MSV 13: The Colin and Phil Problem

Colin and Phil play a game with a dice. Colin Phil

Colin rolls first; if he gets a six, he wins. If Colin does not get a six, then Phil rolls; if he gets a six he wins.

They keep rolling until one of them gets a six. The first to get a six wins! Problem: what are the probabilities of Colin winning, and of Phil winning? Try this!

Let p = P(C wins). Then p = (1/6)+(5/6) 2 (1/6)+(5/6) 4 (1/6)+(5/6) 6 (1/6)+… This is an infinite geometric series with a = 1/6 and r = (5/6) 2. a/(1-r) = (1/6)/(1-(5/6) 2 ) = 6/11 (slightly bigger than ½, as expected.) So P(C wins) = 6/11, P(P wins) = 5/11.

Here is an alternative method. Again, let p = P(Colin wins) Either Colin wins on the first throw, or Phil then finds himself in exactly Colin’s position at the start, with a probability p of winning. So p = 1/6 + (5/6)(1-p). Solving for p gives p = 6/11, as before.

‘What’s that, Phil?’ ‘Woof woof WOOF!’ ‘All right,’ says Colin, ‘I’ll pick a whole number at random from the numbers 1 to m. You pick a number at random from the numbers 1 to n. I win if I get an m before you get an n. We’ll choose m and n so that the game is fair.‘ ‘You think the game is unfair?’

For what values of m and n is the game fair? Try this!

Again, let p = P(C wins). We have p = 1/m + (m-1)(n-1)p/(mn). So mnp = n + mnp – mp – np + p, which gives p = n/(m + n - 1). So p = ½ gives 2n = m + n – 1, or m = n + 1.

So if Colin rolls a dice trying to get a six, while Phil rolls a dice trying to get a five (rolling again if he rolls a six), for example, then the game will be fair.

Investigate what might happen if Albert, Tony and Peter decided to play this game… ‘It’s a three, boys!’

is written by Jonny Griffiths With thanks to Pixabay.com