## Probability Query

For the discussion of math. Duh.

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Hemmers
Posts: 117
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### Probability Query

Happy New Year!

This question was posed in the pub and I would appreciate the opinion of the esteemed denizens of this board.

You are in a game show and are faced with 3 doors. Behind two doors are goats, behind the third is a car.
We shall assume that you are not from a nation where Goat is considered a delicacy and that you wish to win the car.

You are given a choice (doors A, B and C).
After making that choice, the host opens a door behind which is a goat (but which IS NOT the door you chose)
You now have the option to change your choice.

Question: Is it in your interest to change?

The argument for "Yes" is that in the initial decision you have a 1/3 chance of winning a car. With one of the goats exposed you have a 1/2 chance (i.e. better chance) of winning a car, so you should re-choose.

I argue that it makes no difference whether you change your vote and furthermore that the whole scenario is flawed. By changing your vote or not changing your vote you are still making a choice. If you choose Door A and a goat is revealed behind door C, then in the second round you are still making a decision for Door A or B. The fact that you chose Door A in the first round (i.e. you are "sticking with" or "changing" your decision) is irrelevant. You are making an independent decision and it's a 50/50 chance.

The first round of voting is actually pointless because the terms of the scenario state that the host will always open a door with a goat behind it, and it won't be the door that you chose. i.e. if you chose the car, he will open either of the remaining doors. If you chose a goat, he will reveal the other goat.
It comes down to a 50/50 guess with a goat behind one door and a car behind the other.

Xanthir
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### Re: Probability Query

This is the Monte Hall problem. Check the links at the bottom of this sticky post.

Short answer, you're wrong. It's better to switch, as that will let you win 2/3rds of the time. This has been discussed multiple times, though, so if you disagree, read the threads linked in the post I linked above. Your questions will be answered.
(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))

Qaanol
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### Re: Probability Query

Hemmers wrote:After making that choice, the host opens a door behind which is a goat (but which IS NOT the door you chose)

Did you, before the game began, know for a fact that the host was definitely going to do this? It matters.
wee free kings

lamemaar
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### Re: Probability Query

It is essential that the host knows what is behind each door. He deliberately takes a door that shows a goat. The probability for the door that you initially chose remains 1/3, and the probabilities for the two other doors together remains 2/3. Now that one of these two other doors turns out to have probability 0, it follows that the remaining door has probability 2/3. Switching doubles your chance!

I know how difficult it is to make this clear, so let me try to add some more explanation.
This scenario should not be confused with one in which the host opens one of the two doors at random. If that were the case, he would offer new information that influenced your chance. Remember, immediately after your initial choice you know that there is a chance of 2/3 that the car is behind one of the two doors you have not chosen. You also know that behind at least one of these two doors is a goat. By telling you which door this is (leaving deliberately the door closed that reveals a car, if any), the host does not influence your chance of 1/3.

If you are not yet convinced, imagine a scenario with, say, ten doors instead of three. In that case the chance that your door is the right one is only 10%, so it is far more likely that the car is behind one of the nine other doors. Then you know that at least 8 of these 9 other doors will show a goat. Revealing which 8 doors these are does not increase your chance of 10%: each time the host opens one of these 8 doors, the chances of those of these nine doors not yet opened increase, but yours does not. If these 8 doors have been opened, the there is a chance of 9/10 that the host deliberately skipped the door that he has not opened because it would have shown the car, so that door is very suspect. There is still a chance of 10% that this door will show a goat and that your chosen door will show a car, but I hope you agree that it would be wise to switch if you were given the opportunity.

t1mm01994
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Location: San Francisco.. Wait up, I'll tell you some tales!

### Re: Probability Query

Qaanol wrote:
Hemmers wrote:After making that choice, the host opens a door behind which is a goat (but which IS NOT the door you chose)

Did you, before the game began, know for a fact that the host was definitely going to do this? It matters.

