Probability puzzle

katxt

Well-Known Member
#1
A beats B in a two way race 2/3 of the time.
A beats C in a two way race 3/4 of the time.
How often will A win a three way race between A, B and C?
 

katxt

Well-Known Member
#2
No takers?
It looks like it should be 2/3*3/4 = 1/2 but there's something funny here.
If A, B and C are evenly matched, A beats B in a two way race 1/2 of the time and A beats C in a two way race 1/2 of the time.
But the probability of A beating B and C in a three way race is not 1/2*1/2 = 1/4. It is clearly 1/3 because they are all equally likely to win.
So how about the problem in the first post?
 

hlsmith

Less is more. Stay pure. Stay poor.
#3
I kind of want to change it to a dice rolling problem since my mind won't let me imagine race conditions can be exactly replicated. However - this problem doesn't discuss ties. Hmmm
 

katxt

Well-Known Member
#5
I kind of want to change it to a dice rolling problem
You can imagine a 22 faced dice. (I can't bring myself to say "die".) 12 faces labelled A, 6 labelled B, and 4 labelled C. The dice is rolled until one of the players gets their letter. In two way matches, A beats B 2/3 of the time and A beats C 3/4 of the time. In a three way match, A wins 12/22 of the time.
But I have to agree with Dason. Other scenarios give different results.
 

Dason

Ambassador to the humans
#6
Yeah imagine A's time is uniform(0,1). B always gets a time of 2/3 and C always gets a time of 3/4. This meets the pairwise requirements as well but A wins overall 2/3 of the times.
 

katxt

Well-Known Member
#9
No you're right. I read it (I don't know why) as scores rather than times as you stated. Low times are indeed better. Cheers. kat