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The Monty Hall problem

For conversation purposes, I'd like to call the fact that Monty shows you the goat behind a door you don't pick, and offers you a choice to switch, the Monty Gambit. Now:
theprestige said:
There's two goats and one car. I have 2/3 chance of picking a goat. Monty shows me the other goat. So now I'm pretty sure I've found both goats. The one remaining door probably has the car. That's about as far as my math goes, or needs to go, to solve this problem.

The rest is up to Monty.
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This is not true. The reason why it's not true is that your being offered the Monty Gambit may depend on whether you initially picked the goat. Think of this not as a mere probability exercise, but rather as a game you're playing with Monty. And you don't know what Monty's motive are.

Evil Monty wants you to lose. If you initially picked the goat, Evil Monty is certainly not going to give you a chance to win. So you're not going to get a gambit offer. You just lose. But there's a 1/3 chance you picked the car instead, much to Evil Monty's dismay. In that case, he has nothing to "lose" by offering you the gambit.

Good Monty is the exact opposite. If you started with the car, Good Monty certainly won't let you risk losing it, so he offers no gambit. You just get a car. But in the 2/3 of the cases where you picked a goat, Good Monty shows you the other goat.

If your strategy is to always switch when offered the gambit, Evil Monty will ensure you'll always lose, and Good Monty will ensure you'll always win.

The rule is to always switch, unless Monty is wearing a black hat, has a curly mustache, or has a Persian cat in his lap.
 
I've always accepted that this works (actually, I know it works) because it simply means that my first choice had at 66% chance of being wrong. Not sure if I'm oversimplifying it or not, but that's how I process it.
 
Pythonic said:
I've always accepted that this works (actually, I know it works) because it simply means that my first choice had at 66% chance of being wrong.
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That's the problem. The initial probability that you picked a goat is 66%. But having a gambit offered to you changes the probability; and you don't know how because you don't know Monty.

With Fair Monty, you have three choices:
  • goat-and-gambit
  • goat-and-gambit
  • car-and-gambit
Fair Monty always offers a gambit, so being offered a gambit does mean you had a 66% chance of initially selecting the goat, and your choices for the gambit are:
  • decline(goat)-accept(car)
  • decline(goat)-accept(car)
  • decline(car)-accept(goat)

But Evil Monty is an entirely different game. Your choices with Evil Monty are:
  • goat
  • goat
  • car-and-gambit
With Evil Monty, your initial chance of selecting a goat is still 66%. But being offered a gambit by Evil Monty means something entirely different--100% of the time that Evil Monty give you a gambit, it means you picked a car (if you didn't pick a car, you just don't get a gambit--you go home with one of the goats). So Evil Monty's only gambit choice for you is this:
  • decline(car)-accept(goat)

If you're lucky enough to see a gambit, and you don't know whether Monty is Fair or Evil, you don't know whether to swap or not. With Fair Monty your chance of getting a car after swapping is 2/3. But with Evil Monty your chance of getting a car after swapping is 0.
 
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The problem is complicated even if Monty always shows a goat and offers the switch, because the probability also depends on how he chooses which door to open when he has a choice. (If you pick a goat, he has no choice, as there's only one goat left to show you, but if you pick the car, he can show you either of the two goats.)

You picked door 1. Monty opened door 2. The probability that door 1 now hides the car is p / (p + 1), where p is the probability that Monty opens door 2 supposing he has a choice. The usual answer of 1/3 is obtained only if we assume that p = 1/2.
 
69dodge said:
The problem is complicated even if Monty always shows a goat and offers the switch, because the probability also depends on how he chooses which door to open when he has a choice.
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If Monty always picks a goat door and offers a swap, then 2/3 of the time you picked a goat, and he eliminates the other goat. In those 2/3 of the time, if you swap, you get a car.

The remaining 1/3 of the time you picked a car. There are two goats remaining--let's say it's a billy and a nanny. If Monty always shows you the billy, you'll always get a nanny that 1/3 of the time. If Monty always shows you the nanny, you'll always get a billy that 1/3 of the time. If Monty shows you a billy with probability b, you'll wind up with a nanny with probability b that 1/3 of the time and a billy with probability 1-b that 1/3 of the time.

Either way, it's only 1/3 of the time Monty has a choice, and you're winding up with a goat. The only consequence of Monty's choosing one or the other goat is whether you wind up with a billy or a nanny; in 100% of the 1/3 cases where you picked a car and swapped, you're getting some sort of goat, regardless of the probability of Monty picking a particular one.
 
yy2bggggs said:
Evil Monty wants you to lose. If you initially picked the goat, Evil Monty is certainly not going to give you a chance to win. So you're not going to get a gambit offer. You just lose. But there's a 1/3 chance you picked the car instead, much to Evil Monty's dismay. In that case, he has nothing to "lose" by offering you the gambit.

Good Monty is the exact opposite. If you started with the car, Good Monty certainly won't let you risk losing it, so he offers no gambit. You just get a car. But in the 2/3 of the cases where you picked a goat, Good Monty shows you the other goat.
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You have these exactly backwards. Good Monty has gotten the goats from the Humane Society and wants to give them away. Really Good Monty will give you the car with a goat, a dog and a mangy cat in the trunk.
 
69dodge said:
The problem is complicated even if Monty always shows a goat and offers the switch, because the probability also depends on how he chooses which door to open when he has a choice.
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Not it doesn't. He only has a choice if you picked a car (which happens 1/3rd of the time), and then the choice doesn't matter, because either way switching still gives you a goat. His choice has no consequence, so it doesn't change any odds.

Test all the permutations. But count based on YOUR choices, not his, because it's YOUR choices that can make a difference.
 
