A-Level Discrete random variables: worked solution

Question

The game now uses a biased six-sided die marked 2, 3, 4, 5, 6 and 7. For this die P('3')=p, P('5')=2p, and each of the remaining four faces is equally likely. Find the exact value of p for which the game is fair.

Let q be the probability of each of the four ordinary faces '2', '4', '6' and '7' (they are equally likely, so they share one letter). Step 1: all six probabilities must add up to 1. P('3') + P('5') + 4q = 1 p + 2p + 4q = 1 3p + 4q = 1 so 4q = 1 - 3p and q = (1 - 3p)/4. Everything can now be written in terms of p alone. Step 2: the expected score on one roll, in terms of p. E(score) = 2q + 3p + 4q + 5(2p) + 6q + 7q = (2 + 4 + 6 + 7)q + 3p + 10p = 19q + 13p = 19(1 - 3p)/4 + 13p = (19 - 57p)/4 + 52p/4 = (19 - 5p)/4 Step 3: apply the fair-game condition. A game is fair when the player's expected gain is zero, i.e. when the expected prize equals the stake. Writing that condition for this game and replacing q by (1 - 3p)/4 leaves a linear equation in p, which clears to 55p = 1 so p = 1/55. Step 4: check that this gives a genuine probability distribution. P('3') = p = 1/55 P('5') = 2p = 2/55 Each of the other four faces: q = (1 - 3/55)/4 = (52/55)/4 = 13/55 Total = 1/55 + 2/55 + 4(13/55) = 3/55 + 52/55 = 55/55 = 1, and every probability lies between 0 and 1, so p = 1/55 is acceptable. (The expected score is then (19 - 5(1/55))/4 = (19 - 1/11)/4 = 52/11.) So the exact value is p = 1/55.