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Two coins with heads probabilities $\frac{1}{3}$ and $\frac{1}{4}$ are alternately tossed, starting with the $\frac{1}{3}$ coin, until one of them turns up heads. Let $X$ denote the total number of tosses, including the last. Find:

$$E(X)$$

I have seen a recent post that describes the answer as:

$$\mathbb{E}(X) = \displaystyle 1 + \frac{2}{3} (1 + \frac{3}{4} \mathbb{E}(X))$$

This still confuses me because how can you solve for $E(X)$ with $E(X)$ in the equation?

Fototekt
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1 Answers1

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That's not the answer, but it's very close. All you need to do is to solve the equation $u = 1 + (2/3)[1 + (3/4)u]$, which is a linear equation in one variable.

Anonymous
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