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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

#71

Post by jdechko »

I completely agree with the math on this (math is math). However here are my observations on why this is still such an interesting problem.

1) All of these explanations end up taking this to the extreme. (Imagine playing 10,000 games. Or imagine 100 doors). The game is purposefully set up as 3 doors and one ”game”. The decision the player makes is final and they don’t get to see “averages over time”.

2) Confirmation bias (there’s probably a more correct term, but I’m going with this). A player picks the door and then switches, knowing their chance of winning is 66% by switching. But it’s also 33% losing. Switch and lose, and psychologically you feel you made the wrong choice. People who don’t understand the math will tell you that you made the wrong choice. I think that can cause a lot of people to second guess themselves, even if they know the problem.

3) Fortunately, the stakes are fairly low. Unlike some of the proposals, losing only means losing out on a car, not death.

The MHP has a mathematical solution, but it’s also very much a human-nature problem.

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

#72

Post by detrites »

fwlr wrote: post #24

Consider the Honty Mall problem: it’s like the original problem, except after you pick a box, Honty offers you both of the other boxes. It’s much easier to see swapping is better in this problem, and it’s also easier to see that the chance is 2/3 if you swap. Then you just have to show that the Honty Mall problem is equivalent to the Monty Hall problem, by stipulating that Monty will always open a box that’s empty.

Best explanation, thanks! It's so obvious now.

By opening the box he knows to be empty, it manufactures an equivalence to if he'd offered to open the box you chose, vs open the other two together.

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

#73

Post by basch »

jasode wrote: post #40
kasperni wrote: post #15

Earlier quoted context omitted.

I think their explanation is a lot easier to understand "When we pick the original box, we know that the probability that the keys will be in there is 1/3. The probability that the keys will not be in the box you originally chose is 1 - 1/3 = 2/3. Just from this knowledge alone, you could decide that you will always switch, since the probability that the other boxes have the keys is 2/3."

>= 2/3. Just from this knowledge alone, you could decide that you will always switch, since the probability that the other boxes have the keys is 2/3. Your sentence the particular way you worded it is not the correct mathematical model. The player does not get to switch to BOTH OF THE OTHER 2 boxes as an alternative to just the 1st box. Therefore the 2/3rd probability doesn't apply. Where the non-intuitive 2/3rds pro…

You do get to switch to both other doors. One of the two remaining doors is a goat, and it is opened for you. Another way to phrase that is getting to pick both doors and the goat doesnt count against you.

The Monty Hall problem distills to simply "would you like one or two doors, (if the prize is behind any door in the set you pick you win.)"

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

#74

Post by basch »

Jensson wrote: post #61
Metus wrote: post #39

Earlier quoted context omitted.

> 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 co…

>if the host wants you to win then you could win 100% of the time by switching etc.

66% of the time. 33% of the time you picked right first time and lost it by switching. The host never conveys "dont switch."

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

#75

Post by basch »

mojomark wrote: post #49

There's actually an easier way to remove the subtleties of this statistical problem by using a hyperbolic example. For kicks and shiggles we'll up the stakes to be a prize of $1B USD. Consider instead a Monte Hall scenario with 100 boxes (vice just 3), maybe we call this the "deal or no deal" variant of the problem... The user picks 1 box and has a 1/100 chance of selecting the box with the prize. Now, the host opens…

I don't see how "would you like 1/3rd odds or 2/3rd odds of winning" is less easy to understand that adding more numbers and layers to the problem by adding in words like $1Billion and 100 boxes.

If someone is willing to pick 1/3rds odds of winning, the discussion needs to go somewhere else.

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

#76

Post by basch »

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…

The thing is, 1/3rd of games ends with the audience member picking the car. So, probability collapses to 50% in 2/3rds of cases.

It's a completely different problem than the Monty Hall problem, which can never end in the car being discarded.

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

#77

Post by charlieflowers »

wanderingstan wrote: post #21
qumpis wrote: post #3

Earlier quoted context omitted.

Or to just imagine a 1000 boxes with the same problem formulation

This was the one that worked when explaining it to my friends. It gives a mental image of the host opening 998 boxes, leaving only your selected box and one other. From here it’s easier to see that there must be something special about that one box the host left un-opened! (Though even then there were people who clung to the “2 boxes means 1-in-2 chance” fallacy, failing to see that the host has revealed information.…

To me, THAT is the most powerful intuitive description.

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

#78

Post by igetspam »

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…

I'm giving you an up vote for creative writing and nothing else. I'm not claiming anything about your data, just your creative writing skills. Thank you.

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

#79

Post by pontus »

basch wrote: post #76
pontus wrote: post #66

Earlier quoted context omitted.

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 thing is, 1/3rd of games ends with the audience member picking the car. So, probability collapses to 50% in 2/3rds of cases. It's a completely different problem than the Monty Hall problem, which can never end in the car being discarded.

The point I was making is that the problem (MH) is a lot more subtle than people give it credit for. Many of the arguments people make for how the MH problem is "obvious" seem to work for this modified game too unless you're careful with them.

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

#80

Post by pjdemers »

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…

I agree that this additional information is necessary to understand the problem. I've seen that show maybe twice in my lifetime, both times when I was a little kid. I did not know that Monty never opens the door with the prize behind it. Not one time in the history of the show did he ever open the prize door. Knowing that is additional information, and can be used to figure out the problem. But you need to watch the show more than twice to know that.
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