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Race Probablility

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by spanlength » Sat Jun 28, 2008 10:08 am
So x can get a position from 1 to 4 in race.
Hece the probablitity =x is 1st in race * y can get 2nd,3rd , 4th or 5th +x is 2nd in race y can get 3rd or 4th, 5th + 3rd in race * y can be or 4th or 5th + 4th in race * ycan be 5th in race = 1/5*4/5 +1/5* 3/5 +1/5* 2/5+1/5*1/5 =2/5.
Last edited by spanlength on Sat Jun 28, 2008 10:21 am, edited 1 time in total.
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by antyagi » Sat Jun 28, 2008 10:16 am
But then the probablity of Y also needs to be linked.
IF X is 1st Y can be 2,3,4,5
IF X is 2 Y can be 3,4,5

Hence it would be : 1/5 * 4/5 + 1/5*3/5 + 1/5 *2/5 + ....
Thanks,
Ankur Tyagi
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by durgesh79 » Sat Jun 28, 2008 11:01 am
2/5 is not the answer
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by gabriel » Sat Jun 28, 2008 11:27 am
It would be true half of the time. So probability is 1/2
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by Ian Stewart » Sat Jun 28, 2008 11:35 am
I assume the order of finish is random? Then it's 50% that x finishes ahead of y, and 50% that y finishes ahead of x.
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by durgesh79 » Sat Jun 28, 2008 12:36 pm
Right. The answer is 1/2

If X is on 1, then y can take any 4 of the 4 places left -> 1/5 * 4/4
If X is on 2, then y can take any 3 of the 4 places left -> 1/5 * 3/4
If X is on 3, then y can take any 2 of the 4 places left -> 1/5 * 2/4
If X is on 4, then y can take only 1 of the 4 places ledt -> 1/5 * 1/4

Probability = sum of all these = 1/5 * ( 1+3/4+1/2+1/4)
= 1/2

Buti knew that there could be a better way to do it.
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Why

by evansbd » Thu Jul 17, 2008 8:45 am
Ian

Intuitively, can you explain how you know that X always beats Y and vice versa are 50%? If its a two person race, then it would be obvious, but I dont see how that could jump out for a race such as in the question.

Is there a way to do this problem using comibinations?

i.e. There are 5! possibilities for a finishing order, only a percentage of these will result in X beating Y...

Also, if there were more participants, and we were still concerned with X always beating Y, would the answer still be 1/2?
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Why

by evansbd » Fri Jul 18, 2008 6:09 am
repost
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by pepeprepa » Fri Jul 18, 2008 7:39 am
I give you two explanations.

If you do not see it, take an example (with 3 runners X,Y and A)
You want to know who wins most of the time between X and Y
Write all the possible results:
XAY XYA AXY
AYX YAX YXA
You can clearly see that the probability "X is before Y" is 1/2.
You can choose the "n" number of runners you want, it will be the same.

Another explanation:
« A »= A is before B
« B »= B is before A
P(A)+P(B)=1 and so P(A)=1/2 because
*)A cannot have the same rank as B in the race so the sum of the 2 probabilities is 1
*)There is equiprobability of both, noone is doping or cheat more than the other one, so they have the same probability to occur each other. Given the result is 1 we have 2P(A)=1
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