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I have the following problem:
A fair coin is tossed. If it shows head(tail), two(six) dice are rolled. Let S be the sum of the numbers displayed by the dice. What is the expectation of S?

So the result is $E[S] = P('head')E[S|'head'] + P('tail')E[S|'tail'] = 14$
My question is how can I calculate $E[S|'head']$ and $E[S|'tail']$?

Thank you

Anonymous283
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  • Well, what Sum do you expect if you throw two dice? (Hint: what sum do you expect if you throw one die?) – lulu Jul 20 '21 at 22:17
  • Sum for two dice {2,..,12}? – Anonymous283 Jul 20 '21 at 22:25
  • How is that an expected value? What's the answer for a single die? – lulu Jul 20 '21 at 22:36
  • Request clarification: if I read the question correctly, you only throw the dice if the coin shows Heads. Does this mean that if the coin shows Tails, that that is interpreted as throwing $0-0$ on the dice, so $(1/2)$ the time the sum of the dice is $(0)$? Have I interpreted the question correctly? – user2661923 Jul 20 '21 at 23:12
  • @user2661923 That's what the the "head(tail)" and "two(six)" is meant to indicate. Six die are thrown if tails is shown. – Graham Kemp Jul 20 '21 at 23:13
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    @GrahamKemp Okay, thanks. Then the expected number of dice that are thrown is $\left{ ~\left[(1/2) \times 2\right] + \left[(1/2) \times 6\right] ~\right} = 4$. In that case, the comments of lulu are conclusive. – user2661923 Jul 20 '21 at 23:17
  • @lulu Oh i misunderstood the question. So I know that E[S|'head'] = 2 * expectation value for one die, but by the formula E[X|A] = sigma X(w)P({w}|A) for all w in Omega. How can I calculate E[S|'head'] according to the formula? – Anonymous283 Jul 20 '21 at 23:23
  • @Anonymous283 $X(\omega)$ under the condition that the coin shows heads, will be the sum of two die for all outcomes where that happens.$$\begin{align}\Bbb E(S\mid\textsf{head}) &= \sum_{x=1}^6\sum_{y=1}^6 (x+y)\mathsf P(D_1=x, D_2=y\mid \text{Coin}=\textsf{head})\&=\sum_{x=1}^6 x\mathsf P(D_1=x)+\sum_{y=1}^6 y\mathsf P(D_2=y)\&~~\vdots\end{align}$$ – Graham Kemp Jul 20 '21 at 23:34

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Let $D_i$ be the result for throwing a single die. These will be independent and identically distributed, and for all $i$, $\mathbb E(D_i)$ is easy to find by applying the definition of expectation.

$\mathbb E(S\mid\textsf{head})$ is the expected sum of two dice.$$\mathbb E(S\mid\textsf{head}) = \mathbb E(D_1+D_2)$$

$\mathbb E(S\mid \textsf{tail})$ is the expected sum of six dice.$$\mathbb E(S\mid\textsf{head}) = \mathbb E(D_1+D_2+D_3+D_4+D_5+D_6)$$

Both are easy to find by applying the Linearity of Expectation.

Note: Also, since we expect to throw four dice, $\mathsf E(S)$ is the expected sum of four dice.

Graham Kemp
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