that this is the solution. This is the probability that he will eventually have twice as many losses as wins, so the probabilty that he never has twice as many losses as wins is:

1 - [sqrt(5) - 1] / 2 =

[3 - sqrt(5)] / 2 = 38.2%

The first probability of 68.2% is that he will never have more than twice as many losses as wins, and so it is 32.8% that he will never have more than twice as many losses as wins.