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Probability
By Anonymous on Sunday, August 27, 2000 - 05:41 am:

Hi. If anyone can help I need to know how to to solve this problem;

I play a very simple game at a casino where I choose to roll anywhere from 4-10 dice. The cost of playing is $10 per die. When the dice are rolled, I lose my investment if a 6 turns up on any of the dice. The casino designs the payout so that , in the long run, it takes 5% of what is bet by the players.
Find out what the payout for 4 dice, 5 dice , 6 dice, 7 dice, 8 dice, 9 dice and 10 dice will be.

If anyone can help me with this I would appreciate it.
By Brad Rodgers (P1930) on Wednesday, August 30, 2000 - 02:56 am:

Sorry if I'm missing something, but what is given to one if his dice do not roll a six?

Brad

By Tom Hardcastle (P2477) on Wednesday, August 30, 2000 - 03:09 pm:

Ok, lets say we choose to roll x dice. Then the amount that we pay is 10x dollars. 5% of this is 0.5x dollars.

Now the probability that a six does not turn up on any dice is (5/6)x. The probability that a six turns up on at least one dice is 1 - (5/6)x.
Now then; if a six turns up, we pay 10x dollars. If a six doesn't arrive, we receive y dollars. The casino wants to arrange matters so that the expectation is that we will lose 0.5x dollars. So,
y(5/6)x -10x(1-(5/6)x) = -0.5x
y = (10x(1-(5/6)x) - 0.5x)/(5/6)x
Putting in values for x show that, e.g, for four dice, the payout would be 38.7968 dollars if no six turned up. For five dice, the payout is 68.1952 dollars. And so forth.

Tom