www.making-statistics-vital.co.uk. 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
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MSV 13: The Colin and Phil Problem
Colin and Phil play a game with a dice.
Colin rolls first;
if he gets a six,
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?
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?’
‘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?
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
if he rolls a six),
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
is written by Jonny Griffiths