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Flip a coin until either HHT or HTT appears. Is one more likely to appear first? If so, which one and with what probability?

24

HHT is more likely to appear first than HTT. The probability of HHT appearing first is 2/3 and thus the probability of HTT appearing first is 1/3. Indeed, both sequences need H first. Once H appeared, probability of HHT is 1/2 (b/c all you need is one H), and probability of HTT is 1/4 (b/c you need TT). Thus HHT is twice is likely to appear first. So, if the probability that HTT appears first is x, then the probability that HHT appears first is 2x. Since these are disjoint and together exhaust the whole probability space, x+2x=1. Therefore x=1/3.

wbt on

11

Let A be the event that HTT comes before HHT. P{A} = P{A|H}P{H} + P{A|T}P{T} = .5P{A|H} + .5P{A|T} P{A|T} = P{A} therefore, P{A|H} = P{A|T} P{A|H} = P{A|HH}P{H} + P{A|HT}P{T} = (0)(.5) + P{A|HT}(.5) Therefore, 2P{A|H} = P{A|HT} P{A|HT} = P{A|HTT}P{T} + P{A|HTH}P{H} = (1)(.5) + P{A|H}(.5) 2P{A|H} = .5 + P{A|H}(.5) P{A|H} = 1/3 and P{A|H} = P{A}, therefore, P{A} = 1/3 So, HHT is more likely to appear first and it appears first 2/3 of the time.

Wharton Student on

4

Above link is the best solution I have seen for this problem http://dicedcoins.wordpress.com/2012/07/19/flip-hhh-before-htt/

Anon on

3

Here's my answer. Let x = probability of winning after no heads (or a tail). y=probability after just one heads. z=probability after two heads. w=probability after HT. Thus x=(1/2)x+(1/2)y, y=(1/2)z+(1/2)w, z=1/2 + (1/2)z, w=(1/2)y. Therefore, z=1, y=2/3, w=1/3, x=2/3. We wanted x at the beginning, so it is 2/3 that HHT comes up first.

Anonymous on

2

You can also just use Crap's Principle. In the sequence of the 3 flips, only the 2nd and 3rd flips matter. We know the first flip is a H and we don't care about sequences starting with T. So, assuming we start with the following structure: H _ _ If we get a HH_, the game is done and HHT will automatically win (no matter how many more heads we see, the next T is an auto win) If we get a HTT, the game is done If we get a HTH, the game is a draw and we essentially restart the game So: P(HHT) = 1/2 P(HTT) = 1/4 P(Draw) = 1/4 P(HHT wins) = P(HHT) / [1-P(Draw)] = (1/2) / (1-1/4) = 2/3

Anonymous on

3

P(A|HH) = 0 because after a sequence of consecutive heads, you can no longer achieve HTT. The moment you get a tail, you will have the sequence HHT. This the reason HHT is more likely to occur first than HTT.

1

Think of HHT as H(HT) and HTT as (HT)T For every occurrence of (HT) in a sequence of flips, there is a 1/2 chance an H occurred before the (HT) and a 1/2*1/2 chance a T occurred after. Thus, HHT is 2x more likely than HTT. HHT is first 2/3 of the time.

Anonymous on

1

I don't get it. Shouldn't P{A|HH} = P{A} in the same sense that P{A|HTH} = P{A} from both HH and HTH we have get the first H from HTT and so it should be P{A|HH} = P{A|HTH} = P{A} Am i wrong?

George on

1

sorry, i meant: I don't get it. Shouldn't P{A|HH} = P{A|H} in the same sense that P{A|HTH} = P{A|H} from both HH and HTH we have get the first H from HTT and so it should be P{A|HH} = P{A|HTH} = P{A|H} Am i wrong?

Anonymous on

1

P(A|HH) = 0 because after a sequence of consecutive heads, you can no longer achieve HTT. The moment you get a tail, you will have the sequence HHT. This the reason HHT is more likely to occur first than HTT.

0

P{A|H} = P{A|HH}P{H} + P{A|HT}P{T} = (0)(.5) + P{A|HT}(.5) Need help - - why is P{A|HH} = 0 ?

Jilie on

0

Ending up with HHT more likely (with probabilty 2/3).

Nova on

0

HHT is more likely (2/3) probability. People with wrong answers: Did you not Monte Carlo this? It takes 5 minutes to write a program, and you can then easily see that 2/3 is correct empirically.

Jesus F Christ on

0

Apologies, Below*

Anon on

3

You guys seem to be mixing order being relevant and order being irrelevant. If order is relevant (meaning HHT is not the same as HTH) then this has a 1/8 of occuring in the first 3 tosses. Also HTT has a 1/8 chance of occurring in the first 3 tosses, making them equally likely. Now, if order is not relevant. (so HHT and THH are the same), then this has a (3 choose 2) * (1/8) probability of happening in the first 3 tosses. The same goes for HTT (which would be the same as THT etc and others) so this has a (3 choose 2) * 1/8 probability of happening in the first 3 tosses as well. Either way they come out to being equally likely, please comment on my mistake if I am doing something wrong.

Anonymous on