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A friend and I play a game. We each start with two coins. We take it in turns to toss a coin; if it comes down heads, we keep it, if tails, we give it to the other. I always go first, and the game ends when one of us wins by having all four coins. If we play this game 840 times, what is the expected number of games that I would win?

I found the probability of me winning in a single game to be 1/4. I multiplied 840 by 1/4 to have an answer of 210. However, it turns out to be wrong since 360 was written in the key. How was that happened?

Karl Henry Dahao
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1 Answers1

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Say that my probability to win is $p$, and I want to find the value of $p$.

If I get head in the first flip (0.5 chance), my friend can win with $p$ probability or lose with $(1-p)$ probability. $\frac{1-p}{2}$

If i get tail in the first flip (0.5 chance), then my friend and I keep getting heads until eventually he get tail, the game reset to the start and I can win with $p$ probability. $\sum_{i=1}^{\infty}{\frac{p}{4^{i}}}$

$$ \begin{align} p&=\frac{1-p}{2}+\sum_{i=1}^{\infty}{\frac{p}{4^{i}}}\\ &=\frac{1-p}{2}+\frac{p}{3} \end{align} $$

Solve for $p$ and you will get $p=\frac{6}{14}$

acat3
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