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A fair die is tossed and a coin is thrown the number of times as the score shown on the die. If any heads are shown in the throws of the coin, we sop, otherwise, we continue the experiment of tossing the die and coin until at least one head is shown. Find the expected number of throws of the coin before we stop.


Denote $ N: N ^o$ shown on dice : ${1,2,3,4,5,6}$

$ \therefore \mathbf{P} (N=n) = \frac{1}{6} $

$ \mathbf{P} (X=H) = \frac{1}{2}$

$ \mathbf{E}(N=n) = \sum \mathbf{E}(N=n|X=H) \ .\ \mathbf{P}(X=H) $

$ =\frac{1}{2} \sum \mathbf{E}(N=n|X=H) $

Then I am not sure if I am on the right track. The answer is $\frac{448}{107}$

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This is a very interesting question. Thanks for sharing it! Here is my solution:

For the $k$-th tossing of the die, denote $X_k$ as the score shown on the die, and $S_k = 1$ if there is no head appeared since the first tossing until this (the $k$-th) tossing, otherwise $S_k = 0$. Formally,

$$S_k = 1\{\text{no head appeared before $k$-th tossing of the die}\}.$$

So only when $S_k = 1$ shall we toss the die for $k$-th time, otherwise we shall not toss the $k$-th die and stop here (after the $(k-1)$-th tossing of the die). By definition $S_1 = 1$. Then the total number of throwing the coin can be described as

$$T_n = X_1 + \sum_{k = 2}^{\infty}S_{k}X_{k} = \sum_{k = 1}^{\infty}S_{k}X_{k}.$$

Thus

$$\mathbb{E}T_n = \sum_{k = 1}^{\infty}\mathbb{E}(S_{k}X_{k}).$$

Since $S_{k}$ only depends on $X_1,\cdots,X_{k-1}$, it is independent of $X_k$, so

$$\mathbb{E}(S_{k}X_{k}) = \mathbb{E}S_{k} \mathbb{E}X_{k}.$$

Note that

$$\mathbb{E}[S_{k}|X_1,\cdots,X_{k-1}] = \frac{1}{2^{X_1}}\frac{1}{2^{X_2}}\cdots\frac{1}{2^{X_{k-1}}}$$

and $\mathbb{E}\frac{1}{2^{X_i}} = \frac{1}{6}(\frac{1}{2}+\cdots+\frac{1}{64})= \frac{21}{128}$, we have (for $k\geq 1$)

$$ \mathbb{E}S_{k} = \mathbb{E}(\mathbb{E}[S_{k-1}|X_1,\cdots,X_{k-1}]) = \left(\frac{21}{128}\right)^{k-1}.$$

Thus (note that $\mathbb{E}X_k = \frac{7}{2}$)

$$\mathbb{E}T_n = \sum_{k = 1}^{\infty}\mathbb{E}(S_{k}X_{k}) = \frac{7}{2}\sum_{k = 1}^{\infty}\left(\frac{21}{128}\right)^{k-1} = \frac{7}{2}\frac{1}{1-\frac{21}{128}} = \frac{448}{107}.$$

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    $\begingroup$ Here we use the iterated law of expectation: $\mathbb{E}Y =\mathbb{E}(\mathbb{E}[Y|X]) $ and the independence of $X_1,\cdots,X_n,\cdots$ $\endgroup$ – Wanshan Feb 13 '18 at 7:24

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