4

Out of all of the potential sequences of 73 Heads/Tails games, each being Heads or Tails, how many sequences contain 37 tails and 36 heads?

Express the output in terms of factorials.

Because there are sequences of 73 games my initial thought is, that the answer would be $\frac{73!}{36!\cdot 37!}$.

Would this be correct?

callculus42
  • 30,550
MathsUndergrad
  • 195

2 Answers2

1

Yes that is correct. You can basically assume that you have 73 boxes and you have to fill 36 boxes with heads and 37 boxes with tails. Thus you can fill heads first and then fill the remaining ones with tails. Thus the answer is $\dbinom{73}{36}$.

Akshay Arora
  • 130
  • Thank you. If there were to be the 73! possible sequences of heads and tails, would it be true that precisely half of these end with a Tails? Since 73! is even and the odds of picking tails each time is 0.5 it suggests that this would be correct but I am unsure how to explain it? – MathsUndergrad Oct 10 '15 at 14:44
  • Firstly, no there are not $73!$ possible sequences of heads and tails. There are $\dbinom{73}{36}$. Secondly if your question is how many of these sequences will end with a tail, you can simply fix the last tail (that is fill the last box with tail first) and then arrange 72 remaining heads and tails. Thus the answer to that would be $\dbinom{72}{36}$. – Akshay Arora Oct 10 '15 at 14:49
  • Possible sequences in my above comment means pertaining to the constraints specified by you. – Akshay Arora Oct 10 '15 at 14:56
  • Ahh right, I understand. So if we fix the last tail, wouldn't this mean that there are 72C36 out of 73C37 sequences with the last being a tail, giving a fraction of odds 37/73 rather than 1/2? – MathsUndergrad Oct 10 '15 at 15:01
  • Yes, that's correct. – Akshay Arora Oct 10 '15 at 15:02
0

You are correct. The number of coin flip sequences of length $n$ with exactly $k$ heads is $$\binom{n}{k} = \frac{n!}{k!(n-k)!},$$ because that is the number of ways to choose the $k$ positions in which the heads appear. Letting $n = 73$ and $k = 36$ gives you your answer.

Eric Tressler
  • 4,309
  • Thank you. If there were to be the 73! possible sequences of heads and tails, would it be true that precisely half of these end with a Tails? Since 73! is even and the odds of picking tails each time is 0.5 it suggests that this would be correct but I am unsure how to explain it? – MathsUndergrad Oct 10 '15 at 14:42
  • @MathsUndergrad There are not $73!$ possible sequences, but $2^{73}$. If we count the number with $0,1,2,\ldots,73$ heads, we get the identity $\sum_{k=0}^{73} \binom{73}{k} = 2^{73}$. Of these $2^{73}$ sequences, though, exactly half end with tails. It may help you to write all of this out on a couple of sheets of paper for $n=3,4,5$ (I really recommend that you do, it will make this all much less abstract). – Eric Tressler Oct 10 '15 at 14:47