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Hot Hand Reply
I see it now, having worked on this at length (3+ hours), did the same exact thing you did Eric 3x, and having read both papers. It's quite subtle.
To be formal, we're sampling length-4 sequences and computing the expectation of p(H|H) from our sample.
It's easier to visualize if we talk about P(seq) where seq is a particular sequence and P(H|H,seq) is the conditional probability of H following H given a particular sequence. We estimate P(H|H,seq) as C(H follow H)|C(H) in seq. Then P^(H|H) is our estimation of conditional probability of H following H in any sequence, and is equal to SUM over all sequences, P(H|H,SEQ) * P(SEQ). The sum over equiprobable sequences (SUM P(SEQ)) is where your intuition that the sequences need to equally count comes in, but it isn't the whole story.
The problem with your math, and what I also did, lies in cases with multiple H. We both naively summed up H and T from that to all the other cases, and got 0.5. Take for instance THht, where we both got one head and one tail. P(THht) = 1/16 (or 1/14 counting just the eligible sequences) just like every other sequence. The problem with summing down H and T and ignoring P(seq) is that the denominator C(H) here is 2 but in THtT the denominator C(H) is 1, so summing just the counts over the rows misses the fact that they have different denominators.
I'll derive your example according to the sum over products I outlined. I'm "shoving" the sequence probability (1/14) into each outcome,so instead of the average of conditional probabilities, the total probablility is just the sum of these products..comes out the same.
0: TTTT: - X -- conditioned out
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1: TTTH: - X -- conditioned out
1: TTHt: 1t, 0h = P(SEQ) * P(H|H,SEQ) = 1/14 * 0/1 = 0/14 = 0
1: THtT: 1t, 0h = P(SEQ) * P(H|H,SEQ) = 1/14 * 0/1 = 0/14 = 0
1: HtTT: 1t, 0h = P(SEQ) * P(H|H,SEQ) = 1/14 * 0/1 = 0/14 = 0
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2: TTHh: 0t, 1h = P(SEQ) * P(H|H,SEQ) = 1/14 * 1/1 = 1/14 = 0.0714
2: THtH: 1t, 0h = P(SEQ) * P(H|H,SEQ) = 1/14 * 1/1 = 0/14 =0
2: THht: 1t, 1h = P(SEQ) * P(H|H,SEQ) = 1/14 * 1/2 = 1/28 = 0.0357
2: HtTH: 1t, 0h = P(SEQ) * P(H|H,SEQ) = 1/14 * 0/1 = 0/14 = 0
2: HtHt: 2t, 0h = P(SEQ) * P(H|H,SEQ) = 1/14 * 0/1 = 0/14 = 0
2: HhtT: 1t, 1h = P(SEQ) * P(H|H,SEQ) = 1/14 * 1/2 = 1/28 = 0.0357
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3: THhh: 0t, 2h= P(SEQ) * P(H|H,SEQ) = 1/14 * 2/2 = 2/28 = 0.0714
3: HtHh: 1t, 1h= P(SEQ) * P(H|H,SEQ) = 1/14 * 1/2 = 1/28 = 0.0357
3: HhtH: 1t, 1h= P(SEQ) * P(H|H,SEQ) = 1/14 * 1/1 = 1/28 = 0.0357
3: Hhht: 1t, 2h= P(SEQ) * P(H|H,SEQ) = 1/14 * 2/3 = 2/42 = 0.0476
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4: Hhhh: 0t, 3h= P(SEQ) * P(H|H,SEQ) = 1/14 * 3/3 = 3/42 = 0.0714
0.0714 * 3 + 0.0357 * 4 + 0.0476 = 0.4046
Couple of things are going on here...you're conditioning out two sequences that are heavy with tails, which means that the remaining sequences are more likely to have multiple heads. Now, you'd think that would make P(H|H) go up, but each of those sequences is still weighted by the sequence probability, so you're just shoving the cases where H follows H into the same sequences, such that the individual counts of H all "share" the same 1/14 sequence probability (bc P(SEQ) * C(H follow H) / C(H)), while the cases where T follows H are more likely to occur "by themselves".
Not sure this translates into the Hot Hand effect (my own take--Steph Curry has some psychological/physical factor that violates independence to make him streakier when the game is on the line)< but their statistical argument bears out.
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