Probability to win a coin toss game (take 2)

Factual Questions
1

On TV here in Australia they had a show where the presenter told an audience member:

“Let’s play a coin tossing game. Pick 3 consecutive outcomes (eg, HHT, THT, TTT) that you think will occur first.”

(So the person picks THT).

The guy then says:

“Now I’ll pick mine. I’ll take the ‘T’ off the end of your combination to give me TH. I’ll then add either a head or a tail to the front, as long as it doesn’t read the same way forwards as it does backwards. What I’m left with, is TTH”.

The presenter now claims that if they were to toss a coin repeatedly until one of the combinations came up, the presenter would win 75% of the time, REGARDLESS of what the guy from the audience chose as his initial combination. Is this so, and if it is, why is this so?

2

I’m not sure what the set up is here… you’re saying, if the contestant picks ABC, the presenter drops the last result (so has AB) and then puts on front either the H or T that gives a non-symmetric sequence?

I have killed the other thread just to avoid confusion.

3

Before I start my race with Dex to calculate the answer: does the contestent have to pick from the three you mentioned (HHT, THT or TTT), or can it be any sequence of 3 (eg HHH)?

pan

4

Don’t worry - I just figured it out. It’s quite simple. I’ll post it below…

5

Kabbes…

The audience member can pick any 3 combinations he wants.

6

OK. Say the contestant picks ABC (thanks Dex ;)) The presenter then picks (WLOG*) TAB.

Then in order for you to get ABC it has to pass through AB first. But then as long as the presenter has the correct toss before the AB he will win.

To use your example: the presenter has TTH, you have THT.

For you to win, the sequence “TH” must come up. But then if T came before this, the presenter would win and not you. So the presenter has his chance to win before you do. If the presenter doesn’t win then you still only have a 50/50 chance the complete the THT sequence, otherwise the whole thing starts again.

Your chance of a winning sequence is therefore (0.5)^4 (must be HTHT) = 1/16. The presenters is (0.5)^3 (TTH) = 1/8. Therefore the presenters chance of winning (at any time - remember that this game goes on until one of you wins) is twice as much as yours. In order to work out who actually wins you have to do a wee infinite sum. I’ll spare you that, suffice to say that it comes out with him having 3 times the chance of winning as you. I.e. he win 75% of the time.

The whole this falls apart if he picks a palindromic sequence however.

Erk. Clear as mud. Shall I try again?

pan

*Without loss of generality. It means that it could be a head and the same argument would apply.

7

I don’t have time to work it out, but I think it’s pretty much that the presenter has taken the FIRST two outcomes that the contestant wants, and made them the SECOND two outcomes of his winning combination.

Thus, comparing THT (contestant) with TTH (presenter).

Note the subtlety here: the coin is not being tossed in groups of three, but in a continuing sequence. If the coin were tossed in groups of three (“OK, first result is HHT, no one wins; second result is THH, no one wins…”) then the two would have equal chance of coming up.

Instead, as I read the problem, the coin is being tossed and results recorded: HHTTHHTHTH etc.

OK, so the point is that the contestant’s first two outcomes are the presenter’s second two outcomes. So when you get to those outcomes, the presenter may already have won before the contestant gets to the third outcome, if that makes sense.

Let’s use THT (contestant) and TTH (presenter).

So, we may have a few heads before we get to a T.

THen, patterns of next three flips (with the first flip a T) could be:

TTTT
TTTH - presenter wins on fourth flip
TTHT - presenter wins on third flip*
TTHH - presenter wins on third flip
THTT - Contestant wins on third flip
THTH - Contestant wins on third flip
THHH

    • this is the critical one. Note that the contestant would win on the fourth flip, but you never get to the fourth flip, because the presenter has already won (because his LAST TWO flips were same as the contestants FIRST TWO)

The presenter is winning 3 times to the contestant winning twice.

If I expand the results that have no winner after four tosses and add a fifth and sixth toss:
TTTT-TT
TTTT-TH - presenter wins on 6th toss
TTTT-HT - presenter wins on 5th toss*
TTTT-HH - presenter wins on 5th toss

THHH-TT
THHH-TH
THHH-HT
THHH-HH

Again, you see that the case marked *, the contestant would have won on the 6th toss, but it never gets there.

So, in 6 tosses, we have presenter winning 6 times, contestant winning only 2 times.

I haven’t worked out the numbers, but the pattern is pretty clear, how the presenter wins more often, and why.

8

I understand, but do you have to take in to consideration if the contestant gets his combo in the first 3 tosses? THT? Coz then he wouldn’t need to get a HTHT.

9

Yes but 'Hatter the probability of that is only 1/8 (and the presenters chance is the same).

It does slightly move the odds back in the favour of the contestant - the chances for the presenter are slightly less than 75% - but there isn’t that much in it by the time you work out the infinite sum.

And Dex - hah! Beat you to it! Score 1 for the British actuaries against their US counterparts :wink:

pan

10

You only beat me to the punch, kabbes, because of the time zone diff: you got a six hour head start on me.

Yes, the chance of winning in the first three tosses are equal (that’s why it makes a difference whether the game is played three tosses at a time, or in a sequence.)

If you’re having trouble visualizing, imagine that the contestant picks a string of T’s: TTT as his winning sequence. This has the same chance of showing up as any other combination, in the first three tosses.

The presenter will take the combination HTT (dropping the last T of the contestant and add an H in front).

Now, you can see that, once the sequence starts rolling (beyond the first three tosses), the contestant doesn’t win until a string of T’s starts. But when that string does start, he won’t win until the third T… by which time the presenter will already have won on the second T in the string. In this case, if the contestant doesn’t win in the first three tosses, he’ll never win.

I therefore don’t think I quite believe the accuracy of the 75% number, I think it’s more a rough approx than a precise statistical calc, since the exact odds seem to vary somewhat with the sequence selected by the contestant… although still loaded way in the presenter’s favour.

11

It’s not exactly 75%. I did an empirical test, and it came out 74.23% to 25.77%, with over half a million trials.

12

With little to do until the Florida vote is finally(?) tallied, I decided to calc the probabilities of winning for each of the contestants choices.

I’ll list half of the contestant’s choices; the remainder can be reached by interchainging H’s and T’s.

TTT 1/8 = .125
TTH 1/4 = .25
THT 21/64= .328125
THH 1/3 = .333…

So if contestant gets to pick his combination, he does best to go with THH (or HTT). If his choice is determined randomly his total winning chance is 63/256 = .24609375 leaving the presenter with chances of .75390625