Why on earth would that matter, O great Qaanol?

bobbeathome
Posts: 27
Joined: Sat Aug 01, 2009 12:14 am UTC

### Re: Probability Query

t1mm01994 wrote:
Qaanol wrote:
Hemmers wrote:After making that choice, the host opens a door behind which is a goat (but which IS NOT the door you chose)

Did you, before the game began, know for a fact that the host was definitely going to do this? It matters.

Why on earth would that matter, O great Qaanol?

I think it does matter... If the host has the choice of opening a door and giving you a second chance, or just ending the game, then the host has a strategy that brings your chances of winning the car back to 1/2 no matter what your strategy. At least, I seem to recall this.

I'll try to remember how exactly the host's strategy goes.....

Edit: I had no idea what I was talking about, but see Quaanol below.
Last edited by bobbeathome on Sat Jan 01, 2011 10:05 pm UTC, edited 1 time in total.

Qaanol
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### Re: Probability Query

t1mm01994 wrote:
Qaanol wrote:
Hemmers wrote:After making that choice, the host opens a door behind which is a goat (but which IS NOT the door you chose)

Did you, before the game began, know for a fact that the host was definitely going to do this? It matters.

Why on earth would that matter, O great Qaanol?

Suppose instead that the host uses this procedure: after you pick your first door, then if it is a goat he immediately opens it and gives it to you. However, if your first pick is the car, then he opens one of the other two doors, revealing a goat, and offers you the option to switch to the third door. If that is how the host operates, then given that you are in the situation described—where you picked a door and the host then revealed a goat—it follows that switching is a guaranteed way to go from having a car to having a goat. So we need to know what rules the host is following.
wee free kings

t1mm01994
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Location: San Francisco.. Wait up, I'll tell you some tales!

### Re: Probability Query

Qaanol wrote:
t1mm01994 wrote:-snip-
Why on earth would that matter, O great Qaanol?

Suppose instead that the host uses this procedure: after you pick your first door, then if it is a goat he immediately opens it and gives it to you. However, if your first pick is the car, then he opens one of the other two doors, revealing a goat, and offers you the option to switch to the third door. If that is how the host operates, then given that you are in the situation described—where you picked a door and the host then revealed a goat—it follows that switching is a guaranteed way to go from having a car to having a goat. So we need to know what rules the host is following.

Ah, like that.. I understand now

314man
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Location: Ontario

### Re: Probability Query

For anyone that needs help to visualize this should take a more extreme case to see how it works.

Say there are 100 doors, with only 1 door with a car behind it and the rest are goats. When you pick a door, the host will open 98 doors with a goat behind it. There are two doors left, should you switch? Of course since 99/100 you picked a door with a goat.

I find that using the extreme case helps people understand the logic behind getting the correct probability really easily.

Kirby
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Joined: Tue Mar 30, 2010 11:08 pm UTC

### Re: Probability Query

Or, perhaps, think of it like this:

If you initially pick the goat, but switch, in all cases you will get the car (the other goat is revealed, leaving the car as the other door).
If you initially pick the car, but switch, in all cases you will get the goat.

There is a 2/3 chance of getting the goat initially, and a 1/3 chance of getting the car initially. Applying the observations from above, switching will get you the car with 2/3 probability and a goat with 1/3 probability.

If you choose to stay, however, then what you started with is what you get - a car with 1/3 probability and a goat with 2/3 probability. Switching improves your odds of getting the car.

skullturf
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### Re: Probability Query

I know this topic has been discussed thoroughly elsewhere, but I also think it may be helpful to point out that probability questions, especially in scenarios like these, are "really" just about proportions.

Questions such as this essentially have the form, "In what proportion of situations where we 'do this' will the contestant end up with the car by switching?"

Hence the importance of precisely defining what 'do this' means. (E.g. did the host just happen to open a door with a goat, or does he deliberately open an unpicked goat-door every time.)

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