View Full Version : Profit and Probability

08-24-2010, 06:30 AM
Given a game with a probability of 1/x and a payout of y:1 where y=x/2 and a number of players n, where n>x and n%x=0. Build a distribution of profits and indicate the confidence limits for CI=.95 and CI=.90. Assume each player bets the same amount.

So far I understand the the payout is 50%.

And also the the number of players is a multiple of the probability...

This must be just to simplify?
And also should allow us to model the thing on n=x right?.

Consider n=x then:

Each player has a 1/x chance of winning, but profits depend on payouts, and there are chances that more than one person *could* win, in which case your profits would be nil or negative. 2 people winning would mean nil, anything more than that would be negative, and anything less than 1 would be 100% profit.

And, now I'm lost... Not even sure where to start...

Mean Joe
08-24-2010, 01:23 PM
This is the way I understand it...

We'll work out an example: Let x=4 and n=4, and each person bets 1.
ie game probability = 1/4, payout is 2:1, and 4 players.

Each player has 2 possible profits: -1 (probability=3/4), or +2 (probability=1/4).
So the distribution of profits for all 4 players is
+8 (probability=1/256), +5 (probability=12/256), +2 (probability=54/256), -1 (probability=108/256), or -4 (probability=81/256)

I assume that's what they mean by "build a distribution". And when they say "indicate the confidence limits", I guess they mean the CL for all profits.

08-25-2010, 08:29 AM
I think they want the SD, and the profits are for the game owner, not the players...

08-25-2010, 10:22 AM
Let P be the profit for the game owner.
Let N be the number of players win.

Then N \sim \mathrm{Binomial}(n, \frac {1} {x})

P = (-y)N + (1)(n - N) = n - (y+1)N

E[P] = n - (y+1)\frac {n} {x}

Var[P] = (y+1)^2 \frac {n(x-1)} {x^2}

Then you can build the model from this..