Question

I have 2 10-sided die and I roll them both. If the sum of the die is greater than a target value (not less than or equal to) you make the difference in dollars. The target value is 14. Using all the money I make from this game, I also buy a race track. I bought 25 of the nicest horses. I have a 5 lane race track as well. Let A be the amount of money (in cents) you should pay to play the first game described, and let Z be the number of races necessary to determine the 3 fastest horses out of my fleet of 25 horses. What is A*Z?

Answer 1

Solution

Consider the First game (Dice Game)

Assuming payoff is zero when sum of the die is less that or equal to 14

When sum of the die is greater than 14, Payoff = Number on dice 1 + Number on Dice 2 - 14

We get the following pay-off matrix for different combinations of numbers on the two dices

Being a fair dice, P(any number on either dice) = 1/10=0.1

Expected Value of pay-off = Sum of all values in the payoff matrix* P(any number on dice 1)*P(any number on dice 2)

=> 56 * 0.1 *0.1

=>$0.56 or 56 cents

Thus the amount of money you should pay to play the first game (A) = 56 cents

Now lets consider the second game

First, lets make groups of 5 horses and run 5 races. Suppose the 5 groups are A,B,C,D,E and the horses are labelled A1,A2,B1,B2 etc. based on the results of the 5 races (see table of outcomes below)

Race 1-5

Race 6

Next lets make a group of all winners in the first 5 races and have them run a race.

Suppose A1> B1> C1>D1>E1

Thus, A1 is the fastest horse but for second position we could have either B1 or A2

Similarly, for Third Position we could have either B1, B2, A2, A3 or C1

Race 7

Now lets make a group of  B1, B2, A2, A3 and C1 and make them run a race. The fastest horse in this race will be the second fastest overall and the second fastest in this race will be the third fastest overall.

Thus, number of races necessary to determine the 3 fastest horses (Z) = 7

A*Z = 56*7 = 392