1. ## Probability Distribution

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

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

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 3. 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? 4. 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?
That is the second of the calculations of return I gave (they are both for the
player)

RonL