Probability question...

[Tater10]

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)

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.

[gaming_mouse]

[ QUOTE ]
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.

[jason1990]

You could get away with a single integral here, if I'm interpreting the assumptions right:

P(A<B,A<C,A<D)=E[P(A<B,A<C,A<D | A)]=E[f(A)]

f(x)=P(x<B)P(x<C)P(x<D)

[gaming_mouse]

[ QUOTE ]
You could get away with a single integral here, if I'm interpreting the assumptions right:

P(A<B,A<C,A<D)=E[P(A<B,A<C,A<D | A)]=E[f(A)]

f(x)=P(x<B)P(x<C)P(x<D)

[/ QUOTE ]

Jason, I don't follow this. Are you saying you can just multiply the individual probabilities together? I could be wrong, but for some reason that didn't seem correct to me.

Also, would you mind taking a look at that other thread I started, down below?

Thanks,
gm

[jason1990]

Conditioned on A, those events are independent. In general, if

f(x) = P(A<B,A<C,A<D | A=x),

then

P(A<B,A<C,A<D) = E[f(A)].

In this case, since everything is independent, we have

f(x) = P(x<B,x<C,x<D) = P(x<B)P(x<C)P(x<D).

[Siegmund]

Looks to me like Jason is right.

He's *NOT* saying P(A beats B and C)=P(A beats B)*P(A beats C).

He's saying P(min{B,C,D}>x) = P(B>x)P(C>x)P(D>x), so that if we want P(A wins a 4-way race), we integrate f(x)P(B>x)P(C>x)P(D>x) over all x.

This still isn't a pretty integral to do - since for each x, we need need the CDF of a normal distribution 3 times. Just a matter of notation to write it in the above form instead of a quadruple integral, Integrate[f[w,x,y,z], {z,w,Infty}, {y,w,Infty}, {x,w,Infty}, {w,-Infty,Infry}] in Mathematica-ese.

[gaming_mouse]

[ QUOTE ]
Looks to me like Jason is right.

He's *NOT* saying P(A beats B and C)=P(A beats B)*P(A beats C).


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Thank you for that clarification. That was all that was confusing me.

And yes, I was pretty sure already that jason was right and that I just wasn't following his argument /images/graemlins/smile.gif.

[alThor]

[ QUOTE ]
what is the probabilty that Alf wins a race with all 4 of them running?

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I get around 43%. I did a discretized version of what the others referred to as the single integral method, in Excel. I don't think I made any mistake, but you never know.

alThor