# Gambler's Ruin - Chance of winning increases by doubling the stakes

A coin is tossed repeatedly, heads appearing on each toss with probability $p$. A gambler starts with initial fortune k (where $0 < k < N$); he wins one point for each head and loses one point for each tail. If his fortune is ever $0$ he is bankrupted, whilst if it ever reaches $N$ he stops gambling to buy a Jaguar. Suppose that $p < \frac{1}{2}$. Show that the gambler can increase his chance of winning by doubling the stakes. You may assume that $k$ and $N$ are even.

• I think you should offer a bounty, otherwise users may not have enough incentive to spend time on thin kind of questions.

It is well-known that the probability of the gambler winning $N$ dollars before going bankrupt with starting fortune $k$ is
$P_k= \frac{1-(\frac{1-p}{p})^k}{1-(\frac{1-p}{p})^N}, p\ne q.$

This is assuming that every time the gambler bets 1 dollar, and they need $N$ successes to win a total of $N$ dollars.

Now if the bets are doubled, then the probability of winning $N$ dollars is the same as probability of winning $\frac{N}{2}$ dollars with one dollar bets with starting fortune of $\frac{k}{2}$. Thus the probability is
$P'_k=\frac{1-(\frac{1-p}{p})^{\frac{k}{2}}}{1-(\frac{1-p}{p})^{\frac{N}{2}}}.$

It is enough to show that
$P'_k>P_k.$
We have
$P_k= \frac{1-(\frac{1-p}{p})^k}{1-(\frac{1-p}{p})^N}=\frac{1-(\frac{1-p}{p})^{\frac{k}{2}}}{1-(\frac{1-p}{p})^{\frac{N}{2}}}\cdot \frac{1+(\frac{1-p}{p})^{\frac{k}{2}}}{1+(\frac{1-p}{p})^{\frac{N}{2}}}.$
Hence
$P_k=P'_k \big ( \frac{1+(\frac{1-p}{p})^{\frac{k}{2}}}{1+(\frac{1-p}{p})^{\frac{N}{2}}} \big). (*)$
Notice that since $p<\frac{1}{2}$, $$\frac{1-p}{p}>1,$$
thus $N>k$ implies
$1+(\frac{1-p}{p})^{\frac{N}{2}}>1+(\frac{1-p}{p})^{\frac{k}{2}} \Rightarrow \frac{1+(\frac{1-p}{p})^{\frac{k}{2}}}{1+(\frac{1-p}{p})^{\frac{N}{2}}}<1.$
In view of (*) this implies that
$P'_k >P_k.$
The proof is complete!

Erdos
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• Erdos
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