Shell game: probability to earn $1000 Now I'm learning Probability Theory and recently I got stuck with the following problem:
Player decided to play the shell game with the bet size \$1. At the start he has \$10 and he wants to get out from the game with \$1000(including his funds at start). What is the probability that he will earn desired money? Is it more profitable to always go "all in"?
UPD
At each play our player is supposed to make \$1 bet. If he wins he will get \$2(assuming that \$1 is his bet + \$1 is his prize). If he lose then he just lose his \$1
My thoughts:
Let $A_n$ and $B_n$ represent the number of wins and losses respectively(in $n$ plays). Then he will get his \$1000 in $n$ plays iff the following two conditions holds:
$A_n-B_n+10=1000 \tag{1}$
$A_k>B_k-10 \space \forall k: 1 \le k \le n \tag{2}$
The second one doesn't let the player lost his all funds: if at some step he run out of money he won't be able to continue the game.
It is easy to show that $A_n+B_n=n$. Then from the equation $(1)$ follows:
$$
A_n-(n-A_n)+10=1000 \\
A_n=\frac{n+990}{2} \\
B_n=\frac{n-990}{2}
$$
And now we need to mention that $n \ge 990$ since he can't earn \$1000 earlier.
The probability that he will get \$1000 in $n$ plays:
$$
\binom{n}{A_n} \big(\frac{1}{3}\big)^{A_n} \big(\frac{2}{3}\big)^{B_n}
$$
But it is not the true in general since it includes the cases when $(2)$ doesn't hold. How can I  exclude these cases? I believe that if I somehow exclude them then by substituting the expression for $A_n$ and $B_n$ I'll be able to evaluate the series:
$$
\sum_{n=990}^{\infty} \binom{n}{A_n} \big(\frac{1}{3}\big)^{A_n} \big(\frac{2}{3}\big)^{B_n}
$$
and thus, find the answer. Am I right?
Or does there exists simpler way to solve the problem?
 A: You need to specify the chance that the player wins.  The dividing point is $\frac 12$ because that makes it a fair game.  If the player wins less, the house has the advantage.  The player then needs to be lucky and fewer bets means he doesn't need to be so lucky.  He should bet all his money every time until he has $500$ or more, then bet just enough to hit $1000$.  As he is doubling, he gets to $640$ with $6$ wins, then bets $360$.  If he wins the first $7$ tries he has his $1000$.  If he loses one of the first six, he is broke and done.  If he wins six and loses one, he now has $280$ and keeps going.  We could find the series of wins and losses that terminate the game and their probability if we knew the probability of winning one game.
If the winning chance is exactly $\frac 12$ it doesn't matter how the player bets.  As the game is fair, he has $\frac 1{100}$ chance of ending with $1000$ and $\frac{99}{100}$ chance of ending with $0$.  The expected value of the ending condition is the same $10$ he started with.
If the winning chance is greater than $\frac 12$, the game is in the player's favor and he wants to play a lot.  Playing the minimum bet gives him the best chance to win at the price of boredom.  The more plays, the lower the variance of the outcome, which is what the player wants.
