The Paradox of the Three Prisoners
Chances are you may have heard of the famous Monty Hall paradox (you know, the one about the game show with three doors and the car and the goats? Click here for a recap). If so, you were probably caught out at first by the counter-intuitive answer. No shame there — when the problem was published in a 1990 Parade magazine column, even established math professors wrote in condemning the answer as wrong.
Anyhow, if you’re into counter-intuitive probability puzzles (and who isn’t?), then here’s another for you, originally published by Martin Gardner. It’s an absolute classic, based on a problem first introduced by mathematical Frenchman Joseph Bertrand back in 1889. Monsieur Bertrand, for the record, was quite the child prodigy — by the age of seventeen, he had already earned a degree and a doctorate. As you do.
Three prisoners (let’s call them Alan, Ben and Claire) are locked in up in adjacent cells. They’ve all been convicted of identical crimes against the King, and are each facing many, many years in jail. However, His Highness, having a sadistic interest in cruel mind-games, has decided to let one of the prisoners go free the following day. The prisoners know this, but they haven’t been told who the King has chosen.
That information has been given to the Guard, who has been sworn to secrecy. He has been banned from saying which prisoner has been chosen to walk free, for fear of losing his job and all his pension contributions.
As you might expect, the three prisoners are feeling a little anxious. They each know they have a 1-in-3 chance of being freed in the morning, which isn’t that great. Alan decides to try his luck with asking the Guard for more information.
“I know you cannot directly say which prisoner is going to be freed,” he whispers through the bars, so the other prisoners cannot hear. “But there is nothing preventing you from saying the name of one of the prisoners who won’t be freed.”
“Technically, that is correct,” replies the Guard in a hushed voice.
“Right — if Ben is to be freed, say Claire’s name. If Claire is to be freed, say Ben’s name. And if I am to be freed, toss a coin. If heads, say Ben’s name, and if tails, say Claire’s name”.
The guard agrees to this, and turns his back (to hide whether or not he flips the coin).
“Well?” asks Alan.
“Ben will not be freed,” whispers the guard, who then returns to pacing the corridor.
Alan feels slightly relieved. He now knows that either himself or Claire will be set free, and figures his chance of freedom is now 50:50 (which is better than the 1-in-3 chance he calculated earlier).
Plot twist! Claire has overheard everything so far. She figures that she is twice as likely as Alan to be released in the morning. Who is correct?
As you may have guessed, the seemingly counter-intuitive answer is the correct one — Claire is indeed twice as likely as Alan to be set free. Mathematically speaking, the puzzle is pretty much identical to the Monty Hall paradox. The “confounding” factors here are the format of the question, and the role of the coin flip. Let’s look in more detail…
OK, picture it this way: whatever answer the Guard gives, Alan will come to the same conclusion — that his chance of freedom jumps up from 1-in-3 to 50:50.
However, all of the prisoners could each pose a similar question. By Alan’s logic, all three would be correct in thinking they had a 50:50 chance of freedom. That’s clearly impossible! Since they’d all have used identical reasoning, we must conclude that this line of reasoning is wrong.
By hearing Ben’s name, Alan has not received any extra information about his own chance of being freed. He could have already known that at least one of Ben and Claire were not going to be set free — so hearing one of their names tells him nothing new. But how can Claire know she is twice as likely to go free?
Well, let’s recap what we know. Alan has learned nothing new, so his chances must remain 1-in-3. Ben’s chances are confirmed as zero. So, in order to sum all the probabilities to 1 (because we’re reasonably certain that someone will walk free), Claire’s chance of freedom must be 2-in-3 — twice that of Alan’s. Not convinced? Read on…
At the start of the scene, there are three *equally likely* possibilities.
- Alan is to be freed (1-in-3)
- Ben is to be freed (1-in-3)
- Claire is to be freed (1-in-3)
However, Ben is doomed, so let’s scratch out the middle option. This means that one of the following events might happen (let’s say we re-run the whole scenario a hundred times):
Alan is to be freed (this happens 50/100 times)
- the coin lands on “heads”. The Guard says Ben’s name (25/50 times)
- the coin lands on “tails”. The Guard says Claire’s name (25/50 times)
Claire is to be freed (this also happens 50/100 times)
- in this case, the Guard will always say Ben’s name (50/100 times)
Except — we can cross out the events in which the Guard says Claire’s name. This leaves two possible things the Guard might say:
- “Ben” — because Alan goes free AND the coin shows “heads” (25 times)
- “Ben” — because Claire goes free (50 times)
See how the second event is twice as likely as the first? Unlike the first, it doesn’t depend on the result of a coin flip.
The format of the question posed by Alan creates an asymmetry between his and Claire’s situations. Neither hear their name from the Guard, but for fundamentally different reasons. Alan could never have heard his name, whereas Claire could — but didn’t. It is this difference which allows Claire to receive more information from the Guard’s answer than Alan does.
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