Re:Say It Is So: Baseball’s Disgrace
by LeifCarrotson on Tuesday February 25, 2020 @05:47PM
(#22418357)
It's just 2^24 for probability 1/2, right?
Re:Say It Is So: Baseball’s Disgrace
by aplummer on Wednesday February 26, 2020 @01:57AM
(#22420961)
It’s a different calculation Matt is doing. You are calculating the probability of that run happening now. Matt is calculating the expected length of the longest run in 1 million tosses.
Re:Say It Is So: Baseball’s Disgrace
by mattkrause on Wednesday February 26, 2020 @11:06AM
(#22424206)
Exactly! (Thanks!)
The distinction is important because a sequence of ten heads seems 'rare' in isolation. However, it is not particularly unusual when you go looking for it as a subsequence of some bigger set of trials.
I had to dig into this for work, and doing statistics on runs is surprisingly interesting. Suppose you've got a sequence of $n$ events, each of which 'succeeds' with probability $p$. The expected length of the longest run is approximately $\log_{1/p}(n*(1-p) + 0.577 \ln(1/p) - 1/2$. For a fair coin with $p$ = 0.5, this reduces to log_2(n) - 2/3, which is about 19 for one million events. Amazingly, the variance only weakly depends on n, but is about 2 for p=0.5.
Thus, you're probably not going to see a 24 head run in 1M events. I'm excited I got to use this information, as the project I learned it for was a total bust.
More here: Shilling (1990, College Math. J.) https://www.csun.edu/~hcmth031/tlroh.pdf