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.

SWAP

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.

Sons or Daughters?

This is a very simple problem indeed, which well illustrates Bayesian probability, and how it can be very counter-intuitive.

A man has two children. One is a boy. What is the chance that the other is a girl? For the purpose of your calculation, assume that the man, his wife, and his children are all normal, and that half of all births are male, and half female.

None 1/8 1/4 1/3 3/8 1/2 5/8 2/3 3/4 7/8 Certain

WRONG!

Try again. This is fiendishly simple.

CORRECT!

There is a two-thirds chance that the other child is a girl. In other words, the chance that the other child is a girl, is twice the chance that the other child is a boy. How can this be, if half of all births are male, and the chance of a child's being one sex is unaffected by the sex of a sibling? Here's how.

We know that the man has two children in total. There are four possibilities. Here they all are:

He had a boy then another boy
He had a boy and then a girl
He had a girl and then a boy
He had a girl and then another girl

Since half of all births are male, and half female, these are the only four possibilities. There is no third sex. Also, each of these four possibilities is as likely as any of the others, and so we could assign a probability of 1/4 to each of them. If you took a random sample of two-child families, you would find that roughly one quarter of this sample came from each of these four categories.

However, one of these four possibilities can be discounted. We know that the man has one son, therefore the possibility that he had two daughters can be ruled out. There are three possibilities left, which are still each equally probable. We can therefore assign a probability of 1/3 to each. One of these three is that he had a son and then another son. The other two possibilities are that he had one of each sex, and since we know he had one son, then the other must therefore be a daughter. So the chance that the other child is a girl is 2/3, and not 1/2 as you might think.

Another question is this: "A man has two children. The elder one is a boy. What is the chance that the other child is a girl?" Again, we can take a look at the four possibilities, and can logically rule out two of them (girl then girl, and boy then girl), leaving us with two (boy then boy, and girl then boy). So, the answer to this question is 1/2, which is what you would probably expect. Bayesian mathematics deals with applying what we know, to what we don't know, of a known set of possibilities. The exact knowledge we have makes a big difference. Knowing that the elder of the two children is a boy, makes a big difference to this question. There are Bayesian mathematicians who argue that it should be used more in courts of law, which deal with this sort of problem all the time, with very serious consequences. Most juries might think to themselves that a man is probably guilty or probably innocent, based on a judgement of the chances of what they know's being representative of what they don't know. Their assessments of the guilt or innocence of the accused can be wildly innaccurate. With the problem you have just done, most people would say that the chance of the other child's being a girl is the same as the chance of it's being a boy, whereas in truth it is double that. It would be a shame to send an innocent man to prison for a similar inaccuracy.

Amazing Dice

I have a six-sided die, with faces numbered one to six, as normal. I roll it and get a result of four. Are you amazed? How amazed do you feel? I then tell you that I am going to roll it again, and that this time I will roll a three. I roll it and get a three. The chance that any given face will end upmost after a roll is one in six. The chance of my rolling a four is one in six. The chance of my rolling a three is one in six. Therefore, neither roll was more improbable than the other. Would you, do you think, be more amazed at the second roll than the first?

What are the true chances of the two rolls? Pick and click.

1/6 & 1/6 1/6 & 1/36 1/6 & 1/12 1/3 & 1/6 1/6 & 1/24 1/6 & 1/18 1/6 & 1/2 1/1 & 1/6 1/infin. & infin./6 1/36 & 1/120 1/6 & 1/216 1/12 & 1/36

WRONG!

Try again. The main clue is in the question.

CORRECT!

Well done. Simple really, eh?

Let me deal with the second roll first. The chance that I will roll the number I specify is one in six. There are six possibilities, and only one which fulfils the requirement. Therefore the chance that I will roll that number is one in six, so the chances are fairly strongly against, so you would be rational to express some degree of surprise that my prediction came true, but not rational to be astounded, because one in six is not very improbable.

The first roll was very different. I did not specify in advance what I was going to roll. The problem considers that there are six possible outcomes. We do not take into account the possibilities that the die will balance on a corner or be eaten by a passing goat. Each of the six outcomes is as likely as the others. The die is not biased. It is therefore infinity likely that I will get a result of some sort. Four is one example of a result, just as are three, and six, and one. That I got a result of some sort is not at all surprising, as it is infinity likely.

The logic of this simple puzzle is important, because many people don't understand it, and make inaccurate judgements on the world around them as a consequence of this lack of understanding. The best example I know (of many) is the argument used by some people who occasionally make futile attempts to persuade me that a god created the world. Their argument goes like this: the world around us is very complicated. It is fantastically unlikely that the world we see around us is a coincidence, because it is so very improbable. Therefore, a consciousness must have designed it, because the chance that the world would by some random evolution end up just as we see it, is mind-bogglingly unlikely.

I shall now apply the logic of this puzzle, to refute the above argument. The world we see around us does not coincide with anything we can perceive. There is no other world, no model of another world, to which to compare the world we know. For this world to be a coincidence, it would have to have something with which to coincide. We know of no other world. Yes, the world as we know it is very complicated, and it is truly very unlikely indeed for it to have ended up as it is by some random process. However, random processes will always end up creating some result or other. Yes, there are countless billions of worlds which could have been, but there is only one world which is the one we see. The world around us is one result, like the three rolled on the die, but this three was not predicted in advance. The world is a result, not the result. If a god had, thousands of years ago, dictated a book which described the world exactly as it is now, then this would truly be an amazing coincidence, and evidence for divine creation. We would have a model with which the real world might coincide. We have no such model. The world we see is improbable, but so would any other world as big and complicated as this one be. We are not amazed when someone rolls a three on a die without having previously predicted this, nor should we be. Neither should we be amazed that the world we live in so unlikely. All worlds would be.

The Wason Test

This is a puzzle named after its creator.

There are four cards lying on a table in front of you. You know for certain that all the cards have one letter marked on one side, and one number marked on the other. Marked on their upper sides are the following:

A

F

3

4

There is a statement which is this: "If a card has a vowel on one side, then it must have an even number on the other side."

How many cards need you turn over, and which cards, to establish whether the statement is true or false?

None  A   F   3   4  A & F A & 4 F & 3 A & 3 3 & 4 F & 4 A, F, & 3 A, F, & 4 F, 3, & 4 A, 3, & 4 All four

WRONG!

Try the same puzzle again, or if you give up, click here for an explanation.

CORRECT!

Perhaps you found that very easy, but in tests, people in general answer this puzzle correctly about as often as they would do if they were guessing randomly.

For an explanation of this puzzle, click here.

Lincoln's Riddle

This is a riddle which, according to legend, was a favourite of Abraham Lincoln's.

How many legs has a dog, if you call a tail a leg?

 1   2   3   4   5   6   8   9   24  4 + no. of hairs Infinity Other

WRONG!

Try again.

CORRECT!

Well done. Too simple?

Abraham knew an important thing. He knew that things are separate from their names. A dog has four legs. If someone calls its tail a leg, it still has four legs. The number of legs does not alter, merely the names for the parts of its body. The physicist Richard Feynman was a great advocate of this sort of thinking. He got annoyed with people who thought that they understood something, just because they knew a word for it. When you crash your car into a bollard, while wearing no seat belt, you may find yourself going through your windscreen. Why is this? Some people would answer "momentum" and leave it at that, smugly assuming that they knew the answer. In fact, momentum is just a word for the property that objects have, which is that when they are moving, they like to stay moving. Why this should be is another matter entirely from coming up with a name for this observation.

So, this one was a bit of a trick question, perhaps.