# Random Walk or Gambler's Ruin

I am doing a practice problem in chapter 3 (problem 81) in my old probability book "A First Course in Probability" by Sheldon Ross, 5th Edition. Most of the problems have answers so I can check my work, however this one does not. So I am going to state the question and my answer to see what people think the right solution is. Thanks!

Question: A stock market investor owns shares in a stock whose present value is $25. She has decided that she must sell the stock if it either goes down to 10 or up to 40. If each change is either up 1 unit with probability .55 or down 1 unit with probability .45, and the successive changes are independent, what is the probability that she retires a winner? My Answer: I set this problem up as follows. I calculated the probability of a win as $$P(W) = \sum_{n=15,17,19,..} P(W|N=n)P(N=n)$$ where $$N$$ is the number of trials when she sells (this could mean a loss or a win). Notice that the minimum value of $$N$$ is 15 because she must have at least 15 up days in a row or 15 down days in a row in order to divest and retire a winner or a loser. Then the next highest value has to be 17. For example, say she has 14 up days in a row and on the 15th day she has a down day, then essentially she is back at 13 and thus needs 2 more days of ups to get to 15. Hence she needs 17 days, not 16. This trend continues on forever as she can alternate up days and down infinitely, in theory. Now let's examine $$P(W|N=15)P(N=15) = {15 \choose 15} (.55)^{15}$$. Similarly, $$P(W|N=17)P(N=17) = {17 \choose 15} (.55)^{15} (0.45)^{2}$$, and so on. Therefore I have the probability of retiring a winner as: $$P(W) = \sum_{n=15,17,..} {n \choose 15} (0.55)^{15} (0.45)^{n-15}$$ This is reminiscent of a "negative binomial-type" problem. However I cannot simplify this down at all to get an exact number. Am I wrong in my approach? Alternative Solution: Apply the Gambler's ruin formula and treat this as a contest between players A and B, where each starts with 15 dollars and plays a game until one goes broke, where P(A Wins) = 0.55. Then the answer is $$\frac{1 - (.45/.55)^{15}}{1 - (.45/.55)^{30}} \approx 0.95$$. • Your alternative answer, more precisely$\approx 0.953026$, is correct. Your$\sum\limits_{n=15,17,..} {n \choose 15} (0.55)^{15} (0.45)^{n-15}$looks much too small, about$0.006\$ – Henry Jan 4 at 13:37
• Henry -- thanks for answering! Where is my intuition wrong in the first solution? – M. Austin Jan 4 at 13:50
• Please add the self-study tag. – Xi'an Jan 4 at 14:33

Your first formula sums into approximately 0.9. It's not valid because you only account for 15 wins, i.e. $$0.55^{15}$$ term means that the other wins and losses that cancel each other are associated with probability $$0.45$$, which is wrong. If, for example, n is $$17$$, then you need to have $$16$$ wins and $$1$$ loss, i.e. for a particular ordering the probability is $$(0.55)^{16}(0.45)^{1}$$. Even if you correct the exponents, you consider all orderings which is also wrong, but for example if $$n=17$$, we can't have $$16$$ successes and $$1$$ loss afterwards because the experiment ends at $$15^{th}$$ trial, not $$17$$.
Gambler's ruin formula is correct because you can view the situation as two players each having $$15$$ coins, and winning probabilities are as listed.