Probability game

Ok, i was thinking a bit about game theory and probability, and remembered the basic framework of a game (simplified a bit):

0) Before the game, player A (or computer) decides how many offers they will make, and doesn't tell player B. Also, the number of offers B can accept is also made public.
1) At evey offer, player A offers player B either $X, or $Y. (chosen at random, with probabilities m% and n% [m+n=100%])
2) When an offer is made, B can either accept, using up an acceptance but gaining money, or deny.
3) The aim of the game is for B to use their accepts most wisely in order to gain the most amount.

Now, the thing is, if B blindly accepts everything, then some of their accepts might be wasted on the lower amount. However, if they only accept offers of whichever is higher, then they might not have used all their acceptions by the time A's offers have finished.

Note that A is completely random, and is not concerned about losing money.

Now the question is, what strategy does B use? obviously, if $X=$Y, they just accept evey bid cos they can't do better. However, it became much harder to find the winning strategy with different values, and at different probablilites.

Any suggestions?

thanks for reading, i know its long, sorry :p

How does A determine the number of offers it will make?

Ah, yes, i see that determines it somewhat.
Well, i thought of this too, and came up with some possibilities:

1) Completely random (could be 3, could be 12846322946)
2) Poisson (given an 'average')
3) Average + standard deviation
4) etc...

but i guess it is such that (1) is true.

I.e. A choses a number 'out of the hat', irregardless of whether physically offering this many times would last 5 minutes or 10000 millenia.

So, my question, are x and y static, or also random?

I assumed X and Y are given constants (as are m and n)

Quote:

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1) Completely random (could be 3, could be 12846322946)

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But what kind of distribution do you use, if every number in N has the same chance (a uniform distribution), then the avarage number of offers will be inf/2 = inf. In that case B should only accept offers of max(A,Y).

If (3) is true there are two additional variables, μ (mu) and σ (sigma).

OK, yeah, the final thing should be dependant on X, Y, m and n, as twanvl said.

However, as for the number that A picks, i'm not sure what the average has to do with it.
For the purpose of simplification, i guess there could be a limit, like A can make up to 100 offers, with 1% chance of picking any particular number.

However, this adds constraints/cases for m and n, and it gets trickier.
so, if possible, if A could choose any number of offerings that would be good.

Basically, the reason why i cam up with this is because its guessing the future, i was wondering if there was any 'good' approach you could take.

You cannot guess something with infinite bounds.

To know one out of infinity yields a probability of 0(impossible!).

Meaning.... your game stinks. :p

Then again if you consider offers as a variable...

aa=o/2

(Irrelevent if the offers aren't relative to the accepts(sql?))

Average number you can accept will be half that offered, given, $X and $Y are then also equal choices, you'd be best to always accept the greater of x and y.

Thus you'd accept the first offer(we'll call it $X) regardless. If the second offer is less($Y), you refuse and only accept $X, othwerwise only accept the greater $Y.

You may not use them all some of the time, but that's probability for ya.