It Is Harder To Get HT?

Probability Level 2

Alice and Bob play the following game:

A fair coin is tossed repeatedly until they get a consecutive sequence of Heads-Heads or Heads-Tail.
Alice wins if a sequence Heads-Heads shows first.
Bob wins if a sequence Heads-Tails shows first.

Who is more likely to win this game?

Alice Equally likely Bob

1 solution

Brian Moehring
Feb 28, 2017

Assume the first H occurs on the $n$ th flip.

If flip $n+1$ is a H, Alice wins, and if it is a T, Bob wins. Since H and T are equally likely to occur on flip $n+1$ and it is independent of all past flips, the two are equally likely to win.

Your argument seems very clear. But if we were to consider expected values then Bob has better chances. E(HT) = 4, E(HH) = 6.

Siva Bathula - 4 years, 3 months ago

To clarify, what E(HH) refers to is "the expected number of coin tosses to get a HT is 4".

Note that the probability of getting HT or HH on two coin tosses are both equal to $\frac{1}{4}$ . This is also represented as the expected value of the indicator variable which identifies if HT or HH appears.

Yes, there are "mixing effects" that happen because we're not considering all possible series. In particular, for Alice, we're only looking at series which end with a HH, and do not have a HT. This means that just after the first time a H appears, the series would have to end (as explained by Brian in the solution).

Calvin Lin Staff - 4 years, 3 months ago

Can you add the context as to why this is an interesting question to consider?

Calvin Lin Staff - 4 years, 3 months ago

I'm not sure which context you mean. We can of course compute many values coming from the study of stochastic processes, but in the context of this question, none of them really apply.

For me, the most interesting part is that Bob would have a much higher chance of winning if he won upon seeing TH rather than HT.

Brian Moehring - 4 years, 3 months ago

Oh sorry, I was having a conversation with @Brandon Monsen about a similar problem , and was using this to illustrate the misconceptions. As the messages came in one after the other, I mixed up the two of you when I posted my comment.

Calvin Lin Staff - 4 years, 3 months ago

I'll add this to my solution from before to avoid any further confusion from anyone. Did you see my question attached to your solution, though?

I was basically wondering why stating that after an H, Bob will always have a equal to or greater probability than Alice of getting a point if they have not already, due to the fact that Bob is more restricted on what he can flip that doesn't land a point as you outlined in your solution and did the math for.

I thought that was a rough, qualitative approach to probability density, but I seem to be missing something.

Also, am I correct in assuming that the equal chance here is due to there being only one string of flips, which implies that the sequence of flips MUST terminate on the next flip after $H$ with equal chances?

Brandon Monsen - 4 years, 3 months ago

I will reply to the context about your problem on the problem itself.

With regards to this problem, it is because of the mixing effects of ending the game at HH or HT that impact how soon the game ends. This changes the space of events that we're looking at, and thus the probabilities.

For example, Bob no longer gets to win on HHT, because Alice already has won. So, even though in the individual game, Bob is likely to win earlier, in the combined game, Alice has scenarios which prevent a Bob win, and thus Bob ends up winning later (than in the individual game).

Calvin Lin Staff - 4 years, 3 months ago

It Is Harder To Get HT?

1 solution

0 pending reports