1

Solutions are available for numbers of flips expected in order to obtain HT and HH. The numbers differ, yet all single flip outcomes have equal probability. Does anyone have a (possibly qualitative) account of why they differ?

pathlets
  • 13

4 Answers4

3

Whether you are aiming for HT or for HH, in either case you would first have to toss a head. So suppose you toss as many times as necessary to get the first H. You then toss one more time to try to get the second half of your goal sequence HT or HH.

If you are aiming for HH and that last toss failed to be an H, you must have tossed an T. You will now essentially have to start over to get your first H again.

If you are aiming for HT and that last toss failed to be a T, you must have tossed an H. You don't have to start all over, as that failure still gives you the first half of your goal HT.

So if you repeatedly toss a coin, then HT will on average come up earlier than HH.

Jaap Scherphuis
  • 13,660
1

Without doing the actual numbers:

Suppose you want to get HH and have flipped an H. You have a 50% chance of completing with an H and a 50% chance of getting a T and going back to needing the length 2 sequence HH.

Suppose you want to get HT and have flipped an H. You have a 50% chance of completing with a T and a 50% chance of getting an H and then needing the length 1 sequence T.

nickgard
  • 4,116
1

Aiming for HT.   You flip $X_{\sf H}$ times, some count of tails then the first head, then you flip $X_{\sf T}$ times, some count more heads then the next tail, and then you are done.   This is a straight up sum of two Geomentric random variables.   The expected count of tosses is $4$.

$$\mathsf E(X_{\sf HT})=\mathsf E(X_{\sf H}+X_{\sf T}) = 4$$

Aiming for HH.   You flip $X_{\sf H}$ times, some count of tails then the first head, then you flip $1$ more time and either (1) get a tail then start over, or (2) get a head and then you are done.   This is a recursive process, and the expected count of tosses is $6$.

$$\mathsf E(X_{\sf HH}) = \mathsf E(X_{\sf H})+1+\tfrac 12\mathsf E(X_{\sf HH})\\ \mathsf E(X_{\sf HH})~{=2\mathsf E(X_{\sf H})+2 \\= 6}$$

In each case, you toss the coin until the first head, then toss the coin again to either get what you want or don't. What happens if you don't get what you want is where the cases differ. In the first case (HT), you continue until you get what you want; in the second case (HH), you have to start again and that is at least one more loop.

Graham Kemp
  • 129,094
0

This is essentially because if you get H then you're halfway to HH or HT. If you get a T you haven't started any of the patterns.

The reason HH has expectation 6 instead of 4 is that you "lose a symbol" when a T occurs after H (with prob. 1/2) and you need to start again.

Pictorially if there is no HH in n tosses, half of those still end with H, and we can extend it.

$$E(HT) = (E(H)+E(T))*Pr(Toss ~2 ~is ~T)+(E(H)+E(HT))*Pr(Toss ~2 ~is ~H)$$

$$E(HT) = (E(H)+E(T))*(1/2)+(E(H)+E(HT))*(1/2)$$ move $E(HT)$to the left

$$(1/2) E(HT) = (E(H)+E(T))*(1/2)+(E(H))*(1/2)$$

so by a renewal type argument, since E(H)=E(T)=1/Pr(H)=1/Pr(T)=2

E(HT)= 2*(4*(1/2)+ 2* (1/2))=6.

There is also a related question here

kodlu
  • 9,338
  • These responses very helpfully address my request for a simple and plausible account. I think there appears to be a parallel here with the so-called 'principle of restricted choice' at bridge (the card game), the Monty Hall puzzle and the Foshee boy born on a Tuesday puzzle. I wonder if there is an over-arching principle or whether that is just fanciful thinking. – pathlets Sep 01 '17 at 14:33