# A game costs £1 to play, the coin rolls into a random cup if it goes in the WIN cup, you get£3, if LOSE then gets nothing. Lily plays 12 games(gets £6 profit), Mila plays 25(£4 loss). a)use their results to find out the probability of winning once?

### Ciara Daniels

3+ years teaching Maths

Let's name and conquer. We'll write P(m,n) to mean "the probability of getting m or more with n rolls". For example, the probability of winning the game is P(10,3): the probability of getting 10 or more with 3 rolls. We'll also use P(n) for "the probability of getting (exactly) n with 1 die". For example, P(3) is the chance of getting a 3 with one roll. Thinking about the problem a bit, we can find a way to simplify it down to 2 dice instead of 3, like this: P(10,3) = P(1) * P(9,2) + P(2) * P(8,2) + P(3) * P(7,2) + P(4) * P(6,2) + P(5) * P(5,2) + P(6) * P(4,2). This means that for example if the first roll is a 3 (with probability P(3)), then the chance of winning is the same as the chance of getting 7 or more with 2 rolls, and similar for the other terms. We also know that P(1) = P(2) = ... = P(6) = 1/6, because the dice are fair. So we have: P(10,3) = 1/6 * P(9,2) + 1/6 * P(8,2) + 1/6 * P(7,2) + 1/6 * P(6,2) + 1/6 * P(5,2) + 1/6 * P(4,2) = (P(9,2) + P(8,2) + P(7,2) + P(6,2) + P(5,2) + P(4,2)) / 6 Now we have not one problem, but six: what are P(9,2), P(8,2), ...? Fortunately, these are a bit easier. We can apply the same trick as before, transforming the probabilities of 2 dice into sums of probabilities of 1 die. For example for P(9,2): P(9,2) = P(1) * P(8,1) + P(2) * P(7,1) + P(3) * P(6,1) + P(4) * P(5,1) + P(5) * P(4,1) + P(6) * P(3,1) Again, we know P(1) = P(2) = ... = P(6) = 1/6: P(9,2) = 1/6 * P(8,1) + 1/6 * P(7,1) + 1/6 * P(6,1) + 1/6 * P(5,1) + 1/6 * P(4,1) + 1/6 * P(3,1) = (P(8,1) + P(7,1) + P(6,1) + P(5,1) + P(4,1) + P(3,1)) / 6 But we also know that P(8,1) and P(7,1) are 0. After all, you can't roll an 8 or a 7 with one die: P(9,2) = (0 + 0 + P(6,1) + P(5,1) + P(4,1) + P(3,1)) / 6. The other probabilities here we can get simply by counting. For example P(4,1), the "probability of getting 4 or more with 1 die", is 3/6 = 1/2: out of the 6 possible rolls a 4, 5, or 6 is good enough. Putting it all together, we get: P(9,2) = (1/6 + 2/6 + 3/6 + 4/6) / 6 = (1 + 2 + 3 + 4) / 36 = 10 / 36. (I'm not simplifying this because leaving the numerator as 36 makes the next step easier.) If we do the same for P(8,2), P(7,2), P(6,2), P(5,2), and P(4,2), and plug it into the formula for P(10,2), we get (check this for yourself!) : P(10,3) = (10/36 + 15/36 + 21/36 + 26/36 + 30/36 + 33/36) / 6 = (10 + 15 + 21 + 26 + 30 + 33) / 216 = 135 / 216 = 62.5%.

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