Players $A$ and $B$ play a game where whoever loses gives $USD 1$ to the winner. They repeat the game until one player is bankrupt. Suppose Player A has more skill, and wins each game with probability $2/3$. If Player A starts with $USD 1$, and Player B begins with $USD 2$, whats the probability that Player B wins?

Really don't know how to solve this... any links, hints or suggestions?

  • $\begingroup$ Have you studied random walks ? $\endgroup$ – true blue anil Oct 20 '15 at 3:54
  • $\begingroup$ Have just started, looking for a formula to solve the above... $\endgroup$ – mathstudent12321 Oct 20 '15 at 4:00
  • $\begingroup$ So, you mean to say player A wins with a probability of 2/3 and B with a probability of 1/2 ? $\endgroup$ – gar Oct 20 '15 at 8:54
  • $\begingroup$ No A wins with probability 2/3 and B with probability 1/3, because they are playing against each other $\endgroup$ – mathstudent12321 Oct 20 '15 at 9:58
  • $\begingroup$ In that case, your question is answered in wikipedia and many other resources. I was thinking something like during A's turn, A wins with a probability of 2/3 and during B's turn, B has equal chance of winning or losing. That would be much more difficult. $\endgroup$ – gar Oct 20 '15 at 10:55

Assuming no one has won yet, then A will have \$2 on even turns (and cannot lose) and B will have \$2 on odd turns, and can therefore only win on odd turns.

The probability that B wins on then $n^{th}$ odd turn is the probability that A and B exchange wins for $2n-2$ turns and then B wins on an odd turn.

The probability that they exchange wins for 2n-2 turns is $(\frac{2}{3}\frac{1}{3})^{n-1}$ and the probability that B wins on this turn is that times 1/3.

Since the probability that B wins on an even turn is zero, the probability that B wins is the sum for all n's that he wins on the $n^{th}$ odd turn.

P(B wins)$=\sum\limits_{n=1}^\infty (\frac{2}{3}\frac{1}{3})^{n-1}*\frac{1}{3}=\frac{1/3}{1-2/9}=\frac{9}{21}$


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