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A smart prisoner? (Posted on 2004-01-08) Difficulty: 4 of 5
First, I suggest you take a look at this problem, as you may decide that this is very similar! But here's a little twist.
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Three men, Alan, Bob, and Charlie, were in separate cells under sentence of death when the governor decided to pardon one of them. He wrote their names on three slips of paper, shook the slips in a hat, drew out one of them, and telephoned the warden, requesting that the name of the lucky man be kept secret for several days. Rumor of this reached Alan. When the warden made his morning rounds, Alan tried to persuade the warden to tell him who had been pardoned. The warden refused.

"Then tell me," said Alan, "the name of one of the others who will be executed. If Bob is to be pardoned, tell me Charlie. If Charlie is to be pardened, tell me Bob. And if I'm to be pardoned, flip a coin to decide whether to name Bob or Charlie."

"But if you see me flip the coin," replied the wary warden, "you'll know that you're the one pardoned. And if you see that I don't flip a coin, you'll know that it's either you or the person I don't name."

"Then don't tell me now," said Alan. "Tell me tomorrow morning."

The warden, who knew nothing about probability theory, thought it over that night and decided that if he followed the procedure suggested by Alan, it would give Alan no help whatever in estimating his survival chances. So next morning he told Alan that Bob was going to be executed.

After the warden left, Alan smiled to himself at the warden's stupidity. There were now only two equally probable elements in the "sample space" of the problem. Either Charlie would be pardoned or himself, so by all the laws of conditional probability, his chances of survival had gone up from 1/3 to 1/2.

The warden did not know that Alan could communicate with Charlie, in an adjacent cell, by tapping in code on a water pipe. This Alan proceeded to do, explaining to Charlie exactly what he had said to the warden and what the warden had said to him. Charlie was equally overjoyed with the news because he figured, by the same reasoning used by Alan, that his own survival chances has also risen to 1/2.

Did the two men reason correctly? If not, how should each have calculated his chances of being pardoned.

No Solution Yet Submitted by SilverKnight    
Rating: 4.0000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Get a grip | Comment 13 of 22 |
It really tickles me when I read these ‘a coin has no memory’ advocates, or read cries of ‘what’s happened, has happened and no …’
We’re on a puzzle site and things, more than often, have a tendency to require some thought.
I don’t even understand why people even post
‘why, it’s gotta be 50:50, since it doesn’t matter…..’
answers – especially after numerous posts to the contrary have been, well, posted.
What kind of self-confidence must you have to hear half a dozen people say something different to your thoughts, simply to cite a principle we all learnt aged 12 and then claim they’re all wrong?
I’m not against speaking out or having conviction for your answer, but when the problem itself is fairly straightforward???

Charlie has the greater chance of the pardon.
It isn’t a puzzle if he doesn’t.
Why? There are several ways of looking at this.

View 1 (I prefer this one)
Since Alan never receives ‘Alan’ as an answer, he’s totally in the dark about the chances of himself being pardoned. The answer is irrelevant to Alan and his chances of a pardon remain 1/3. Since all probabilities (of survival) sum to 1 the remaining 2/3 must be in Charlie’s favour since Bob is going to be executed (prob’ survival =0)

View 2 (the way the problem initially hit me)
Probability of hearing ‘Bob’ if Charlie is to be pardoned =1 (given in the question)
Probability of hearing ‘Bob’ if Alan is to be pardoned = 0.5 (given in the question – coin toss)
It’s right there – the probability of the warden saying Bob is to be executed is twice as likely if Charlie is to be pardoned.

View 3
Imagine 1000 prisoners. Alan (prisoner # 1000) says “if it is prisoner 1, tell me 2-999 (the rest other than Alan) are going to be executed - if it is prisoner 2, tell me 1, 3-999 are going to be executed……..if it is me to be executed pick a prisoner at random and tell me they’re going to be executed”
There is only 1/1000 chance the warden will respond with news of the execution of a random prisoner – anything else Alan’s doomed.
This viewpoint might not seem intuitive but it’s the natural extension of the problem with more than 3 prisoners/options.

View 4 (Similar to 1)
Imagine the scenario is played out 666 times.
222 times Charlie is to be pardoned and the warden says “Bob is to be executed”
222 times Bob is to be pardoned and the warden says “Charlie is to be executed”
222 times Alan is to be pardoned and the warden says either,
“Bob is to be executed” (111 times) or,
“Charlie is to be executed” (111 times)

Out of the 333 times the warden says
“Bob is to be executed” – as he has done in the question,
222 (2/3) of these are when Charlie is to be pardoned.

Charlie’s expectation of survival initially, of course, is a third – but the fact that Alan has put Charlie in a situation (and here’s why it’s a puzzle) where you’re name has a 50% chance of zero or full survival (between Charlie and Bob) changes things entirely.
On one hand you surrender your 1/3 to Bob on the other you assume his1/3.

There’s been a few of this type of puzzle on this site (‘a something is somethinged and you are then told….’) and some debate as to the motives of the participants and what might happen over repeated ‘plays’. Silverknight has outlined the problem clearly in this case (unlike the one he’s linked to in the question) and there is no debate.

Sorry to go on – I’m a bit drunk.

  Posted by Lee on 2004-01-09 14:26:45
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