In this particular case, you pick door 1 and Monty opens door 2. Therefore, when "test[ing] all the permutations", ignore those where Monty doesn't open door 2. They're not relevant to the case before us.

Or use Bayes's Theorem, which is how I did it:

Let c1 mean "the car is behind door 1", c2 mean "the car is behind door 2", and c3 mean "the car is behind door 3". Let m2 mean "Monty opens door 2".

Let p = P(m2 | c1).

Then, P(c1 | m2) = P(c1) P(m2 | c1) / P(m2)
= (1/3) p / P(m2)
= (1/3) p / [P(m2 | c1) P(c1) + P(m2 | c2) P(c2) + P(m2 | c3) P(c3)]
= (1/3) p / [p (1/3) + (0)(1/3) + (1)(1/3)]
= (1/3) p / [(1/3) (p + 1)]
= p / (p + 1).
 
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theprestige said:
There's two goats and one car. I have 2/3 chance of picking a goat. Monty shows me the other goat. So now I'm pretty sure I've found both goats. The one remaining door probably has the car. That's about as far as my math goes, or needs to go, to solve this problem.

The rest is up to Monty.
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That is a good explanation.

A good thing to tell people is that they do NOT want to pick the car on the first guess (1/3). Switching will then get them a goat. Because picking a goat is more likely (2/3), if they do pick a goat on the first guess, switching will get them a car!

After you tell them they DON'T want to pick the car on the first guess, it usually clicks. It's hard for people to not think: must pick car, must pick car.
 
I can think of three reasons people have such hard time with Monty Haul problem.

1. Conditional probability is something most people never learn. If you understand conditional probability, it is much easier.

2. A lot of people, including mathematically educated ones, tend to think of "probability of an event" as some objective quantity of said event, whereas in reality it is just a measure of one's ignorance. A tossed coin has "traditionally" 50% of landing on either side. But if you knew precisely the coin's velocity, rotation, aerodynamic properties, air density, and air currents, you would be able to tell "heads" or "tails" with 100% certainty. If you had only some of that information, you could say "80% chance of heads" (for example). As your knowledge of a situation changes, "probability of event" changes. And in Monty Haul you receive additional information when the host opens a door and shows you the goat.

3. The situation is incredibly contrived. Where else, outside a game show, you have to make a decision based on a partial information, from a knowledgeable source, who is deliberately withholding part of information? It is just not a common occurrence.
 
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How does this relate to Deal or NoDeal when it comes down to the final box and you are offered the chance to swap your box with the one left?
 
Reno: I don't think it does. In DoND nobody involved in the game knows any of the box contents.
 
yy2bggggs said:
That's the problem. The initial probability that you picked a goat is 66%. But having a gambit offered to you changes the probability; and you don't know how because you don't know Monty.

With Fair Monty, you have three choices:
  • goat-and-gambit
  • goat-and-gambit
  • car-and-gambit
Fair Monty always offers a gambit, so being offered a gambit does mean you had a 66% chance of initially selecting the goat, and your choices for the gambit are:
  • decline(goat)-accept(car)
  • decline(goat)-accept(car)
  • decline(car)-accept(goat)
...
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The wikipedia page has a section on the variants of the problem, as well as explaining the originally intended version of the puzzle.
 
edd said:
Reno: I don't think it does. In DoND nobody involved in the game knows any of the box contents.
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Well, I don't believe Noel Edmunds when he says he doesn't know, just like I don't believe Jeremy Kyle when he says he doesn't know the results of the lie detector test. Besides, Noel Edmundses head is too large for his skinny little shoulders. He'd be better of shaving his face and getting a good haircut - he'd look more in proportion then, I say.

But anyway...even if no-one knows what's in da boxes, surely it would be better to switch yours at the end as you only had a 20/1 shot (or something like that) of picking the box with the big dosh in it?

As you may have guessed, I never really watch these shows, but my missus does and while I potter about the house, I can't help but catch little sections of these highbrow educational shows...
 
But anyway...even if no-one knows what's in da boxes, surely it would be better to switch yours at the end as you only had a 20/1 shot (or something like that) of picking the box with the big dosh in it?
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In most of those cases when you get to the last two, you already saw the big dosh go out of the game. When there's just the two boxes left you're equally likely to have either remaining value.
It has to be this way, as exactly the same logic applies if you were trying to go away with 1p rather than £250,000.
I'm pretty convinced Noel doesn't know, and the banker. It'd be mildly scandalous to be any other way, and the producers seem to be a good bunch who acted well when the pseudorandom number generator thing happened.
 
I remember reading something about a psychic-only edition of the USA version of DoND when the final 2 boxes were left and the psychic swapped boxes. She opened her prize to find 1 cent.
 
edd said:
It'd be mildly scandalous to be any other way, and the producers seem to be a good bunch who acted well when the pseudorandom number generator thing happened.
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Yeh and it'd be mildly scandalous if Blue Peter already had picked a winning contestant before some kids had even auditioned. And it'd be scandalous if ITV charged money for texts after the prize draw had closed and the winner been picked. And it'd be mildly scandalous if Ant n Dec....well, you can see what I mean.

What's this about the psuedorandom number generator thing? Please do tell.
 
Originally, in the UK DoND the amounts in each box were determined using an Excel spreadsheet using the built in random number generator. This random number generator needs seeding with a suitable source, and it wasn't. This lead to a repeated set of box configurations which recurred over the course of some number of shows (not consecutively I believe though).
This was spotted by an eagle-eyed viewer who posted to the Bothers Bar website, and the producers were contacted and the problem found. They immediately switched to using balls drawn from a bag instead (or something like it).
There is no evidence that any contestant was ever aware of the non-random nature of the box values.
 
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