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There is a little problem in the working on this one I have encountered recently. Here it goes: Starting off with \$100, when you toss a coin, if there is a tail, you will get \$10 and lose $10 if it is a head. What is the probability that after 1000 tosses, the amount of money you have is between 90 and 110?

Here is the way I do it. Let $X$ be the difference in amount when you toss a coin. Hence, X could be -10 and 10 with the probability of 1/2 each. The expected value is 0 and the standard deviation is 10. Hence, I go on formulating the sum $S = X_1 + X_2 + \ldots + X_{1000}$ and figure out the $\mu(S) = 0$ and $var(S) = 100000$ and I do not know how to go from here? Any hints will be greatly appreciated.

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  • $\begingroup$ Are you allowed to keep tossing after you hit \$0? And if yes, and you get heads, will you go into debt or stay at \$0? This will significantly affect the answer. $\endgroup$
    – Ilmari Karonen
    Commented Apr 14, 2014 at 1:18
  • $\begingroup$ That is a very insightful question, IImari. For this moment, we assume that the game is simple enough (you can be in debt) unless stop playing at \$0. It is another question to consider here. $\endgroup$
    – Viet Hoang Quoc
    Commented Apr 14, 2014 at 1:23

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The exact probability can easily be computed with the binomial distribution.

Let $X$ be a binomial random variable corresponding to the number of tails obtained in the 1000 tosses. You could do the following:
a) Determine the number of tails out of 1000 tosses that would result in the final amount of money being between 90 and 110. Denote $a$ the minimum number of tails, and $b$ the maximum number of tails.
b) The requested probability is $\Pr(a \leq X \leq b)$, which can be computed using the probability mass function of the binomial distribution.

Hint: can the final amount be 90? And 110?

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  • $\begingroup$ That is an excellent approach. Cheers. $\endgroup$
    – Viet Hoang Quoc
    Commented Apr 14, 2014 at 0:55
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You should use Central Limit Theorem and deduce your answer from normal approximation: http://www.randomservices.org/random/sample/CLT.html

Basically you sum by CLT is just a normal distribution with the mean and variance you have already calculated.

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    $\begingroup$ +1 The Binomial distribution gives $0.0252250$ while this Normal approximation gives $0.0252271$: they agree to over four significant figures. $\endgroup$
    – whuber ♦
    Commented Apr 13, 2014 at 15:36

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