Nair pointed out in the comments yesterday that some triplets of heads and tails are more common than others. At first glance, that seems impossible – each toss is independent of every other toss, and any combination of heads and tails must be as likely as any other combination of heads and tails.

Here’s a post from Futility Closet, on the other hand, that tries to explain why HHT is a better choice than HTH. There seems to be in fact a general strategy for choosing these triplets: flip the middle coin, add what you get to the beginning and drop the last one. (This turns HTH into HHT.)

I wanted to see if I could actually verify this on a computer. I tried two approaches: First, I ‘tossed a coin’ a million times, noted down each toss, and searched for HTH and HHT in the resulting array. Second, I tossed a coin repeatedly, noted down each toss, and stopped as soon as either the combination HTH or the combination HHT appeared.

In the first case, not surprisingly, the number of combinations, triplets, of either type are of very similar frequency – each one-eighth of the total number of coin-tosses. The surprise comes with the second experiment: The combination HHT is more frequent than the combination HTH – the average number of tosses of the coin before an HHT appears is 8. The number for HTH is 10.

My explanation for this is also how I wrote my code for the experiments. When you want an HHT to turn up, if your third toss is an H instead of a T, you are still two-thirds of the way to your target. If, on the contrary, you wanted HTH to turn up, and your third coin-toss turned up a T instead of an H, you’d have to start all the way from the beginning.

Can somebody here think of a better explanation? I’d like to hear it. If somebody wants to see the code I wrote for these experiments, ping me and I will send it to you. Try these experiments on your own and see if you can confirm that my answers are right. If you do, please do tell me.

[End. Fini. Kaputski. Coin-Toss!]

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In the link for a post you provided… lets say I choose HHT. As mentioned in the link, if u choose THH, the chances of u winning is more than mine ??

croor send me your matlab codes..

@Mickey-I, yes; that’s the idea. It always works, apparently. It is, however, important to remember what the problem at hand is:

If you start tossing your coin repeatedly and stop when you get one of the two hits, how often will you end up with one over the other?

I’ve tried to answer this by counting how many tosses on average each of the outcomes requires.

@Mickey-II, I will email the code to you.

If the idea always works, then we would have the following

HTH < HHT < THH < TTH < HTT <

HHT.Is there a contradiction? Or am I missing something?

As far as I can make out, HTH < HHT, THH < TTH, and HTT < HHT are all true.

Did you check the link from Futility Closet I've posted? The page has a link at the bottom called Non-transitive dice.

I haven’t checked this for coin-tosses, though. Could something like that work here too, d’you think?

As for the HHT vs HTH problem. Let P(n) and Q(n) be the probabilities of obtaining a HHT and a HTH respectively in an infinite sequence, for the first time, at the nth place (n, n+1 n+2 places, so to speak). Note that P(n) = 1/8 – P(1)∩P(n) – P(2)∩P(n) – … – P(n-3)∩P(n). On the other hand, Q(n) = 1/8 – Q(1)∩Q(n) – Q(2)∩Q(n) – … – Q(n-3)∩Q(n) – Q(n-2)∩Q(n). The extra term

Q(n-2)∩Q(n)makes the probability Q(n) smaller.Now, the concept on non-transitive inequalities is very interesting! I’ll read into that.

That’s what I just said! For anything but (n-2), the intersection is null. For (n-2), one of them has an intersection; the other doesn’t. Nicely done!

That said wouldn’t the occurance of HTH in a million tosses be more common that that of HHT. (because of sequences like HTHTH where HTH occurs twice in a length-5 sequence). Please send me your matlab codes too.

That’s what I thought too. Doesn’t seem that way from the search program, though. I just sent it to you. Would you check?

Turns out the explanations for HTH < HHT for the first occurrence are not complete. For instance, we cannot compare the scenarios, TTH and THH in the same fashion. Check out this video for a simple and complete explanation.

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