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Wednesday, December 14, 2011
Math fun: series odds
You have a 66.7% chance of winning any game. You win the series if you win 3 games before your opponent wins 4 games. What are your chances of winning the series?
Try it yourself. Presuming I did this correctly:
So, your chances of winning 3 and losing 0 is .667^3.
Your chances of LWWW, WLWW, WWLW is 3 times .667^3 * .333^1.
Your chances of winning 3 and losing 2 is “4 choose 2” times .667^3 * .333^2.
“4 choose 2” means 4 factorial divided by (4-2) factorial divided by 2 factorial, or 6.
Your chances of winning 3 and losing 3 is “5 choose 3” times .667^3 * .333^3. “5 choose 3” is 10.
Presuming I did all that correctly, then I get 90.0%.
You can see the pattern as follows:
p = success rate
q = failure rate
w = number of wins needed
x = number of losses allowed
p^w
times
q^x
times
(w+x-1)! / x! / (w-1)!
(where x = 0, 1, 2, 3)
***
The key point is the “minus 1”, because the series ends always on the 3rd win. So, the various combinations is always limited to 2 wins and whatever possible losses (0 through 3).
***
How does this apply to baseball? Dave was asking how often would a batter strike out, if the pitcher has the talent level to throw a strike 66.7% of the time, and the batter takes every pitch.
So, the answer is 90% strikeouts, and 10% walks.