## The Monty Hall Dilemma

You've watched a TV game-show many times, and have become familiar with the format. You appear on the show and become that week's winner. Three boxes are put in front of you. In one, is the star prize. You must pick which box to open, and you will claim the contents of the one you pick.

You pick one, and the host says "Are you sure that's the one you want? There's still time to change your mind." He always does this, so this doesn't surprise you. He only does that to raise the tension for the purpose of entertainment. You stick to your guns.

"Okay," says the host, "I'll just open one of the other two boxes, and then we'll see if you want to change your mind." He opens one of the two boxes which you did not pick, and it is empty. Again, this is nothing unfamiliar. "There are two boxes left," he says. "Do you want to change now, or stick with the one you picked?"

Should you stick, or swap? Does it make no difference?

STICK### You chose to stick

Bad choice

While neither sticking nor swapping will guarantee the star prize, if you swap, you will DOUBLE your chances of winning. This may strike you as odd. Most people insist that there is a fifty:fifty chance of winning regardless whether you stick or you swap. This is not so. I'll try and explain why.

You start with three possibilities, which I shall call X,Y and Z. You pick X. The chance that X is the star prize is one in three. There is therefore a two in three chance that you have picked the wrong box. You are twice as likely to have picked the wrong box as you are to have picked the right box.

There are two boxes that you didn't pick: Y and Z. It is absolutely certain that either Y or Z does not contain the star prize. There is only one box with the star prize in it, so one of the two unpicked boxes MUST be empty.

The game-show host reveals that Y is empty. You now have to chose whether to stick with X, which has a one-in-three chance of being the star prize, or to swap to Z, which has the one in three chance which it started with PLUS the one-in-three chance of Y, which has been eliminated. You were probably wrong to have picked X, but that Y was empty is now a certainty. If you were probably wrong to have picked X, then one of Y and Z was probably the right choice. That remains true now that one of the two unpicked boxes has been eliminated from the decision. If you swap, you have a two-thirds chance of winning, if you stick, you have a one-third chance.

If you ever find yourself in this situation, follow my advice, and swap. If you then lose, don't blame me. You will just have been unlucky. There may still be some of you that doubt me. In an effort to convince you, you can try out the MONTY HALL SIMULATOR written by Christian Ankerstjerne after a visit to this page, and kindly donated to this site.

There is a section on this in the biography of the mathematician Paul Ehrlish. It's called the "Monty Hall Dilemma", after a game show in the USA where the actual dilemma was staged. When publicised in a newspaper column, it caused a huge amount of publicity from mathematicians complaining that the swap solution was wrong! The world of Bayesian mathematics still divides mathematicians into those in it and those not. It does work, though. Honest.

If you are interested in probability, you make like to have a butchers' at the work I've done on calculating the probability of die roll results, which appears in my Crossfire wargaming section.

### You chose to swap.

Well done.

While neither sticking nor swapping will guarantee the star prize, if you swap, you will DOUBLE your chances of winning. This may strike you as odd. Most people insist that there is a fifty:fifty chance of winning regardless whether you stick or you swap. This is not so. I'll try and explain why.

You start with three possibilities, which I shall call X,Y and Z. You pick X. The chance that X is the star prize is one in three. There is therefore a two in three chance that you have picked the wrong box. You are twice as likely to have picked the wrong box as you are to have picked the right box.

There are two boxes that you didn't pick: Y and Z. It is absolutely certain that either Y or Z does not contain the star prize. There is only one box with the star prize in it, so one of the two unpicked boxes MUST be empty.

The game-show host reveals that Y is empty. You now have to chose whether to stick with X, which has a one-in-three chance of being the star prize, or to swap to Z, which has the one in three chance which it started with PLUS the one-in-three chance of Y, which has been eliminated. You were probably wrong to have picked X, but that Y was empty is now a certainty. If you were probably wrong to have picked X, then one of Y and Z was probably the right choice. That remains true now that one of the two unpicked boxes has been eliminated from the decision. If you swap, you have a two-thirds chance of winning, if you stick, you have a one-third chance.

If you ever find yourself in this situation, follow my advice, and swap. If you then lose, don't blame me. You will just have been unlucky. There may still be some of you that doubt me. In an effort to convince you, you can try out the MONTY HALL SIMULATOR written by Christian Ankerstjerne after a visit to this page, and kindly donated to this site.

There is a section on this in the biography of the mathematician Paul Ehrlish. It's called the "Monty Hall Dilemma", after a game show in the USA where the actual dilemma was staged. When publicised in a newspaper column, it caused a huge amount of publicity from mathematicians complaining that the swap solution was wrong! The world of Bayesian mathematics still divides mathematicians into those in it and those not. It does work, though. Honest.

If you are interested in probability, you make like to have a butchers' at the work I've done on calculating the probability of die roll results, which appears in my Crossfire wargaming section.