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Thinking about the counterintuitive Monty Hall Problem (stick or switch?), revisited in this ME question, I thought I would issue a challenge:

Give in one (perhaps long) sentence a convincing explanation of why switching is twice as likely to lead to winning as sticking.

Assume the game assumptions are pre-stated and clear.

The probabilities are not even close, so there should be a convincing explanation after all the discussion of this topic, even though "1,000 Ph.D."s got it wrong (in 1990 when it first went viral).

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    $\begingroup$ Possibly some of the answers in: How to explain Monty Hall problem when they just don't get it In fact, looking it over my answer is pretty close to one given there. $\endgroup$
    – quid ♦
    Aug 27, 2016 at 0:25
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    $\begingroup$ I would like to mention this incredible sentence of the Wikipedia article: "Pigeons repeatedly exposed to the problem show that they rapidly learn always to switch, unlike humans (Herbranson and Schroeder, 2010)". $\endgroup$
    – Benoît Kloeckner
    Aug 31, 2016 at 9:05
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    $\begingroup$ Also, I like to recall when this problem is mentioned that the underlying assumptions are crucial. If the presenter where to open a random door, possibly revealing the car, and we only assume that in the present case it turned out that he revealed a goat, then the answer would be totally different. $\endgroup$
    – Benoît Kloeckner
    Aug 31, 2016 at 9:17

3 Answers 3

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If you "stay" then you win when the prize is behind the one door your originally selected, yet when you "switch" you win when the prize is behind one of the two doors you originally did not select.

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  • $\begingroup$ I am not sure this what you were looking for, if it isn't let me know and I'll remove it. $\endgroup$
    – quid ♦
    Aug 27, 2016 at 0:20
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    $\begingroup$ Your explanation is excellent, emphasizing the two doors you didn't select. The counterintuitiveness arises from thinking that the two unopened doors are an independent trial, each half likely to win. $\endgroup$
    – Joseph O'Rourke
    Aug 27, 2016 at 0:44
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    $\begingroup$ Basically the same answer, but in other words: at first, you have a chance of 2/3 to choose the wrong door, so switching will bring you to the right door. That's how I often explain it. $\endgroup$
    – Philipp Imhof
    Aug 27, 2016 at 5:49
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My preferred explanation -

The key thing to understand is that MH knows the correct door. Say there were a thousand doors. Your chance of choosing the correct door is 1/1000. Now, MH has 999 doors, and after opening 998, there's one left. In effect, he has reduced all the chance, the .999 to that one door. Now, by switching, your chance of success is 999/1000 because your chance of being right (pre-switch) was always 1/1000, and wrong, 999/1000.

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  • $\begingroup$ This is absolutely true, but I find it dangerous. Students often do not.know what parameters of a problem they can safely modify without changing the mathematics behind it. $\endgroup$
    – Philipp Imhof
    Aug 28, 2016 at 20:17
  • $\begingroup$ Understood. And I would agree with you. The question asked for a one-liner type of answer and I broke that rule. Had I gone on even further I would've continued to reduce the number to then arrive at the 3 door problem. Questions like this or somewhat contrived and students need to keep an open mind for how to approach any problem. $\endgroup$
    – JTP - Apologise to Monica
    Aug 28, 2016 at 20:20
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The sticking strategy does not use the additional information revealed by the presenter, and thus cannot have more chance of winning than if the presenter would open no door, which is 1/3.

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