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.

- 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).
- 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).
- 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%.
- 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!