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The probability that $n$ dice are rolled is $1/2^n$. Let $S_N$ denote the sum of the numbers shown on $N$ dice. Find the probability of the following events:

a) $S_N = 4$, knowing that $N$ is even;
b) $N = 2$, given that $S_N = 3$;
c) $N = 2$, given that $S_N = 4$ and the rst die showed the number $1$;
d) the largest number shown on any die is $4$, knowing that $N = 3$.

Can somebody please help me with this?

For point a, we have two cases: $N=2$ and $N=4$. When $N=2$ we can get $(1,3)$, $(2,2)$, and $(3,1)$ and $N=4$ $(1,1,1,1)$ for favourable cases.

However, I don't know where the $1/2^n$ part comes in? The total number of cases should be $6^N$, right? So $6^2 + 6^4 = 1332$. Then, the number of favorable cases/total number of cases should give $6/1332$?

I think I'm missing something important here, please help me out! Thanks

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Are $n$ and $N$ the same? They seem to be. If so, please make them the same. – Ross Millikan Oct 21 '12 at 16:45
up vote 1 down vote accepted

The first sentence means that the number of dice rolled is a random variable $N$ such that $\Bbb P(N=n)$, the probability of rolling $n$ dice, is $\frac1{2^n}$. Thus, the probability of rolling one die is $\frac12$, the probability of rolling two dice is $\frac1{2^2}=\frac14$, and so on. Since $\sum_{n\ge 1}\frac1{2^n}=1$, this makes sense.

(a) If $S_n=4$, we can’t possibly have more then $4$ dice, so if $n$ is even, $n$ must (as you said) be $2$ or $4$, and the favorable rolls are exactly the ones that you specified. We want $\Bbb P(S_N=4\mid N\text{ is even})$, which by definition is

$$\Bbb P(S_N=4\mid N\text{ is even})=\frac{\Bbb P(S_N=4\text{ and }N\text{ is even})}{\Bbb P(N\text{ is even})}\;,$$

so we should try to calculate the probabilities on the righthand side.

$$\begin{align*} \Bbb P(N\text{ is even})&=\sum_{k\ge 1}\Bbb P(N=2k)=\sum_{k\ge 1}\frac1{2^{2k}}\\ &=\sum_{k\ge 1}\frac1{4^k}=\frac{\frac14}{1-\frac14}\\ &=\frac13\;. \end{align*}$$

If $X_k$ is the random variable representing the number on the $k$-th die,

$$\begin{align*} \Bbb P(S_N=4\text{ and }N\text{ is even})&=\Bbb P(S_N=4\text{ and }N=2)+\Bbb P(S_N=4\text{ and }N=4)\\ &=\Bbb P\big(N=2\text{ and }X_1+X_2=4\big)\\ &\qquad+\Bbb P(N=4\text{ and }X_1=X_2=X_3=X_4=1)\\ &=\frac1{2^2}\cdot\left(3\left(\frac16\right)^2\right)+\frac1{2^4}\cdot\left(\frac16\right)^4\\ &=\frac14\cdot\frac3{36}+\frac1{16}\cdot\frac1{1296}\\ &=\frac1{48}+\frac1{20736}\\ &=\frac{433}{20736}\;. \end{align*}$$

Thus, $$\Bbb P(S_N=4\mid N\text{ is even})=\frac{\frac{433}{20736}}{\frac13}=\frac{433}{6912}\approx0.0626\;.$$

This should get you started; I’ll add a bit more later.

Added: For (b) we want $\Bbb P(N=2\mid S_N=3)$. If $S_N=3$, there are three possibilities: $N=1$ and $X_1=3$; $N=2$ and either $X_1=1$ and $X_2=2$, or $X_1=2$ and $X_2=1$; or $N=3$ and $X_1=X_2=X_3=1$. Thus,

$$\begin{align*} \Bbb P(S_N=3)&=\Bbb P(N=1\text{ and }S_N=3)+\Bbb P(N=2\text{ and }S_N=3)+\Bbb P(N=3\text{ and }S_N=3)\\ &=\frac12\cdot\frac16+\frac1{2^2}\cdot\frac2{36}+\frac1{2^3}\left(\frac16\right)^3\\ &=\frac{169}{1728}\;. \end{align*}$$

The term $\dfrac1{2^2}\cdot\dfrac2{36}=\dfrac1{72}$ corresponds to the cases in which $N=2$: it’s the probability that $S_N=3$ and $N=2$. It’s share of the total probability of the event $S_N=3$ is therefore

$$\Bbb P(N=2\mid S_N=3)=\frac{\frac1{72}}{\frac{169}{1728}}=\frac{1728}{72\cdot169}=\frac{24}{169}\approx0.1420\;.$$

I’ll leave the other two to you, but feel free to ask questions if you get completely stuck.

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thank you guys, you are being really helpful. The thing with the small and big 'n' confused me in the first place, I'm glad I came and asked here, thanks! – CloneXpert Oct 21 '12 at 18:33
Very good job on explaining the solution. However, I think you missed one of the cases when you are writing P(N=2 and (X1=1 and X2=3) or (X1=3 and X2=1)). Should I add the case when both of dies are 2, such as ... or (X1=2 and X2=2)? – CloneXpert Oct 21 '12 at 19:26
@CloneXpert: You’re right, I did; I’ll fix that in a few minutes. – Brian M. Scott Oct 21 '12 at 19:28

Hints: For a, assuming $n$ and $N$ are the same, the statement about the number of dice rolled says there was $\frac 12$ chance that one die was rolled, $\frac 14$ chance that two were rolled, $\frac 18$ that three were rolled and so on. Then we are told that $N$ is even. The a priori probability of that is $\sum_{i=1}^\infty \frac 1{2^{2i}}=\frac 13$, so now the chance that two were rolled is $\frac 34$, the chance that four were rolled is $\frac 3{16}$, and so on. Now the chance we got $4$ is $\frac 34 P(\text{get 4 given we rolled two dice})+\frac 3{16}P(\text{get 4 given we rolled four dice})$

For b, there are only three choices for $S_N=3$, you can roll a single 3 or roll two dice and get (1,2) or (2,1)

For d, you can ignore the probability distribution on the number of dice, as we are given it.

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I suspect that $N$ is the random variable and $n$ the value. – Brian M. Scott Oct 21 '12 at 17:07

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