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.

1 Answer

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 

It is enough to show that 
We have 
\[P_k= \frac{1-(\frac{1-p}{p})^k}{1-(\frac{1-p}{p})^N}=\frac{1-(\frac{1-p}{p})^{\frac{k}{2}}}{1-(\frac{1-p}{p})^{\frac{N}{2}}}\cdot \frac{1+(\frac{1-p}{p})^{\frac{k}{2}}}{1+(\frac{1-p}{p})^{\frac{N}{2}}}.\]
\[P_k=P'_k \big ( \frac{1+(\frac{1-p}{p})^{\frac{k}{2}}}{1+(\frac{1-p}{p})^{\frac{N}{2}}} \big).     (*)\]
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 Erdos
  • Erdos Erdos

    This took me over 30 minutes to answer. You may want to offer a bounty for your questions, otherwise you may not get a response.

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