If you roll a fair 6-sided die and then flip a fair coin that number of times, what is the probability that you will get at least two heads?

Question

My idea is to use disjoint events and calculating the probability of getting at least two heads for each number rolled. For example, if I roll a 3, I would calculate the probability with the expression $(\frac{1}{6}) (\frac{1}{2})^3 \binom{3}{2} + (\frac{1}{6}) (\frac{1}{2})^3\binom{3}{3})= \frac{1}{12}$ and then add up the probabilities of getting at least two for each rolls, since the events are disjoint, summing to $\frac{67}{128}$. Is this a valid solution? Is there a better approach to solving this problem?

Answer 1

It is valid. You find that P(at least 2 heads|die=1) = 0, P(at least 2 heads|die=2)=1/4, P(at least 2 heads|die=3)=1/2, P(at least 2 heads|die=4)=11/16, P(at least 2 heads|die=5)=13/16, and P(at least 2 heads|die=6)=57/64. Then 1/6*(0+1/4+1/2+11/16+13/16+57/64)=67/128.

There is a way you can numerically approximate the answer and that is to use simulation. You can write code to run 10000 rolls of the die to calculate the probability that you get at least 2 heads. Then do this 100 times, and on each iteration gets a probability of getting at least 2 heads. The mean of these 100 probabilities is a cross-validated probability of 0.523544. We can check that $\frac{67}{128}\approx0.5234375$, which is very close.

Answer 2

You can compute this much simpler. The probability that you get not more than one head out of $n$ flips is $\frac{n+1}{2^n}$. Therefore the probability in question is: $$ \frac16\sum_{n=1}^6\left(1-\frac{n+1}{2^n}\right). $$

Answer 3

There is a 1/6 chance you will roll a 1. Then you CAN'T flip two heads. The probability is 0.

There is a 1/6 chance you will roll a 2. The probability of flipping two head in two flips is (1/2)(1/2)= 1/4. The probability is (1/6)(1/4)= 1/24.

There is a 1/6 chance you will roll a 3. The probability of flipping two heads,and then a tail, HHT, (1/2)(1/2)(1/2)= 1/8. But the probability of HTH or THH is the same. The probability of two heads in three flips is 3/8. The probability is (1/6)(3/8)= 1/16.

There is a 1/6 chance you will roll a 4. The probability of flipping two heads,and then two tails, HHTT, (1/2)(1/2)(1/2)(1/2)= 1/16. There are $\frac{4!}{2!2!}= 6$ permutations, HHTT, HTHT, THHT, THTH, and THHT, all having the same probability, 1/16. The probability of two heads in four flips is 6/16= 3/8. The probability is (1/6)(3/8)= 1/16.

There is a 1/6 chance you will roll a 5. The probability of flipping two heads, and then three tails is $(1/2)^5= 1/32$. There are $\frac{5!}{2!3!}= 10$ permutations (I won't write them all) so the probability of two heads and three tails in any order is $\frac{10}{32}= \frac{5}{16}$. The probability is (1/6)(5/16)= 5/96.

There is a 1/6 chance you will roll a 6. The probability of flipping two heads then four tails is $(1/2)^6= 1/64$. There are $\frac{6!}{2!4!}= 15$ permutations so the probability of two heads and four tails in any order is [tex]\frac{15}{64}[/tex]. The probability is (1/6)(15/64)= 5/32.

If you roll a die then flip a coin that number of times the probability you get exactly two heads is 1/24+ 1/16+ 1/16+ 5/96+ 5/32= 4/96+ 6/96+ 6/96+ 5/96+ 15/96= 36/96= 3/8.