The Fair Stall Dice Game
The mathematics club of a school in Kalaburagi runs a dice stall at the annual fair. A player pays ₹20 to roll two fair six-sided dice once. The stall pays ₹50 if the sum is 7, pays ₹30 if the sum is 11, and pays nothing otherwise. The organisers expect about 300 plays during the day. The treasurer wants to know the probability of each winning sum, the expected payout per play, and the expected profit over 300 plays. Some members argue that the ticket price should be reduced to ₹15 to attract more students; the treasurer wants to know whether that still meets a target expected profit of at least ₹2,000. Students must use the sample space, not intuition, and must state the minimum whole-rupee ticket price that meets the target.
1. Find the probability of rolling a sum of 7, a sum of 11, and of winning something.
Read the worked answer
Two dice give 36 equally likely ordered outcomes.
Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 outcomes, so P(7) = 6/36 = 1/6.
Sum 11: (5,6), (6,5) = 2 outcomes, so P(11) = 2/36 = 1/18.
P(win something) = P(7 or 11) = 8/36 = 2/9.
2. Calculate the expected payout per play and the expected profit over 300 plays at the ₹20 ticket price.
Read the worked answer
Expected payout = 50(1/6) + 30(1/18) = 50/6 + 30/18 = 25/3 + 5/3 = 30/3 = ₹10 per play.
Revenue per play = ₹20, so expected profit per play = 20 − 10 = ₹10.
For 300 plays, expected profit = 300 × 10 = ₹3,000.
3. Find the minimum whole-rupee ticket price that gives an expected profit of at least ₹2,000 over 300 plays. Is ₹15 enough?
Read the worked answer
Let the ticket price be p. Expected profit per play = p − 10. Need 300(p − 10) ≥ 2000 p − 10 ≥ 2000/300 = 6.666... p ≥ 16.666...
The minimum whole-rupee price is ₹17. At ₹17, expected profit = 300 × 7 = ₹2,100, which meets the target. At ₹15, expected profit = 300 × 5 = ₹1,500, which is below ₹2,000. This is an expected value, not a guarantee for a single fair day.