# Expected sum of money

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. • 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. – Ilmari Karonen Apr 14 '14 at 1:18 • 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. – Viet Hoang Quoc Apr 14 '14 at 1:23

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?

• That is an excellent approach. Cheers. – Viet Hoang Quoc Apr 14 '14 at 0:55

You should use Central Limit Theorem and deduce your answer from normal approximation: http://www.math.uah.edu/stat/sample/CLT.html

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

• +1 The Binomial distribution gives $0.0252250$ while this Normal approximation gives $0.0252271$: they agree to over four significant figures. – whuber Apr 13 '14 at 15:36