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MSV 13: The Colin and Phil Problem

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MSV 13: The Colin and Phil Problem

Phil

Colin

Colin and Phil play a game with a dice.

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.

‘Woof woofWOOF!’

‘What’s that, Phil?’

‘You think the game is unfair?’

‘All right,’ says Colin, ‘I’ll pick a whole number at random from the numbers1to 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.‘

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!’

With thanks to Pixabay.com

www.making-statistics-vital.co.uk

is written by Jonny Griffiths

hello@jonny-griffiths.net