# Problem similar to Gambler's Ruin

I am working on a problem that is similar to the gambler's ruin problem.

We start with a bet of $$\1$$.

• With probability $$p$$ the game is won and we win the amount of the bet, i.e., if we bet $$\1$$, we win $$\2$$.

• With probability $$1-p =: q$$ the game is lost.

If we win the game, we take our earnings and stop playing. If we lose we double our bet and keep on playing until we lose.

What is the expected amount we bet in total?

I would argue that the formula for the total bet looks like this, but I'm not sure if I made a mistake somewhere, as it would give negative bet values for a small $$p$$.

$$\sum \limits_{k=1}^{\infty}(2^k-1)p(1-p)^{(k-1)}=\frac{2p}{2p-1}-1$$

Many thanks for your time and help!

• I'm not sure we mean the same thing by "the winnings are twice the amount of the bet." If I start with one dollar, and win the first bet, do I now have two dollars or three dollars? – saulspatz Mar 8 at 15:47
• 2 dollar, how would you say that in english, so i can clarify it in the text – Ang Mar 8 at 16:00
• "You win the amount of the bet" is what I would say, but I'd also give an example to be sure it's clear. – saulspatz Mar 8 at 16:02

## 1 Answer

You can split your sum into two separate sums: $$E(p)=\sum \limits_{k=1}^{\infty}(2^k-1)p(1-p)^{(k-1)}=p\sum \limits_{k=1}^{\infty}2^k(1-p)^{k-1} - p\sum \limits_{k=1}^{\infty}(1-p)^{k-1} \\ = 2p\sum \limits_{k=0}^{\infty}2^{k}(1-p)^{k} - p \sum \limits_{k=0}^{\infty}(1-p)^{k}$$ The sum $$\sum\limits_{k=0}^{\infty}2^{k}(1-p)^{k}$$ is convergent only, if $$p>\frac{1}{2}$$. For $$p \leq 2$$ we have $$\sum\limits_{k=0}^{\infty}2^{k}(1-p)^{k} =\infty$$ The second sum is equal $$\frac{1}{p}$$ for any $$p\neq 0$$

Therefore we have: $$E(p)=\begin{cases}\frac{1}{2p-1}, & p>\frac{1}{2}\\ \infty,& p\leq \frac{1}{2}\end{cases}$$

• many thanks, just one question, why is it defined for p=1/2? i get that it diverges as p approaches 1/2, can we therefor say its infinity at P=1/2? – Ang Mar 8 at 16:06
• @Ang for $p=1/2$ we have $\sum 1$, which is definitely $\infty$ – Jaroslaw Matlak Mar 8 at 16:20
• ok i see, i was just asking bc it would become 0 in the denominator...., but i see now - many thanks! – Ang Mar 8 at 16:25
• You could simplify $\frac{2p}{2p-1}-1$ to just $\frac1{2p-1}$. – Théophile Mar 8 at 16:26