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Disclaimer, I'm a computer person, not a math person, so surely I'm not going to use the proper math lingo. I apologize in advance. Feel free to edit to make this question more readable to others. This is a 2 part question:

  1. What is the probability of achieving a goal that has a probability of $1/n$, with $n$ tries, as $n$ goes to infinity?

  2. The actual question I'm trying to solve is related to a hypothetical poker game, and I believe its answer hinges on the answer to question 1 above. The game is as follows: Player 1 $(P_1)$ begins with a score of $x$, Player 2 $(P_2)$ begins with a score of $N*x$. Both players "go all in" on the flip of a coin comparison, and wager the amount of the lower player's score. If $P_1$ gets down to 0, he starts with $x$ again. We continue until $P_2$ has 0. What is the expected number of times this game can be played until $P_2$ hits 0, in terms of $x$ and $N$?

I wrote a simple computer program to illustrate the game, and played it with x = 1 and N = 100. After 20 or so tries, here was one result that was short enough to paste here:

Starting Position - $P_1:1, P_2:100$

Hand Count: 1 $P_1:2, P_2:99$

Hand Count: 2 $P_1:4, P_2:97$

Hand Count: 3 $P_1:0, P_2:101$

Hand Count: 4 $P_1:0, P_2:102$

Hand Count: 5 $P_1:2, P_2:101$

Hand Count: 6 $P_1:4, P_2:99$

Hand Count: 7 $P_1:0, P_2:103$

Hand Count: 8 $P_1:2, P_2:102$

Hand Count: 9 $P_1:4, P_2:100$

Hand Count: 10 $P_1:0, P_2:104$

Hand Count: 11 $P_1:2, P_2:103$

Hand Count: 12 $P_1:4, P_2:101$

Hand Count: 13 $P_1:8, P_2:97$

Hand Count: 14 $P_1:16, P_2:89$

Hand Count: 15 $P_1:32, P_2:73$

Hand Count: 16 $P_1:64, P_2:41$

Hand Count: 17 $P_1:105, P_2:0$

RESULTS:

$P_1$ stated with 1. $P_1$ Ended with 105.

$P_2$ started with 100. $P_2$ Ended with 0.

Total Hands played until $P_1$ busts $P_2$: 17

Total amount $P_1$ invested: 5

Notes: Hand #4 shows the result after Player 1 performed a "rebuy" of 1 and immediately lost. Hand #5 show the result after $P_1$ performed a rebuy of 1 and won, etc. In order for Player 1 to ever win this game, he must win 7 in a row, and I'm thinking he has about 128 tries to do it before he then has to hit 8 in a row, then 256 tries to achive the $1/256$ before getting 512 tries at $1/512$, etc.

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2 Answers 2

The probability of not succeeding in question 1 is $\left(1-\frac1n\right)^n$ and $$\lim_{n\to\infty}\left(1-\frac1n\right)^n=e^{-1}$$ is "well-known", thus giving a limit probability of $\approx 63.212\,\%$.

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Thanks! I edited my question and removed my guess, which was pointless for me to include to begin with. –  TTT Apr 13 '13 at 17:16
    
Do you think the answer to question 2 is related to Euler's number, or should I strip that assumption as well? –  TTT Apr 13 '13 at 17:28
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The value of $x$ will have no effect on the probability (since P2's money is just a multiple and the rebuy is $x$ again), but $N$ obviously will.

Let's just set $x=1$ and $X_n$ as the number of times P1 lost by the $n$th try, so $X_n$~$Bin(n,\frac{1}{2})$. We then have $N+1+X_n$ money in the game. To win, P1 will have to get a winning streak of lenght $log_2(N+1+X_n)-1$ (base $2$ because he doubles every time, $-1$ because the oppenent loses that amount, so he has to win one game less).

So let's say $Y_n=1$ iff the $n$th try starts a winning streak and $0$ elsewise. So $P(Y_n=1)$ is $\frac{1}{2}(\frac{1}{2})^{log_2(N+1+X_n)-1} =\frac{1}{N+1+X_n}$ under the condition that P1 did not win beforehand.

It gets ugly from then on, so we will do some approximations. You can start with approximating $X_n$ by its expected value of $\frac{n}{2}$. Then $P(Y_n=1)=[\prod_{j=1}^{n-1}(1-\frac{1}{N+1+\frac{j}{2}})](\frac{1}{N+1+\frac{n}{2}})$ and thus, roughly, $$E[M]=\sum_{n=1}^{\infty}nP(Y_n=1)=\sum_{n=1}^{\infty}n[\prod_{j=1}^{n-1}(1-\frac{1}{N+1+\frac{j}{2}})](\frac{1}{N+1+\frac{n}{2}})$$ for the expected number of tries $M$ until P1 starts a winning streak.

This is just a quick glance at the problem with some possible errors and the approximations may cause problems, but I hope it goes into the right direction.

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