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This Question is from DEGROOT's "PROBABILITY AND STATISTICS".

Consider the following three different possible conditions in the gambler’s ruin problem: (a) The initial fortune of gambler A is two dollars, and the initial fortune of gambler B is one dollar. (b) The initial fortune of gambler A is 20 dollars, and the initial fortune of gambler B is 10 dollars. (c) The initial fortune of gambler A is 200 dollars, and the initial fortune of gambler B is 100 dollars. Suppose that $p < \frac{1}{2}$. For which of these three conditions is there the greatest probability that gambler A will win the initial fortune of gambler B before he loses his own initial fortune?


The instructor manual says about solution here:

If the initial fortune of gambler A is $i$ dollars, then for conditions (a),(b) and (c), the initial fortune of gambler B is $\frac{i}{2}$ dollars.Hence $k=\frac{3i}{2}$.If we let $r=\frac{q}{p}>1$, then it follows that the probability that A will win under conditions (a),(b) and (c) is $$\frac{r^i-1}{r^{(\frac{3i}{2})}-1}=\frac{1-{(\frac{1}{r^i})}}{r^{(\frac{i}{2})}-{(\frac{1}{r^i})}}.$$ If $i$ and $j$ are positive integers with $i<j$, it follows that $$\frac{1-\big(\frac{1}{r^j}\big)}{r^\frac{j}{2}-\big(\frac{1}{r^j}\big)}<\frac{1-\big(\frac{1}{r^j}\big)}{r^\frac{i}{2}-\big(\frac{1}{r^j}\big)}<\frac{1-\big(\frac{1}{r^i}\big)}{r^\frac{i}{2}-\big(\frac{1}{r^i}\big)}.$$ Thus, the larger the initial fortune of gambler A is, the smaller is his probability of winning. Therefore he has the largest probability of winning under condition (a).


But,as I asked here, the given inequality is false in general. So, how to have this question's answer?

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  • $\begingroup$ Note: to refer to a book of 400+ pages without mentioning the page you are interested in is at best naive. $\endgroup$ – Did Oct 9 '13 at 13:46
  • $\begingroup$ Very sorry.See page 41, Section 2.5,Solution 3 of the instructor's manual to the DeGroot's book. $\endgroup$ – Silent Oct 9 '13 at 13:50
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The conclusion of the exercise holds if the quantity $$ u=\frac{r^{2n}-1}{r^{3n}-1} $$ decreases with $n$, where $n$ is a positive integer and $r\gt1$. In the situation of the exercise, $n$ is the initial fortune of B, $2n$ is the initial fortune of A, and $r=(1-p)/p$.

Note that $$ u=\frac{r^n+1}{r^{2n}+r^n+1}=1-\frac1{1+\frac1{r^n}+\left(\frac1{r^n}\right)^2}. $$ When $n$ increases, $\frac1{r^n}$ decreases hence $1+\frac1{r^n}+\left(\frac1{r^n}\right)^2$ decreases hence $u$ decreases, as claimed in the book.

(This is fortunate since, as you explain, the second inequality claimed in the solution you quote, is in fact wrong.)

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  • $\begingroup$ Thanks a lot, Sir. At last I have the answer! $\endgroup$ – Silent Oct 9 '13 at 14:03

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