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How to explain the Monty Hall problem to a disbeliever

michalpaszkiewicz.co.uk

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Re: How to explain the Monty Hall problem to a disbeliever

#61

Post by Jensson »

Metus wrote: post #39
adhesive_wombat wrote: post #5

The thing that is often de-emphasised in the presentation of the problem, in order to make it seem more mysteriously paradoxical, is that the presenter knows where the car is and this knowledge is always used perfectly. If the question always ended with "remember: Monty knows where the car is and will use this information", it would be more obvious. Imagine a universe with many simultaneous Monty Hall clones playing…

> The thing that is often de-emphasised in the presentation of the problem, in order to make it seem more mysteriously paradoxical, is that the presenter knows where the car is and this knowledge is always used perfectly. If the question always ended with "remember: Monty knows where the car is and will use this information", it would be more obvious. The associated line of reasoning resolved the paradox for me. If I…

> If I switch the choice, I react to the new information

This assumes that him opening a door is new information. It isn't unless you make an assumption about the game masters intentions.

If the host doesn't want you to win then you will lose 100% of the time if you switch, if the host wants you to win then you could win 100% of the time by switching etc. But people think that these assumptions are "obvious" so of course you must make them, but that makes the riddle bad.

If the riddle doesn't say that the host always opens an empty door after you pick a door, then the typical solution isn't correct. People does the typical "X happened once" and assumes it means "X always happens", that is a typical naive assumption but no, just because it happened once doesn't mean it always happens, you can't make that assumption here unless stated in the riddle.

Re: How to explain the Monty Hall problem to a disbeliever

#62

Post by em500 »

mcphage wrote: post #60
em500 wrote: post #57

Earlier quoted context omitted.

The puzzle is is commonly stated is not how any TV show ever worked. In a real (repeated) TV show the host behavior will be non-deterministic, and will sometimes be benign, sometimes adverse, to increase the suspense and ratings. If the host follows a deterministic set of rules (as is usually implicitly assumed in the puzzle version), the optimal strategy is pretty easy to work out.

> If the host follows a deterministic set of rules (as is usually implicitly assumed in the puzzle version), the optimal strategy is pretty easy to work out. And yet it is the subject of endless discussions. The optimum strategy is pretty easy to work out, yet not widely believed.

You can browser the countless HN or even Reddit threads and it's always filled immediately with one of the few standard explanations (which are all in the Monty Hall Wikipedia article). None of the "endless discussions" actually disputes the strategy or nobody needs to be convinced. It's all about whoever things they can come up with the most "intuitive" explanation. At this point I'm starting to thing that it's more a myth than reality that it's "not widely believed".

Re: How to explain the Monty Hall problem to a disbeliever

#63

Post by Jensson »

mcphage wrote: post #59
mountainlks wrote: post #52

Earlier quoted context omitted.

The problem I always struggle with is the premise that presenter not only knows which box is the prize, but also does not want you to win. In other words, there should never be a scenario where the show host gives you an opportunity to switch boxes unless you have already chosen the prize box, in which your choice should be to not change boxes. I guess I just naturally assumed that game shows don't want contestants t…

> there should never be a scenario where the show host gives you an opportunity to switch boxes unless you have already chosen the prize box The host always gives you an opportunity to switch boxes.

That is an assumption though, the problem usually doesn't say that.

Re: How to explain the Monty Hall problem to a disbeliever

#64

Post by celeritascelery »

v64 wrote: post #7

I love the Monty Hall problem because it's so unintuitive that even Paul Erdos struggled to believe it for a while. > Vazsonyi ran the program 100,000 times. Erdos watched the results of the simulation. The simulation results indicated that by switching, the odds of winning are indeed two out of three. Finally, he was grudgingly convinced that switching was better. He did not like it but seeing was believing. He coul…

That was how I had to convince myself. I wrote a python program to simulate it a million times, Thinking for sure I was going to disprove everyone. But the result clearly showed 2/3 probability of a good outcome if you switched. This let my mind switch from “this can’t be true” to “this is true, now I need to intuit why”.

Re: How to explain the Monty Hall problem to a disbeliever

#65

Post by kthejoker2 »

The easiest way is to convince them is with real money on the line.

Like the article, take a deck of cards. Ask them to pick a card for you without looking. Set it aside.

Tell them if they have the queen of diamonds, you buy lunch, otherwise they buy lunch. Ask if they want to swap decks.

After they inevitably say yes, go through the 51 other cards and turn over 20 cards that aren't the queen of diamonds. Ask if they're sure they'd like to switch.

