Good! Then at least you see that the odds of winning the game are asymmetrical — against most people’s intuition.

One of the problems behind the hot hand fallacy fallacy is the strong intuition that everything to do with coin flipping has always even odds. You are entirely correct that ‘the probability of a head following a head’ is 50/50.

The question here (which is relevant in the context of the hot hand hypothesis) is different. We want to know, in a given finite sequence, how likely it is that we flip heads, provided the previous flip was heads too.

How does that work out for 3-flip sequences? Say I flip a coin three times, and I tell you that (at least) one of the first two flips is heads. This eliminates sequences 1 and 2 from the table. My question to you is, what is the chance that, given this heads outcome, the next flip is also heads?

You don’t know which of sequences 3–8 I produced, so each of the six is equally likely. Now let’s go through each of the six, and work out the answer to the question.

  • In sequence 3, it is 0 (the next flip to the only heads produced tails).
  • In sequence 4, it is 1 (the next flip is heads)
  • In sequence 5 and 6 it is 0 (in both cases the next flip is tails)
  • In sequence 7 it is 0.5 (the first of the heads is followed by heads, the second by tails)
  • In sequence 8 it is 1 (both heads are followed by heads)

So on average, the chance of heads-followed-by-heads in this sequence of three flips is 2.5/6.

How can this surprising (because it is not 50/50) ratio be explained? Miller and Sanjurjo give an intuitive sketch in their paper, which goes something like this.

  1. The number of recorded flips (H or T) is not the same for each sequence. The more H in the first two flips, the more recorded flips (zero flips in sequences 1–2, one in sequences 3–6, and two in sequences 7–8).
  2. We record more Hs in the sequences that contain more Hs, but there are fewer such sequences (7 and 8, compared to 3–6 with just one H).
  3. So while every flip in sequences 3–6 counts for 100% of the total recorded flips, those in sequences 7–8 only count for 50%.
  4. So while it is true that 4 out of 8 recorded flips are H, 3 of those flips only have a ‘weight’ of 50%; for T there is only one such lightweight flip (in sequence 7).

This is why the game I proposed is not a fair one.

Hope this helps!

Written by

Accidental behavioural economist in search of wisdom. Uses insights from (behavioural) economics in organization development. On Twitter as @koenfucius

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