Question

Q3) Ahmed plays a game where he tosses two balanced 4 sided-dice, each with faces labeled by 1, 2, 3 and 4. He wins 2 points if the sum is 4. He wins 1 point if the sum is greater than 4. He loses k points if the sum is less than 4.

i. Find the probability distribution sum.

ii. Find the value of k which achieves the fairness of game. (i.e. The fairness is achieved if the game is not biased neither to loss or win)

Answer 1

I) probabillity distribution table

Let X be the points won .

Note : negative shows loss of points.

Total number of possible outcomes = 6*6 = 36

X	2	1	-k
P(X)	3/36 = 1/12	30/36 = 5/6	3/36 = 1/12
Comment	he gets a sum of 4 on pair of die Possible cases are (1,3)(3,1)(2,2)	he gets a sum of more than 4 Possible cases are (1,4)(4,1)(1,5)(5,1)(1,6)(6,1)(2,3)(3,2)(2,4)(4,2)(2,5)(5,2)(2,6)(6,2)(3,3)(3,4)(4,3)(3,5)(5,3)(3,6)(6,3)(4,4)(4,5)(5,4)(4,6)(6,4)(5,5)(5,6)(6,5)(6,6) There are 30 such cases where you can get a sum of more than 4.	he gets a sum of less than 4 Possible cases are (1,1)(1,2)(2,1)

Check: total probability of all outcomes = 1/12 + 5/6 + 1/12 = 1

hence probabillity distribution table is valid.

II) for a fair game, the expected value of game should be zero

that is expected winning = expected loss

Here, expected value of game = X. P(X)

= 2*(1/12) + 1*(5/6) +(-k) *(1/12)

= 1/6 + 5/6 - k/12

= 1 -( k/12)

Equating expected value of game to zero, we get

k = 12

Hence for fairness of game , k should be equal to 12.