What is the probability that the number of heads is equal to the number showing on the die?

For this question, I know how to calculate the sample space, but I'm not sure how to use permutations or combinations. Here is what I have so far. Can anyone please help me out?

Suppose we roll one fair six-sided die, and flip six coins. What is the probability that the number of heads is equal to the number showing on the die?

$S = 6 + 2^6 = 70$

$P(1H, 1) = \frac{7!}{5!2!*70}$

$P(2H, 2) = \frac{7!}{4!3!70}$

$P(3H, 3) = \frac{7!}{4!3!*70}$

$P(4H, 4) = \frac{7!}{5!2!*70}$

$P(5H, 5) = \frac{7!}{6!*70}$

$P(6H, 6) = \frac{7!}{70}$

$Total = P(1H, 1) + P(2H, 2) + P(3H, 3) +P(4H, 4) +P(5H, 5) + P(6H, 6)$

• For a start, $S=6\cdot 2^6 = 384$. Then you can't arrange the die into the coins, so you should be choosing your head-facing coins from $6$, not $7$ – Joffan May 12 '18 at 22:35

2 Answers

The sample space is the product of the sample spaces for the dice and the coins. If you order the coins the space is $6 \cdot 2^6=384$

Now consider the number of ways to get each total of heads from $1$ to $6$. They should add to $63$ because you don't consider the $1$ way to get no heads. For each of them there is one favorable roll of the die, so there are $63$ favorable combinations. The probability is then $$\frac {63}{384}=\frac {21}{128}$$

• So the way to get heads would be total subtract the one way to get no heads? – dg123 May 12 '18 at 23:38
• Can you explain more about where the 63 came from? – dg123 May 12 '18 at 23:40
• There are $2^6=64$ tosses of six coins if you consider order. One of those has no heads and cannot match the die. – Ross Millikan May 13 '18 at 0:01
• So then what about the probability for rolling the die? – dg123 May 13 '18 at 2:28
• An alternative presentation is $\dfrac{63}{64} \times \dfrac16$ for the probability you get at least $1$ head, multiplied by the conditional probability that the die then matches the number of heads – Henry May 13 '18 at 10:42

You seem to be using Binomial coefficients $\large 7\choose k$ for different values of $k$, which would be relevant if you were repeating some binary experiment $7$ times identically and independently, which you are not. One of your seven experiments is drastically different from the others. It is independent from the other six though, which means that you may compute probabilities separately and multiply them in the end ($P(A\cap B)=P(A)\cdot P(B)$ for independent events $A,B$).

Bottom line: use the binomial approach for the coin and then multiply with the probability of getting a given number when you roll a dice.