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May 19, 2013

Today I’m going to discuss the well-known Monty Hall problem of puzzling probabilities; it’s so famous that you can Google it and read plenty of better explanations than my own…

So you’re still here? Excellent!

Imagine, if you will, the shell game of street performers (con men). Three shells (or cups) are placed on a cardboard box and a pea (or ball) is placed under one of them; the shells are shuffled and you’re supposed to guess where the pea ends up.

Now let’s pretend that this is actually a fair game rather than some street con; one of the shells does contain the pea. You’ll find this fair game wherever that altruistic bookie I mentioned in Risk-free gambling with math! can be found. Say the street performer is actually The Flash, so you completely lose track of where the pea as soon as he starts shuffling it.

In front of you is a rather smug looking superhero and three shells; if you can tell him which one the pea is under then you’ll win $100. You reach for one shell and watch his face; he doesn’t flinch. You reach for another, but still he shows nothing – The Flash has a good poker face. So essentially you’re just going to have to choose one at random and hope. Once you’ve picked one, The Flash decides to make it a little more interesting by turning over another shell, revealing nothing under it.

At this point, there are two shells left that the pea could be under. The Flash now offers you the option of switching which shell you’ve chosen before revealing which one has the pea under it. So do you stick with your gut instinct? Or do you switch to the other shell now? Does it even matter? Surely is must be equally likely to be under either one right?


Would you believe that you’re twice as likely to win if you switch to the other shell?

When you picked a shell, there was a 1/3 chance that it was right and a 2/3 chance that it was wrong, agreed? So what if the rules of the game were that you could now either keep the one shell that you’ve picked, or instead look under the other two shells? Since there is a 2/3 chance that it’s under one of the 2 remaining shells, you’re definitely better off switching now! Now here’s the kicker… Of the two remaining shells, at least one of them has no pea; when The Flash turns up one of them to show it’s got nothing under it, the probability that it’s under one of the two that you didn’t select is still 2/3! That is, the one that you’ve selected still has a 1/3 chance of having the pea, but the one that you didn’t select has a 2/3 chance of having the pea (since the one you didn’t select now is no different to selecting the other two, before The Flash removed it).

This is a little bizarre to think about, but if you imagine 100 shells with 1 pea; after you pick 1, The Flash removes 98 of the remaining 99 and shows that all 98 of them have no pea. The flash can do this every single time regardless of what’s under your shell, so it can’t make your shell be any more likely to have the pea than it did originally – there’s still only a 1/100 chance of being right, but the one remaining one that you didn’t select almost certainly has it!


From → Math

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  1. Probabability Paradox* — part 2 | Casual Calculations

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