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.
(In reply to
Yep by zaphod)
You are effectively saying that even though we know that Bob is definitely going to be executed (ie zero survival chance) he still has a 1 in 3 of surviving. Nonsense!
That's like saying that even though I was born a girl that there's still a 50:50 chance of me being a boy.
By providing more information the valid 'sample space' has changed. Therefore the probabilities of the outcomes change.
Think about it this way: I'm feeling particularly generous and make a game where you receive £100 if a fair die lands on 1,2,3,4 or 5, but you must pay £1 if it lands on 6. Even more generously I even offer to roll the die and show you the outcome before you decide whether or not to play. Now suppose I roll a 6, would you play?

Posted by fwaff
on 20040109 08:49:04 