Since you have equal chances of getting any of the the 6 numbers, the expected return would just be the average of all the 6 payouts.

(2+2+2) + (3+6+9) / 6 = $4 IMO

The second part,

Again keeping equal probability of getting any of the 6 numbers,

for every 6 rolls, the club would have to calculate spending $24.

We have a total of 75 participants, so total expected expense would be = (24 * 75)/6 = $300

Since they charge $5 per head, total revenue = 75*5 = 375

So expected profit would be = 375 - 300 = $75 IMO

~ Edited ~

amit28it wrote:At a social club function, a dice game is played where on a single roll of a six-sided die the following payouts are made:
$2 for an odd number, $3 for a 2, $6 for a 4 and $9 for a 6

a.what is the expected return for a single roll of the die?
b. if the club charges $5 for each roll, how much money would the club expect to make if 75 people played the game once each?

Great explanation, Shankar - the expected return is going to be the probability of each return multiplied by that return, summed up. And since the probabilities are all the same, you can just average out the returns.

The only mistake there - the returns are 2, 2, and 2 (for the odd rolls) and 3, 6, and 9 (for the 2, 4, and 6 numbers), so actually the average is (2+2+2+3+6+9)/6, or 24/6 = 4. So the expected return is 4, and if the club plans to charge $5 it expects to make a dollar per roll, or $75 on 75 rolls.

So...your method was right but you just plugged in a couple wrong values (you inputted the numbers on the dice for the even rolls...not the expected payouts for each).

Thanks Brian,

read it in a hurry and confused the values..

Thanks Shankar.............for your explanation.It helped me lot to solve my question .