Can anyone help me with the probability distribution of the following game

http://img453.imageshack.us/img453/5170/mathgameks7.jpg

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- November 5th 2007, 04:56 PMInfinitiProbability Distribution
Can anyone help me with the probability distribution of the following game

http://img453.imageshack.us/img453/5170/mathgameks7.jpg - November 5th 2007, 07:49 PMCaptainBlack
Each result 1, 2, 3, and 4 has an equal chance of occuring and that is 1/4.

So the player wins with probility 1/4 and loses with probability 3/4.

His expected return is:

r = (4)(1/4) + (0)(3/4) - 2 = -1

That is he gets $4 a quarter of the time $0 3/4 of the time and always pays

$2 to play. We could treat the cost to play differently to get:

r = (2)(1/4) + (-2)(3/4) = -1.

RonL - November 5th 2007, 08:11 PMInfiniti
Is this for the player or house?

What do you make of this?

Chance of 3: 1/4

Payoff for 3: $2

Chance of non-3: 3/4

Payoff for non-3: -$2

Expected return: 1/4 * $2 + 3/4 * -$2

Is it correct? (Handshake) - November 5th 2007, 08:30 PMCaptainBlack