Wikipedia:Reference desk/Archives/Mathematics/2010 July 25

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July 25

The probability that Bayes theorem is correct

Bayes theorem states that $P(H|E)={\frac {P(E|H)\,P(H)}{P(E)}}$, and it is often used to evaluate the probability that a particular hypothesis is correct when presented with some evidence. However, how do we know if bayes theorem is correct? Obviously we can't use bayes theorem to calculate the probability that bayes theorem is correct!--220.253.172.214 (talk) 05:36, 25 July 2010 (UTC)

We have a proof of Bayes theorem and hence it is correct-Shahab (talk) 11:43, 25 July 2010 (UTC)

Ah, but a true Bayesian thinker would attempt to estimate probabilities for the proof being incorect and its error being overlooked by everyone who has looked at it so far ... and this seems like a tricky problem because you can't use Bayes' theorem in these estimations. 83.134.169.103 (talk) 12:00, 25 July 2010 (UTC)

You don't need a precise concept of probability to be very certain about an empirical fact. For example I see what appears to be a computer in front of me and so I conclude there is a computer in front of me, all without estimating any probabilities or doing any calculations. Rckrone (talk) 15:32, 25 July 2010 (UTC)

The symmetric formulation $P(H\cap E)=P(H|E)\times P(E)=P(E|H)\times P(H)$ includes the definition of the conditional probability $P(H|E)$ and holds true even when $P(E)=0$ and $P(H|E)$ is undefined. Bo Jacoby (talk) 21:36, 25 July 2010 (UTC).

Nash Equilibrium for infinite game?

So I have a zero-sum game that looks something like this:

State X:

Y	1
1	0

State Y:

X	-1
-1	0

So if both players choose action 1, the game switches from State X to State Y, back and forth forever. But if either player knows the opponent will choose 1, he will choose action 2 when the state is favorable to him and win. Is there a way to find a Nash equilibrium for a game like this? I was able to show that if both players choose each action with probability 1/2, the expected value is an infinite geometric series with a finite sum (0.4). i can add that proof it will help. But I don't know how to find an equilibrium strategy for something like this. Is there a name for this kind of game? Does it have a nash equlibrium?

thanks,

Emeraldemon (talk) 13:49, 25 July 2010 (UTC)

How about just saying the first player chooses their action 1 with probability p and the second player chooses their action 1 with probability q and the value to the first player from symmetry if both choose action 1 is -v where v is the value of the game to them. Figure out an equation for v and p and q must be 0 or 1 or else give a maximum or minimum of v. Then it's just a calculus question and checking the possibilities. Dmcq (talk) 15:19, 25 July 2010 (UTC)

And according to my calculations, the result is $p=q=v={\sqrt {2}}-1\approx 0.4142$. This means that player 1 can gain a little by being less likely to choose action 1. -- Meni Rosenfeld (talk) 11:24, 26 July 2010 (UTC)