Coin toss problem. Which is more likely: THT or TTH?

Question

A coin is tossed until the combination THT or TTH occurs. On average which of these combinations is more likely to occur first?

Answer 1

The solution is correct, please remember that T and H are independent therefore you can multiply probabilities.

Answer 2

The order doesn't matter at all.

Any defined three toss sequence is this likely:

(1/2) x (1/2) x (1/2) = 1/8 = .125

The probability that the first sequence or second sequence occurs is just that like is .125

Either sequence is equally likely, .5

Answer 3

If you flip a coin n times, then each sequence s_1, s_2....s_n of T's and H's has exctly the same probability: (1/2)^n. You have independent experiments, each one with probabilty 1/2.

Hence, any sequence s_p, s_(p+1), ...s_q, appearing in any position when you flip a coin n ≥ q times, is as likely as any other of such sequences.  

But, in my humble opinion,  your question is not precise. The adverb "until" doesn't apply to your experiment, because the sample space is infinite, not finite. That is, the random variable that describes the number of tosses until you get THT (or TTH) can take any number in {3, 4, 5....,n........}.

Answer 4

If THT appears first, it *must* be preceded by an H, since TTH occurs firsts in TTHT.  So in fact we need the string HTHT, whereas TTH can be preceded by either an H or T, so the ratio is indeed 2:1.

Steve

Answer 5

They're equally likely. They both have a 12.5% probability of happening:

0.5*0.5*0.5 = 0.125

Answer 6

The correct answer has been given, and with good explanation...I just want to point out another follow-up for those interested:

http://answers.yahoo.com/question/index;...

Answer 7

I'll be a contrarian.

TTH is 2:1 more likely.

When you first hit a "T", you start.

There is a 25% chance of getting an HT combo, BINGO - THT

There is a 25% chance of getting an HH combo,

There is a 25% chance of getting a TH combo, BINGO -TTH

Lastly, there is a 25% chance of a TT combo.

But if you have a TT combo, you are guaranteed that the next "H" will give you a TTH combo. BINGO -TTH

Therefore the odds are 25% v 50%

Interesting.

************ Addendum********

If you take a random sampling of any three tosses, the odds are identical, i.e., 1/8th each.

But if you Take a random "T" and then ask which comes next in the series, a THT or a TTH, the TTH will predominate 2 to 1 for the reasons I outlined above.  (I ran a small spreadsheet which confirmed this reasoning).

This shows how easy it is to s***w up on your sampling if you are not careful.

Answer 8

Give Remo the pixie doll!  It's called conditional probablility.

Well done.

Answer 9

I think the point here is that either of these combinations can start after the first flip. But THT (which the Y!A automatic nagging system thinks I misspelled) has to be preceded by an H or it doesn't count, because if it is preceded by a T, there's a TTH (and we stop) before we complete it.

Next day: I see I miswrote a couple of codes in the next two paragraphs, which probably accounts for the thumbs-downs. I am therefore correcting those.

There's a two-flip preamble (TH) that would prevent a TTH from being completed, but since that requires two specific results in a row, it's less likely to appear.

The upshot of this is that TTH is more likely to occur first.

Follow-up: A small excerpt from the output of a "Monte Carlo" simulation (edited)

Simulation 4:

  THT: 333677

  TTH: 666323

Results by run length:

3:  125113 THT  124563 TTH

4:  62372 THT  125343 TTH

5:  31253 THT  93454 TTH

6:  31335 THT  77998 TTH

7:  23689 THT  62502 TTH

8:  15566 THT  46858 TTH

9:  11687 THT  35268 TTH

10:  8902 THT  26461 TTH

  :

40:  7 TTH

41:  1 TTH

43:  1 THT  1 TTH

44:  2 TTH

47:  1 TTH

48:  1 TTH

50:  1 THT

The 2:1 overall ratio for TTH over THT seems to be about right. Notice that the chances are even for runs that end in 3 flips, but if they end in 4 flips, TTH has the 2:1 advantage. There must be some unconsidered factor here, because the arguments presented suggest that the chances should be equal for 3-flip runs (one quarter of the total) and 2:1 favoring TTH otherwise.

But that would produce an overall probability of

1/4 * 1/2 + 3/4 * 2/3 = 5/8 in favor of TTH, whereas my Monte Carlo runs (over a score of them involving a million trials each) show a consistent 2/3 probability in favor of TTH.

I suppose my next programming amusement should be calculating the exact probabilities, down to where the likelihood of extending the run is vanishingly small. I haven't approached 60 flips in any of those millions of trials, so that calculation should be manageable.

Answer 10

Dr. D. Just to clarify, the reasoning and answer you have provided to my question is incorrect. One sequence will appear before the other on average. They are NOT the same.

Also how is this any different than the question I asked (other than the obvious)

Edit: Ok you are correct, it does make a slight difference. Still, the concept is the same, and the answer is the same. Just because you switched TTH from THH, this does not affect the average (why? reason this out)

The answer is still the same as the problem I gave. On average it takes 10 flips to get THT and 8 flips to get TTH.

Answer 11

To figure out the probability of THT it's (1/2)^3

To figure out the probability of TTH it's also (1/2)^3

So the probability of each occuring is 1/8=12.5%

Answer 12

Definitely TTH...

It's clear that in each sequence we can drop any initial H's. So we can assume, without loss of generality, that the first toss was T. Now we have 50% for TT and 50% for TH, branching further to 25% for TTT, 25% for TTH (stop), 25% for THT (stop) and 25% for THH and so on.

If the second toss was T, we have 50% of ending with a TTH. In the other 50%, we have TTT which is facing the same consequence and so on. In this branch, it is clear we will end with a TTH and it can happen at any time.

If the second toss was H, we also have 50% of ending, now with THT. The other 50%, however, make up a THH which, after a brief look, takes us to the situation at the very beginning because now all the history is useless for anything. Notice the huge difference of repeating the good conditions for the rising of a THT, which makes it much less probable to be the first of the two words.

I should do that more mathematically, but I think the idea is clear.