Recently I came across the Penney's game as I encountered the following probability question with answer.
Given a fair coin. If we tossed the coin thrice, then the probability of obtaining $HHT$ before $HTH$ is $\frac{2}{3}.$
Just after the question above, I encounter another similar probability question with answer.
Given a fair coin. If we tossed the coin twice and obtained either $HH$ or $TT$, we tossed twice again. If we obtained either $HT$ or $TH$, we stopped tossing. Then the probability of obtaining $HT$ and $TH$ are the same, that is, $\frac{1}{2}.$
Source: what is the probability of BOB winning the game?
My thought:
In the first question, if we disregard the first $H$, then we are actually seeking the probability of obtaining $HT$ before $TH,$ which seems to me is the second question.
I do not understand why the answers to both questions are different.
Any hint is appreciated.
Best Answer
With 2 coins it is of course perfectly symmetrical, so it makes sense it would be $\frac{1}{2}$ for each.
But why, you ask, would this change for $HHT$ vs $HTH$, if both sequences are the same for the first $H$, and then are followed up by the same $HT$ vs $TH$ difference?
Here's the difference: after you get that first $H$, the $HHT$ is bound to win if you get $H$ immediately after that first $H$, whereas if you get a $T$ at that point, then either $HTH$ is going to win ... or you get another $T$, and now the game 'resets'. So, the $HHT$ has an inherent advantage.
With the $HT$ vs $TH$ no such asymmetry will ever be obtained, because after the first throw the winner is immediately known.