Remove 20 more cards and repeat. Then 9 more (leaving you with 2 cards.) Ask again, turn over one last card, ask one more time. (This last iteration is the actual Monty Hall problem.)

The key thing is they should understand now that Monty Hall knows where the queen of diamonds/ car is and turns over other cards/goats precisely because he knows they don't change the odds of the original choice, but many people incorrectly believe that it does.

Re: How to explain the Monty Hall problem to a disbeliever

#66

Post by pontus »

darekkay wrote: post #38
pontus wrote: post #17

There's another version of the Monte Hall problem that highlights why this is such a counterintuitive problem. Imagine that after you pick your box, Monte Hall invites an audience member up on stage and instructs them to choose one of the remaining two doors to open. This audience member doesn't know anything at all and just randomly picks one of the two doors. When their door is opened we see that it's empty. You're…

Correct me if I'm wrong, but in your particular example (spectator opens an empty door and I am asked if I want to switch), nothing changes in regards to the original Monty Hall problem. If a spectator opens a random remaining door, one of two things can happen: - a car is revealed, I lose immediately (there is no option to switch anymore) - no car is revealed, which means I again have 2/3 chances when switching, not…

No, if the spectator happens to open an empty door, the probability collapses to 50/50. That's the point I'm making for how counterintuitive this is. Here's the full set of cases. Let's say that you select door 1 and spectator opens door 2 (all other cases are permutations of this):

Door1 Door2 Door3

Car Empty Empty

Empty Car Empty

Empty Empty Car

Let's suppose your strategy is to stick with your original choice. In the first case above then you get the car. In the second case the audience member stumbles upon the car and you lose. In the third case you lose because you stick with an empty door. All three cases are equally likely and since the second one ends, and you know that your game didn't end, you know that you're either in case 1 or case 2. Your chance is thus 50/50.

The issue is that in the classic MH problem, cases 2 and 3 are collapsed into one outcome (MH opens an empty door), but that's not true here.

More mathematically, you should ask yourself p(Door1 | Game Did Not End).

Using Bayes we see

p(Door1 | GDNE) = p(GDNE | Door1) * p(Door1) / p(GDNE).

p(GDNE | Door1) = 1 p(Door1) = 1/3 p(GDNE) = p(GDNE|Door1)p(Door1) + p(GDNE|Not Door1)p(Not Door1) = 11/3 + 1/22/3 = 2/3.

This, p(Door1 | GDNE) = 1 * (1/3) / (2/3) = 1/2.

Re: How to explain the Monty Hall problem to a disbeliever

#68

Post by pontus »

n4r9 wrote: post #48
darekkay wrote: post #38

Earlier quoted context omitted.

Correct me if I'm wrong, but in your particular example (spectator opens an empty door and I am asked if I want to switch), nothing changes in regards to the original Monty Hall problem. If a spectator opens a random remaining door, one of two things can happen: - a car is revealed, I lose immediately (there is no option to switch anymore) - no car is revealed, which means I again have 2/3 chances when switching, not…

I think you're correct. Whether or not the spectator had a non-zero chance of opening the non-empty door is irrelevant after the fact.

No. The fact that they happened to not open the door with the car tells you something. Take it to the extreme of 100 doors. If the spectator randomly opens 98 doors and doesn't randomly stumble upon the car you should take that as evidence that you might have the car yourself. This extra evidence in favor of staying with your door exactly cancels the original MH advantage of switching.

Re: How to explain the Monty Hall problem to a disbeliever

#70

Post by darekkay »

pontus wrote: post #66
darekkay wrote: post #38

Earlier quoted context omitted.

Correct me if I'm wrong, but in your particular example (spectator opens an empty door and I am asked if I want to switch), nothing changes in regards to the original Monty Hall problem. If a spectator opens a random remaining door, one of two things can happen: - a car is revealed, I lose immediately (there is no option to switch anymore) - no car is revealed, which means I again have 2/3 chances when switching, not…

No, if the spectator happens to open an empty door, the probability collapses to 50/50. That's the point I'm making for how counterintuitive this is. Here's the full set of cases. Let's say that you select door 1 and spectator opens door 2 (all other cases are permutations of this): Door1 Door2 Door3 Car Empty Empty Empty Car Empty Empty Empty Car Let's suppose your strategy is to stick with your original choice. In…

Thanks for the explanation! As with the original MH problem, your "100 doors" example helped me better understand your variation of the problem.
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