There is a quiz show with three boxes...two contain nothing and one contains {$1,000,000.} The contestant is asked to choose one of the boxes, which he does. Of the two remaining boxes that he didn't choose, one is removed which contains nothing. There are now 2 boxes left: one contains {$1,000,000} and the other contains nothing. Assuming the contestant wants to win the {$1,000,000,} is it best to: a)keep the box he chose to begin with, b)change to the other box, or c) it makes no difference to the odds what he does ... and why?

Kainantu

The box that has been removed is inconsequential for now the choice is between remaining 2 boxes. Should the money in the money-box be a cheque, then he has an even chance to get the million dollars,so 'it makes no difference to the odds what he does ...' However, if one of the boxes has real money then the contestant wound know which box has the {$1,000,000} in it by the weight of the original one he is holding/chosen so he would choose the heavier one, thus improbving his chances by 100% Jul 04 02, 7:18 AM

Barrow boy

There was an original one in three chance that the contestant chose the correct box, but there is also a one in three chance that the other box now remaining contains the money. The odds are therefore even that either of the boxes contains the money. So (c) applies - it's a 50-50 choice. Jul 04 02, 7:20 AM

Bryce

By doing (b) he gives himself a 2/3s chance of getting the big money. By remaining with the original box he chose it is only a 1/3 chance. Jul 04 02, 9:54 AM

Tabby Tom

The answer is (b). He should change his choice. The point is that there are two boxes containing nothing, and the one which is removed may be either of them. Suppose the contestant chooses Box A, If the prize is in fact in Box A, and he switches, he will get nothing. If the prize is in Box B, Box C will be taken away, and if he switches, he'll get the prize. If he sticks with A, he'll lose. If the prize is in Box C, then Box B will be taken away. If he switches, he'll win. If he sticks with A, he'll lose. So the chance of his being wrong is not reduced by taking one of the boxes away. If you don't believe this, try a practrical experiment (using, for example, the traditional pea and thimbles). This seems to be a recent variant of an old poser about three cards. One card is white on both sides, one is black on both sides, one is black on one side and white on the other. One of the cards is laid on a table, and the side facing up is black. What are the chances that the other side is also black? Again the answer is 2 chances in 3, because the side you see may be the black side of the black-and-white card, or EITHER side of the all-black card. Jul 04 02, 10:05 AM

ajdale

This question is commonly known as the 'Monty Hall' problem because it's based on the gameshow 'Lets Make a Deal' hosted by Monty Hall. If you search for Monty Hall on the web you'll find there's been much debate about his problem... However, the correct answer is that by changing your mind you double your chances of winning. Jul 04 02, 8:53 PM