P[A beating B and C] < min(p,q)
A, B and C are in three way race - If the probability of A beating B is p, and the probability of A beating C is q, what is the probability of A beating B and C?
It sound like it should be pq. But if if the three are equally good, then p = 0.5, q = 0.5 But the probability of A winning is 1/3, not 1/4.
What is the general answer for p and q?
P[A beating B and C] < min(p,q)
All things are known because we want to believe in them.
Thanks Mean Joe. But I was hoping for some way of actually calculating the answer. Or are you saying that there is no general answer? kat
I don't think there is a general answer, as I'd also have to consider the probability of B beating C.
Even if I knew that, these kinds of questions are tough.
All things are known because we want to believe in them.
Consider your original problem with the the racers that are equally good. I put a magic spell on b and c such that they always finish in a tie with each other. In this situation you can see that the probability that A bears both of them is just 0.5. We already saw a case where the probability is 1/3. So from this we can see that at the very least we don't have enough info to get a definitive conclusion.
I don't have emotions and sometimes that makes me very sad.
I like that concrete example, Dason, and having seen that it's not hard to arrange for any value between 1/2 and 1/3. Cheers, kat
Or visually, for the equal likelihood case, you can imagine a 1x1x1 cube for AxBxC with a random point inside. The volume where A beats both B and C is a tetrahedron of volume 1/3.
Yes, in short, the right-hand side of the three expressions associated with the last equation I provided will yield probability density functions that provide expectations of: 1/2, 1/2, and 2/3, respectively (from left to right).
Hence, given these expectations, we have: 1/2 + 1/2 - 2/3 = 1/3 as the expected value; which is what you should be looking for, katxt.
Tweet |