# Conditional probability - Random number of dice

Can someone help me with this question ?

A random number N of dice is thrown. Let Ai be the event that N = i, and assume that P(Ai) = 1/(2)^i, i >= 1. The sum of the scores is S. Find the probability that:

(a) S = 4, given N is even;
(b) the largest number shown by any die is r, where S is unknown.

EnumaElish
Homework Helper
(a) $$P(S=4 \text{ and N is even})/P(\text{N is even})=$$
$$P\left(\left\{N=2 \text{ and }(S=1+3 \text{ or }S=2+2 \text{ or }S=3+1)\right\}\text{ OR }\{N=4\text{ and }S=1+1+1+1\}\right)\left/ \sum_{k=1}^\infty 2^{-2k}\right.$$

If the world is just, then the denominator should add up to 1/2. Because P(odd) = P(even) and P(odd) + P(even) = 1, so P(odd) = P(even) = 1/2.

So calculating P(S=4 and N is even)/P(N is even) is a matter of calculating the numerator and then multiplying it with 2.

NateTG
Homework Helper
For part b:

The chance that $r$ is the largest value shown is going to be
$$\sum_{i=1}^{\infty} P(A_i) \times p(r,i)$$
where
$$p(r,i)[/itex] is the probability that $r$ is the largest value shown by $i$ dice. Now, the chance of all of the dice being less than or equal to $r$, assuming that $r\in{1,2,3,4,5,6}$ is going to be [tex]\left(\frac{r}{6}\right)^{i}[/itex] and of those [tex]\left(\frac{r-1}{6}\right)^{i}[/itex] don't contain any $r$ So we have [tex]\sum_{i=1}^{\infty} \frac{r^i-(r-1)^i}{12^i}=\sum_{i=1}^{\infty}\left(\frac{r}{12}\right)^i - \sum_{i=1}^{\infty}\left(\frac{r-1}{12}\right)^i=\frac{r}{12-r}-\frac{r-1}{13-r}=\frac{13r-r^2-12r+r^2+12-r}{156-25r+r^2}=\frac{12}{r^2-25r+156}$$
So
$$r=6 \rightarrow \frac{12}{156+36-150}= \frac{12}{42} = \frac{2}{7}$$
$$r=5 \rightarrow \frac{12}{156+25-125}= \frac{12}{56} = \frac{3}{14}$$
$$r=4 \rightarrow \frac{12}{156+16-100}= \frac{12}{72} = \frac{1}{6}$$
$$r=3 \rightarrow \frac{12}{156+9-75}=\frac{12}{90}= \frac{2}{15}$$
$$r=2 \rightarrow \frac{12}{156+4-50}=\frac{12}{110}=\frac{6}{55}$$
$$r=1 \rightarrow \frac{12}{156+1-25}=\frac{12}{132}=\frac{1}{11}$$

The question of the part (a) is P(S = 4 | N is even) and not P(S = 4 and N is even | N is even).

Galileo
$$P(A|C)=\frac{P(A \cap C)}{P(C)}$$
Which is, by the way, trivially equal to $P(A\cap C|C)$.