Expected length of longest winning streak in a 162-game baseball season

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SUMMARY

The expected length of the longest winning streak for a .500 baseball team during a 162-game season can be approximated using probabilistic models. A discussion highlighted that the probability of achieving an 11-game winning streak is approximately 7%. The analysis involved flipping a coin N times, where the probability of heads (winning) is 0.5, and considered the dependencies between games. The conclusion indicates that while the independence of games is not absolute, the probability calculations remain valid for estimating winning streaks.

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ACG
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Hi!

The Red Sox just finished an 11-game winning streak, which made me wonder. Do most teams experience winning streaks this long over the course of a season? What is the expected value of the length of the longest winning streak for a .500 team over the course of a season?

In general: if you flip a coin N times with a probability of getting a head of P, what is the expected length of the longest series of consecutive heads embedded in the results of the N flips?

ACG
 
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To a good first approximation, there are (n - k + 1) independent chances with probability 1/2^k each to get a winning streak of length k in n games. (Of course they're not independent at all, but it's close enough.)

So the chance of getting an 11-game winning streak in 162 games is about 7%.
 
Correct. However, as you know they're far from independent. If game 11 is a loss (50% chance) then there's no way for the set of games ranging from 2 to 11 to be the start of an 11-game streak. That doesn't sound like it will work.

ACG
 
ACG said:
Correct. However, as you know they're far from independent. If game 11 is a loss (50% chance) then there's no way for the set of games ranging from 2 to 11 to be the start of an 11-game streak. That doesn't sound like it will work.

So in that case you overstate the probability of #2 to #11 by 1/2^11 (1/2^11 rather than 0). But if you know that the 11th game is a win, then you're understating the probability of #2 to #11 by 1/2^11 (1/2^11 rather than 1/2^10). So I don't think it's actually a problem!
 

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