Thread: Probability question... View Single Post
#2
04-30-2005, 08:16 PM
 gaming_mouse Senior Member Join Date: Oct 2004 Location: my hero is sfer Posts: 2,480
Re: Probability question...

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40 yard dash statistics below.

Alf has a Avg of 4.42 with a stddev of .04
Bob has a Avg of 4.43 with a stddev of .04
Charlie a Avg of 4.45 with a stddev of .04
Dave's Avg is 4.47 with a stddev of .04

Is the probability that Alf beats Bob = Normsdist(.01/.04), or 59.87%? (.7733 for Alf vs Charlie, .89435 for Alf vs Dave)

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the random variable (Alf - Bob) ~ N(.01,(2*.04^2))

ie, normal with mean .01 and SD = sqrt(2*.04^2) = .057

Chance he wins is thus 0.5696

[ QUOTE ]
If that is so, what is the probabilty that Alf wins a race with all 4 of them running? Do i have to monte carlo this? I am thinking i cant take .5987*.7733*.89435 to get to an answer.

[/ QUOTE ]

That method won't work, as the chance that Alf beats Charlie GIVEN that he has already beaten Bob is not the same as the chance that he beats Charlie. You need to find the joint distribution of all 4 variables, which I believe is multi-variate normal. Then you need to do some triple integration, I believe.

Of course, Monte Carlo would work too.