I came upon an interesting table in the book "The Hot Hand: The Mystery and Science of Streaks" by Ben Cohen:
In this calculation, we are considering 3-flip sequences and are measuring the percentage of the time a head occurs directly after another head, and show that it is 42% and not 50% like we may expect. Mathematically, I initially thought the calculation was the following:
$$
\begin{aligned}
P(\textrm{next flip H} \mid \textrm{3-flip sequence is not TTT or TTH, previous flip is H}) = &\frac{1}{6}\frac{2}{2} \textrm{(for HHH)} + \frac{1}{6}\frac{1}{2} \textrm{(for HHT)} + \\&\frac{1}{6}\frac{0}{1} \textrm{(for HTH)} + \frac{1}{6}\frac{1}{1} \textrm{(for THH)} +\\&\frac{1}{6}\frac{0}{1} \textrm{(for HTT)} + \frac{1}{6}\frac{0}{1} \textrm{(for THT)}\\=&\frac{5}{12} = 41.667\%
\end{aligned}
$$
However, after thinking about this further, I am clearly not keeping track of the events properly. If we define $A = \textrm{next flip H}$, $B = \textrm{3-flip sequence is not TTT or TTH}$, $C = \textrm{previous flip is H}$, and we know that
$$P(\textrm{next flip H} \mid \textrm{3-flip sequence is not TTT or TTH, previous flip is H}) = P(A \mid B, C)$$
We get:
$$
\begin{aligned}
P(A, C \mid B) = &\frac{1}{6}\frac{2}{2} \textrm{(for HHH)} + \frac{1}{6}\frac{1}{2} \textrm{(for HHT)} + \frac{1}{6}\frac{0}{2} \textrm{(for HTH)} + \frac{1}{6}\frac{1}{2} \textrm{(for THH)} +\\&\frac{1}{6}\frac{0}{2} \textrm{(for HTT)} + \frac{1}{6}\frac{0}{2} \textrm{(for THT)} = \frac{1}{3}\\
P(C \mid B) = &\frac{1}{6}\frac{2}{2} \textrm{(for HHH)} + \frac{1}{6}\frac{2}{2} \textrm{(for HHT)} + \frac{1}{6}\frac{1}{2} \textrm{(for HTH)} + \frac{1}{6}\frac{1}{2} \textrm{(for THH)} +\\&\frac{1}{6}\frac{1}{2} \textrm{(for HTT)} + \frac{1}{6}\frac{1}{2} \textrm{(for THT)} = \frac{2}{3}\\
P(A \mid B, C) = \frac{P(A, C \mid B)}{P(C \mid B)} = &\frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \neq \frac{5}{12}
\end{aligned}
$$
My question is what does the $41.667\%$ represent in conditional probabilities, and what does it actually communicate to us about the hot hand? Intuitively, I understand why the probability expressed is not $50\%$ (there is already a great post explaining this much), but I have not been able to articulate the probability it is representing.
EDIT: There seems to be confusion on what I am hoping to learn here. I am not trying to find $P(\textrm{next flip H} \mid \textrm{3-flip sequence is not TTT or TTH, previous flip is H})$ or $P(A, C \mid B)$ in probability. I am trying to understand what the $41.667\%$ is actually representing. The author's argument hinges on this calculation (which I expressed on the right hand side of the first block of calculations) not being equal to $50\%$, and I am confused what the $41.667\%$ actually represents.
Best Answer
$\require{cancel}$ If you're analyzing some data trying to empirically determine whether "streaks" really exist or not, 5/12 is the probability of getting a post-streak success in the "boring world" where successes behave like independent coin tosses. If you weren't being careful, you might assume the boring-world probability of a post-streak success is exactly 50%. It's actually less, because of the way after-the-fact filtering for streaks biases the choice of what comes after the streak.
Establishing the correct baseline number 5/12 is important when you return to examining your data. After all, if you compute the same statistics on your data and you get a number larger than 5/12 --- say 50% --- you should correctly interpret this as fairly significant evidence against the boring world hypothesis and evidence in favor of the hot-hand hypothesis.
If you would like to represent 5/12 as a conditional probability, let:
The observation is that, in the boring world, $$\Pr(\widehat H_1 \widehat H_2 | A,\,BH_1,\,\widehat{B}) = \frac{5}{12}$$
This is because of the filtering effect of having $BH_1$ as given— filtering for triples with potential 2-streaks in them.
We can calculate this conditional probability using Bayes' law (Note: Instead of writing $BH_1$ and $\widehat{B}\widehat{H}_1$, I'll leave the $B$ implied and just write $h_1$ and $\widehat{h}_1$ so as to make it easier to read. You could also suppress $A$, which appears as given in all terms.):
$$\Pr(h_1 | A) \cdot \Pr(\widehat{h}_2\widehat{h}_1 | h_1\, A) = \Pr(\widehat{h}_2\widehat{h}_1 | A) \cdot \Pr(h_1\,| \widehat{h}_2\widehat{h}_1\, A) $$
We can straightforwardly compute three of these conditional probabilities, solving for the one we want $\Pr(\widehat{h}_2\widehat{h}_1 | h_1\, A)$.
Now,
So
$$\frac{1}{2}\cdot \Pr(\widehat{h}_2\widehat{h}_1 |\, h_1 A) = \frac{1}{4}\cdot \frac{5}{6}$$
$$\Pr(\widehat{h}_2\widehat{h}_1 |\, h_1 A) = \frac{5}{12}$$
which we wanted to establish.