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Comment by ColinWright | Hacker News Reader

In presentations/workshops I now find vanilla-flavoured Monty Hall to be of no real value - nearly everyone says "SWITCH!!" without thinking. So I offer this variant.

I have 10 cups, and under one is a prize. I get a victim - ahem - volunteer to select two, and place markers on them.

I then say clearly that I will remove 6 of the other cups and, to retain the mystery, will not reveal the prize. There are now four cups, the prize is under one, and two of them are the volunteer's original choice.

And now the offer: They may retain their original two, or they may surrender their two and select only one of the other, currently unselected two.

So:

* What would you do, and why?

* What interesting further variant can you think of?

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There's a 20% chance that the volunteer was initially correct. In the 80% case, one of the two remaining cups has the price beneath it, so choosing one of the remaining cups gets you a 40% chance of payout instead of your initial 20%. Switching doubles your chance of payout.

This should be straightforward to anyone who actually understands the Monty Hall problem.

A very educational variant I've heard is to up the number of doors from 3 to 10000.

Now that Monty is closing off 9998 doors, it makes the leap of logic much less difficult. Clearly the one door he didn't close off is special.

Personally, I find that completely uninformative. Others find it helpful and I won't deny their experience, but I find that no more convincing that proper arguments about the 3 door version.

My variant was not intended to assist explaining the original, it's a variant intended to make people think again. It often exposes non-understanding in those who will blindly and automatically say "Switch".

Here's another variant. 12 doors, I let you choose 3. I then open 4, leaving your chosen 3 and 5 others.

I let you keep your original choice, or give them up to choose only one of the others. Should you switch now?

No, In that

P(3/12) = 25%

The remaining doors have 75% probability and choosing 1/5 only gives you 15% chance of winning.

Correct.

Explaining that to muggles, though, is non-trivial.

> I then say clearly that I will remove 6 of the other cups and, to retain the mystery, will not reveal the prize

Do you say 'I will now remove 6 empty cups?', or is it believed that you removed 6 unmarked cups at random?

I can't understand how you can quote my text and still ask that question. Seriously, given that I say I won't reveal the prize, how could anyone believe that I would be removing 6 cups at random?

"I am not able rightly to apprehend the kind of confusion of ideas that could provoke such a question." --Charles Babbage

I misread it, as:

> I then say clearly that I will remove 6 of the other cups. And, to retain the mystery, I will not reveal the prize.

It wasn't a criticism.

I just wanted to draw attention to the point that in the random game (I.e. player marks two, you reveal 6 random unmarked, if no prize is revealed, player decides to switch to one or to keep two) it is the better strategy to never switch.

Do you think it would be an interesting variant to play the game like that?

As written, it is ambiguous.

You say you don’t reveal the prize, but one can remove six cups at random without revealing their contents.

“Removing cups” is not the same as “removing empty cups.”

If the rules of the game permit you to remove the cup with the prize, it is significant. I think the math and optimal strategy changes.

> You say you don’t reveal the prize, but one can remove six cups at random without revealing their contents.

One can, but one cannot guarantee it.

Actually, one can. One can choose 6 at random from the ones that are neither selected, nor hide the prize.

My point is that my phrasing is, I believe, sufficient to allow the listener to determine that I will turn six cups without revealing the prize. In particular I say that I will remove 6 of the other cups and ... will not reveal the prize.

I honestly don't see how that can be interpreted in any way other than to say that I will deliberately not turn a cup that reveals the prize.

I think I should still switch, and for the same reason as in the original problem. The fact that you chose not to remove those two cups makes it more likely that one of them contain the prize.

Now try the variant in another comment ...

12 doors, I let you choose 3. I then open 4, leaving your selected 3, and 5 others.

I let you keep your original choice, or give them all up to choose only one of the other five. Should you switch now?

My intuition would be that the evidence in favour of the 5, given by the fact you chose not to reveal them, doesn't come close to outweighing the 3-to-1 advantage granted by sticking.

Let me check with the actual calculations. If I stick I have a 3/12 chance of winning. If I switch and I originally did pick the winning cup then I always lose, and if I didn't then I win with probability 1/5. So my overall probability of winning if I switch is 3/12×0 + 9/12×1/5. So sticking gets me 1/4 and switching gets me 3/20. Sticking is better, but by a smaller margin than I expected.

An interesting variation is "what if you toss a coin to decide whether to switch or not".

I find it weird that that resets the odds back to 50:50

thats not weird at all: 1/2 * 1/3 + 1/2 * 2/3 = 1/6 + 2/6 = 3/6 = 1/2

Edit: Thanks for pointing out the typesetting issue!

You need spaces around the splats, otherwise they are taken as formatting. I assume you mean 1/2 * 1/3 + 1/2 * 2/3 = 1/6+2/6 = 3/6 = 1/2. You can also reformat that as:

    1/2 * 1/3 + 1/2 * 2/3
       = 1/6 + 2/6
       = 3/6
       = 1/2

Thanks to the pair of you for explaining something I pointed out to you in the first place. :)

I said it was weird, not that I didn't understand it.

I suspect this thought process is why numerate people get thrown by this puzzle.

I wasn't explaining, I was just talking about the formatting which was originally wrong, and is now corrected.

And sure you can find it weird, lots of people do, but the reply by DoctorOetker was trying to point out reasoning which, when properly internalised, can make it feel less weird. Some people - possibly you included - never lose the sense of weirdness. I have.

But in truth, sometimes we never really understand things, we just get used to them. For me, maybe this is one of them.

I'm not annoyed at all and added the smiley to try and soften my post. I got the Monty puzzle straight away (brought to my attention by Cecil Adams when I followed the SDMB) but I find it interesting to understand why so many intelligent folk get it wrong.

As for understanding; the stack exchange question on whether it's coincidence that the value of G is approximately pi squared is one of these things I bear in mind whenever I think I understand something. : )

Edit: and if I'm not mistaken, you once spent far more time than I would have the patience for explaining how this worked to a rather offensive young person. I admired your patience there. It was more than I had and had better results.

I think people get it wrong because they assume that with two options the odds are the same, and you can't know anything to make it otherwise. That's why it can help to explain about where the knowledge is.

One reason why g is close to pi^2 is related to the original idea to define the metre as being the pendulum length required to give a 1 second half-tick. If that's your definition of the metre then g is exactly pi^2. So in some sense it's not entirely coincidence. I should go and find the stack exchange discussion, but I don't have time just now.

And thank you for the compliment about my patience - I appreciate it.

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