A probability question

[jack]

I've attempted to formalise mathemetically the most interesting question in http://cartesiandaemon.livejournal.com/179726.html. It should have a solution now, I'm not sure if a trivial one or an unfindable one.

Consider Pa and Po in [0,1]. Have a series of trials (independent random variables taking values 0 or 1) X1,X2,..., succeeding with probabilities x1,x2,..., and an event H that the trials are in the subset of possible results S∈{0,1}^N.

(We should add the constraint that being in S is determined by some finite initial segment of the sequence, or is so almost certainly, but can do so later if necessary.)

Each xi must be either (a) equal to Po, or (b) a function of Pa (with values in [0,1]).

Question: Can we choose x1,x2... and S thusly such that for any Pa and Po:

P(H)=Po if Pa=Po
P(H)<Po if Pa>Po

?

Comments

[cartesiandaemon.livejournal.com]

Has something gone wrong with the formatting?

Oops. A lessthan escaped escaping; it should be fixed, thank you.

Suppose the person whose job it is to choose the new rules is Really Stupid. Then there's nothing you can do that will *enable* him, never mind *incentivize* him, to make the new rules do the right thing with the winning probability.

True. I was stipulating the players could reasonably estimate the probability, maybe that assumption didn't make it into this post.

each player is supposed to end up believing that the final probabilities are just right.

Something like that. The problem is more complicated than the other examples, but they're definitely what it made me think of. They're supposed to think that

(a) their own estimate is accurate
(b) their opponent should have made their estimate accurately or suffer disadvantage
(c) hence either (c1) they agree on a probability, (so are happy because this can be used to mod the game in amusing ways such as introducing new rules mid-game that don't change the chance of winning) or
(c2) they disagree about the probability of winnig and do play something complicated that each thinks gives themselves a greater chance of winning (so are happy).

Having explained why I think the problem might not be solvable, here's a solution

Oohm thank you! That's exactly what I wanted.

It *does* break what I asked for because in the two-game scenario there aren't two clear outcomes that can be assigned the same payoffs as winning and losing in the original, but it's better because you answered the condition which I really wanted and to what I was able to describe was only an approximation, since however you assing the payoff for a draw it should still work.

How I see this being used is the players are half way through a game, and one of them draws a card that says "You may cast any spell equivalently powerful to [example]". Normally it would be impossible to decide what's fair, but here the first player can choose a spell, however complicated, that they consider equivalent, and if the other player disagrees he can challenge the first player to play two games like this, with some scaling factor to account for whichever player is ahead[1]

[1] I *think* such a factor works, I haven't checked. Coming up with that sort of thing was the problem originally.

f you find this unsatisfactory, then clarifying what's unsatisfactory about it may possibly make it clearer just what's being asked for...

No, thank you for responding!

BTW, hi, nice to meet you online. Can I ask if I know you in person at all?

[gjm11.livejournal.com]

Hi. I don't think you know me in person. I'm Gareth McCaughan, another mathmoidal cantabrigiensis working for a local technology company but (o tempora o wossname) somewhat older than you :-). We have a number of LJ-friends in common; for all I know we may have met at a party or something, but I regret that I don't remember the occasion if so.

[cartesiandaemon.livejournal.com]

Pleased to meet you!