Basher Bill, Lightfinger Lou and Con-man Charlie are prisoners with apparently equally good records who have all applied for parole. The parole board has decided to release two of the three, and the prisoners know this, but not which two.
Prisoner Bill is friendly with a warder who knows which two are to be released. Bill feels that it would be unethical(!) to ask if he, Bill, is to be released, but wonders about asking for the name of one prisoner other than himself who is to be released.
However, he decides against it for the following reason.
‘If I don’t ask, my chances of release are 2 in 3. If I do ask, and the warder says Lou will be released, my chances go down to 1 in 2, because either Lou and Charlie or Lou and I will will be released.’
What do you think about this line of reasoning?
Solution
Some familiarity with the workings of probability will be a help with this question!
It is clear that the probability of Bill’s release is 2 in 3 or 2/3.
Let us now include the warder’s response. In this type of probability question it is important to list the possible ‘events’, and to combine the probabilities for these. It turns out that there are now four possible situations or ‘events’, and we must associate a probability with each of these. The events are:
(1) Bill and Lou are released, and the warder says ‘Lou’. In this case the warder must say ‘Lou’ – there is no choice. Hence the probability of this event is 1 in 3 or 1/3.
(2) Bill and Charlie are released, and the warder says ‘Charlie’. Again, in this case the warder must say ‘Charlie’ – there is no choice. Hence the probability of this event is 1 in 3 or 1/3.
(3) Lou and Charlie are released, and the warder says ‘Lou’. Here the warder has a choice: he can nominate either Lou or Charlie with equal probability. Hence the probability of this event is 1/2 the previous probabilities: i.e. 1/2 x 1/3 = 1/6.
(4) Lou and Charlie are released, and the warder says ‘Charlie’. Arguing as in (3) the probability here is again 1/2 x 1/3 = 1/6.
If now the warder says ‘Lou will be released’, either (1) or (3) happens. So the chances of Bill’s release are 1/3 from 1/3 + 1/6. That is, 1/3/(1/3 + 1/6) = 2/3 – the same probability as before.
16. PRISONER’S 
