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This should be known as "gambler's ruin". In a game, at each step, you can win 1\$ or lose 1\$. Let $Z_i$ be a variable that can assume as values 1 or -1. Let $$ X_n=\sum_{i=0}^n Z_i . $$

Can you show me in details how to calculate $P(X_n \geq a)$ for a certain $a>0$? I thought that it was the case to use the cumulative binomial distribution, but I tried to compare my results with the data I have and they did not match.

As second question, I would appreciate just a little hint on how to compute that probability with excel.

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2 Answers 2

You do want to use the binomial distribution. If you have probability $p$ of winning and $1-p$ of losing on each of $n$ games, then $W$, the number of times you win, is

$$W\sim B(n,p)$$

which is equivalent to saying that

$$P(W=w) = {n\choose w}p^w(1-p)^{n-w}$$

This is related to $X$ by the formula

$$X=W-(n-W)=2W-n$$

which gives you

$$P(X=x)=P(2W-n=x)=P(W=(x+n)/2)={n\choose m} p^m(1-p)^{n-m}\quad\textrm{ where }m=(x+n)/2$$

To compute $P(X>a)$ you simply sum:

$$P(X>a) = \sum_{x>a}P(X=x)$$

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It is indeed a binomial distribution if the $Z_i$s are independent bernoulli variables. You should just sum from a to n to get $P(X_n\geq a)$.

EDIT : see my comment below for excel and @Chris Taylor answer for the math.

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The $Z_i$ aren't quite bernoulli. A bernoulli rv takes values 0 or 1, whereas here the $Z_i$ take values -1 or 1. –  Chris Taylor Jun 26 '12 at 8:31
    
Sorry for that, I read the problem to quickly. I shall just add for clarity that you need to have x+n even to get non-zero values in the formula above. –  David Jun 26 '12 at 8:47
    
On excel, depending on the value of a and n it is equivalent to the survival of a binomial distribution but not in a. If you want $P(X\geq a)$ and not $P(X>a)$ you can use : if $(a+n)/2$ even then $P(X\geq a)$ = 1-BINOMDIST((a+n)/2-1,n,p,TRUE) and if $(a+n)/2$ odd then something like 1-BINOMDIST(TRUNC((a+n)/2),n,p,TRUE). I didn't check this so it is subject to errors. –  David Jun 26 '12 at 9:13

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