Probability Question (fair dice)

Question

Roll a fair (six-sided) die, and let $N$ be the result. Now toss $2N$ fair coins and let $M$ be the number of heads. i) Find $\mathbb{E}[M]$. ii)Find $\mathbb{P}(M = 7)$.

I have given a shot at this question and my understanding for part i) is the following: $\mathbb{E}[N]=1/6(1+2+3+4+5+6)=3.5$

$Y=2N \implies \mathbb{E}[Y]=2\mathbb{E}[N]=2(3.5)=7$ Therefore, $\mathbb{E}[M]=7*0.5=3.5$ however, for part ii), I'm unsure of how to go about it. would we treat it as a conditional probability with $N$? Any feedback/help would be great!

Answer 1

Given that there are only 6 possibilities for N, we can just brute force our way through all 6 solutions.

N = 1, 2 or 3

This means that the number of coins flipped is 2, 4 or 6. This is less than 7, so the probability of *M equalling 7* is 0.

N = 4

Now 8 coins are flipped. 7 coins are flipped to heads in exactly 8!/(7! * 1!) (= 8) scenarios, and given that there are 2**8 = 256 ways to flip 8 coins, this gives an 8/256, or 1/32, chance of it occuring.

N = 5

Now 10 coins are flipped. 7 coins are flipped to heads in exactly 10!/(7! * 3!) (120) ways, and given there are 2**10 = 1024 ways to flip 10 coins, this gives a 120/1024, or 15/128, chance of it occuring.

N = 6

Now 12 coins are flipped. 7 coins are flipped to heads in exactly 12!/(7! * 5!) (792) ways, and given there are 2**12 = 4096 ways to flip 12 coins, this gives a 792/4096, or 99/512, chance of it occurring.

Now all that remains is to take the average of all these values:

(0+0+0+(1/32)+(15/128)+(99/512))/6 = 0.05696614583, or roughly 6%

Hope this helps! I'm personally not sure how to solve in the case where the die has a huge/arbitrary number of faces though, seems like an interesting problem to try.

Answer 2

Let $H_{j} \sim \mathrm{Ber}(\frac{1}{2})$ be the success of the $j^{\mathrm{th}}$ head flip and note that $M = \sum_{j=1}^{2N} H_{j}$. Also, note that $M \, | \, N \sim \mathrm{Bin}(2N, \frac{1}{2})$.

(a) As you've already found, apply the law of total expectation:

\begin{align*} \mathbb{E}[M] &= \mathbb{E} [\mathbb{E}[M|N]] = \mathbb{E} \left [\mathbb{E} \left [\sum_{j=1}^{2N} H_{j} | N \right ] \right ] = \mathbb{E} \left [\sum_{j=1}^{2N} \mathbb{E} [H_{j} | N ] \right ] = \mathbb{E} [2N \mathbb{E}[H_{1}]] = \mathbb{E}[N] = \frac{7}{2}. \end{align*}

(b) Similar to (a), apply the law of total probability:

\begin{align*} \mathbb{P}(M = 7) &= \sum_{j=1}^{6} \mathbb{P}(M=7 \, | \, N=j) \mathbb{P}(N=j) \\ &= \frac{1}{6} \sum_{j=4}^{6} \mathbb{P}(M=7 \, | \, 2N=2j) \\ &= \frac{1}{6}\left (C_{7}^{8}\, \frac{1}{2^{8}} + C_{7}^{10} \frac{1}{2^{10}} + C_{7}^{12} \frac{1}{2^{12}} \right ) \\ &= \frac{175}{3072}. \end{align*}