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Monty Haul Problem... for Goats! - The Chronicles of Jack [entries|archive|friends|userinfo]

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Monty Haul Problem... for Goats! [May. 9th, 2010|02:14 am]
This is my explanation of the famous Monty Haul problem, which is a commonly used exercise in how probability. It's not intuitive, but if once you get it you'll have a much firmer understanding of probability.

Say you're on a game show, and you're at the final round where you get to select a prize. There are three doors to choose from; two of the doors have decrepit old vehicles that you have a hard time hauling away for scrap value. These cars are so old that it looks more like the show's crew is using the contestants as a means of hazardous material disposal. We're talking 1978 here, totally worthless. The other door, however, contains something far more valuable than a junky old car: a magical goat. This goat has the power to attract (human) members of the preferred sex to its owner, kind of like a siberian husky that doesn't need walks and you can milk it.

The message here is: Two lousy prizes, and one greatly superior prize.

The catch is that you don't know what door has what prize, and you can only take home the prize behind one door.

So you pick a door, any door. You have no idea what door has that awesome goat, so you just pick a door at random. You chance of getting that goat (and therefore all the ladies/men) are 1 in 3.

But before your choice is finalized, a twist is introduced. (What devilry!) The host reveals one of the doors that you didn't pick, and there's something behind it that might have been a Volvo before something filled it with pinecones and made a home of it. You're now given the option of switching to the other unrevealed door. What do you do, WHAT DO YOU DO??!?!

Your new situation is this: You've picked a door, and you know where one of the wrong doors is. You need to decide whether or not to switch doors and why.

What is the probability/chance of walking away with that awesome goat if you switch/don't switch.


Switch. Always.

There's a few ways to think of this. First, when the door is revealed and you're given the choice to switch, you're basically choosing between the door you have and the better of the two doors remaining. That's like picking both of the other doors at once! Picking two doors has to be better than picking one door if you don't know which door is the best one. In fact, the other door has twice your current door's chance of containing the magical goat.

The door you currently have was picked at a 1/3 chance of containing a goat. Since there are two bum prizes out there, there's always going to be one to reveal through a door you didn't pick, so that doesn't really tell you anything about the door you have picked. Since you have no other information about your selected door, the probability of it having the magical goat is still sitting at 1/3.

Since probabilities always add up to 1, the rest of the probability has to be where? On the other door, which has 1 - 1/3 = 2/3, which matches with our previous statement about the switched door being twice as likely to have the good prize.

The other more classical solution is to look at each situation. First, assume you picked door A because it doesn't really matter. There is a 1/3, 1/3, 1/3 chance of the magic goat being behind doors A, B, or C respectively.

If the goat is behind door A, the host can reveal either door B or C at 50/50 random to reveal a junker.
If the goat is behind door B, the host reveals door C.
If the goat is behind door C, the host reveals door B.

All those probabilities are:
Goat in A, door B revealed: 1/3 * 1/2 = 1/6
Goat in A, door C revealed: 1/3 * 1/2 = 1/6
Goat in B, door C revealed: 1/3 = 2/6
Goat in C, door B revealed: 1/3 = 2/6

Now let's say Door B is revealed (exactly the same time of thing happens if C is revealed)
Goat in A, door B revealed: 1/6
Goat in C, door B revealed: 2/6

Once again, we see that there's twice the chance that the goat is behind C as behind A. Since A is the door you started with, you should switch to C.

Take a look at the probabilities for door C revealed to see the symmetry.

Finally, one last way to look at this:

If at the start of the show, you were given the option of choosing either door A, or BOTH doors B AND C, which would you choose? You'd always choose B and C, because you're twice as likely to get an awesome goat. Even after the host reveals that one of the two doors you picked has a bum prize, who cares? They can't both contain good prizes, there's only one good prize. The remaining door still carries all the chance of containing the magic goat that the two doors together did because you already know that one of the two doors you picked had to be a loser; that door now contains a 2/3 chance of getting you your magic goat.

From: (Anonymous)
2011-01-29 10:14 pm (UTC)



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