Suppose that a game of chance is played with a pair of fair 10-sided dice (with

Question

Suppose that a game of chance is played with a pair of fair 10-sided dice (with the sides numbered 1 to 10). In the game, you can pick any number from 1 to 10 and the two dice are then “rolled” in a cage. If $1 is bet and exactly one of the number that you picked is rolled you win $1, and if both of the dice are the number that you picked you win $20 (in each of those cases you also get your initial $1 bet back). If none of your number winds up being rolled you lose your $1 bet. Suppose that you play this game 8 times and pick the same number each time.

a) What is the probability that doubles of YOUR number (both dice come up your number) does not occur in the 8 rolls?

b) What is your total expected win or loss? Indicate in your answer both the amount (rounded to the nearest $0.01 if necessary) and whether it is a win or loss.

Explanation / Answer

Total sample points are 100 as it is a 10-sided dice

Now the probability of appearing exactly same number on two dice = 1/100

Let x be the number we think then number of cases are when it appears in exactly one of the dice include :

(x,1), (x,2), (x,3), (x,4), (x,5),...,(x,10) and also (1,x), (2,x)...,(10,x)

But here note that on case is not counted when x is on both dice as we want on exactly one of the dice

Hence total 19 cases are there

Hence probability of getting the number we picked come on exactly one dice = 19/100

(a) The probability that doubles does not occur in 8 rolls is given by :

8C0(p)0(1-p)8

Here p = probability of getting doubles = 1/100

=> 8C0(1/100)0(99/100)8 = (99/100)8 = 0.923

(b) Expected value of game = x*P(x)

x is the variable for winning $x

Expected value of game = 1(19/100) + 20(1/100)-1(1-20/100)

=> Expected value = 19/100+20/100-80/100 = -41/100 = -0.41

Hence negative expected value means its s loss of 41 cents

Now in 8 rolls expected loss = 8*-0.41 = -3.28

So total expected loss = $3.28

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