# Probability Distribution

• Nov 5th 2007, 04:56 PM
Infiniti
Probability Distribution
Can anyone help me with the probability distribution of the following game

http://img453.imageshack.us/img453/5170/mathgameks7.jpg
• Nov 5th 2007, 07:49 PM
CaptainBlack
Quote:

Originally Posted by Infiniti
Can anyone help me with the probability distribution of the following game

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

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
• Nov 5th 2007, 08:11 PM
Infiniti
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)
• Nov 5th 2007, 08:30 PM
CaptainBlack
Quote:

Originally Posted by Infiniti
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)

That is the second of the calculations of return I gave (they are both for the
player)

RonL