Rod Liddle
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This is the Monty Hall dilemma and it was back in the news last week, picked up in the weekend’s broadsheets, and has been rolling around the blogs ever since. It resurfaces once in a while and everybody is always dutifully astonished, not to mention utterly disbelieving whenever it emerges in one or another guise. On this latest occasion it was presented in a slightly different (although pretty familiar) form, at the start of an article by Alex Bellos in New Scientist.
He was quoting a puzzle designer called Gary Foshee who was addressing a symposium in Atlanta and who began his speech with this conundrum: ‘I have two children. One is a boy born on Tuesday. What is the probability that I have two boys?’ This is the Monty Hall dilemma, and indeed my card puzzle, slightly rewritten — the same sort of principles apply. In this case, bizarrely, the information that the boy was born on a Tuesday (or any other day of the week) is crucial to the calculation and not, as you might imagine, a red herring. Without that information, you can calculate the odds by looking at the combinations of two children it is possible to have — (gg, gb, bg, bb). As we already know that one child is a boy we can eliminate (gg) — and therefore the odds of two boys are one in three. When you add in the information that the boy was born on a Tuesday, however, the probability changes from one in three to 13/27, or almost 50 per cent — a huge difference. Once you have listed the equally likely possibilities of children, together with days of the week, you end up with 27 possible permutations, 13 of which are two boys.
Now, I think this is a less startling outcome than my card trick or the Monty Hall dilemma — which I promise I’ll come to — because when you strip it down it is not hugely counterintuitive. By which I mean that if I have one boy and asked you what the likelihood of my next child being a boy (ignoring all genetic factors) you would probably say 50 per cent — which would not be far wrong, as it turned out. But in my card game, the answer is genuinely counter-intuitive. Your wife, or catamite, has chosen a card. You have revealed a joker to her, or him. That leaves two cards on the table in front of her, one of which must be a joker, the other of which must be the ace. So it’s surely a 50/50 choice? But the fact is, your wife will be roughly twice as likely to win her biscuit if she switches. Her chances of winning will be about one in three if she stays with her original choice, and about two in three if she changes her mind. The reason I suggested playing this game is that almost nobody believes this outcome to be true.
So, Monty Hall. Monty was the compère of a US gameshow called Let’s Make A Deal, back in the 1960s and 1970s, in which this apparently strange conundrum arose. In 1990 Maralyn von Savant, a columnist for Parade magazine, posed a question to her readership based upon the workings of this show.
She said: ‘Suppose the contestants on a game show are given the choice of three doors: behind one door is a car; behind the others, goats. After a contestant picks a door the host, who knows what’s behind all of the doors, opens one of the other doors to reveal a goat. He then says to the contestant, “do you want to switch to the other unopened door?” Is it to the contestant’s advantage to make the switch?’
She said, incontestably, that it was. And then her answer was contested, with great vigour and anger, by the nation’s mathematicians and one or two Nobel prize-winners, who all argued — as intuition would suggest — that it was 50/50. Almost 1,000 maths PhD chaps wrote in castigating her logic. One of the leading mathematicians of the 20th century, the eccentric Hungarian Paul Erdos, furiously denounced her answer as ‘impossible’ and refused to accept it even when he had seen the mathematical proof. He needed to be sat down in front of a vast series of computer simulations — a faster and less biscuit-based equivalent of my card game — before he conceded that he was wrong, and von Savant was right. And the explanation is not terribly abstruse or difficult, if you go back to my card game; if your wife has guessed her card correctly in the first place and refuses to switch — then she wins, but there is only a one in three chance of that happening. The chances of her having guessed wrongly, meanwhile, are two out of three, so twice as likely as the lucky guess. So she should switch and, much more likely than not, enjoy her bourbon.
Does any of this matter, beyond being of faint excitement to grey-skinned stats nerds? I think it does. We are bombarded with statistics every day and they are mostly misleading, contain false correlations, misunderstand the idea of ‘sample space’. Sometimes this is done deliberately, but more often it is the consequence of ignorance or the chimera of intuition. Nothing is quite what it seems; likelihoods can, with the addition of more information, suddenly become unlikelihoods. For a better understanding of this business I’d direct you to two excellent books — The Drunkard’s Walk, by Leonard Mlodinow and The Tiger That Isn’t, by Michael Blastland and Andrew Dilnot.
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