Cue PH brainiacs - logic puzzle!

[gasblaster]

You are on a TV game show. The main prize is a car. You have to choose between three doors. Behind one of the doors is the car. Behind the other two doors are goats. You are asked to choose your door. You choose your door, but the host does not open it. The host then opens one of the other doors (that you did not pick) to reveal a goat. The host then gives you one final chance to change your mind about which door you want. Your choice is to stick with your original choice, or to pick the other unopened door. What should you do? And why?




this is taken from an amazing book called 'the curious incident of the dog in the night-time' by mark haddon, vintage, 2004

[mybrainhurts]

Just checked....I'm bally well NOT on a TV game show

[Mikey G]

Stick with your original choice, if you picked the door with the goat the host would have opened it and said bad luck.

well thats my logic anyway

[Mr Clive]

I would say 50% but I remember from last time this was posted that its something else and I couldnt get my head around it.

[axeman30]

This dilemma clearly does not apply to Welsh PH fraternity who would be happy in either case, unless of course there's a 4th door with a sheep behind it.

[scared but happy]

Change your mind.....I remember reading this somewhere before. Its about statisics and laws of averages.

[xiphias]

This is some obscure statistical logic that my tutor said I wouldn't really ever need at adegree level. We did this puzzle but I can't remember the answer.

[boiler]

You should switch doors....

To start with you have a 1 of 3 chance of selecting the right door (and therefore a 2 out of 3 chance of selecting the wrong door).

If you choose a bad door to begin with, the other bad door is uncovered, then you must end up with the right door, and this should happen 2 out of 3 times.

If you right door and change, you will end up with the wrong door 1/3 times.

If you don't change, you keep the original odds.

I think.

[Mrs Trackside]

boiler said:

You should switch doors.... To start with you have a 1 of 3 chance of selecting the right door (and therefore a 2 out of 3 chance of selecting the wrong door). If you choose a bad door to begin with, the other bad door is uncovered, then you must end up with the right door, and this should happen 2 out of 3 times. If you right door and change, you will end up with the wrong door 1/3 times. If you don't change, you keep the original odds. I think.

What?

[mybrainhurts]

axeman30 said:

This dilemma clearly does not apply to Welsh PH fraternity who would be happy in either case, unless of course there's a 4th door with a sheep behind it.

Ooooh...dangerous

Bit of advice....

Beware large hairy men singing for no apparent reason..........

[boiler]

Mrs Trackside said:

What?

OK, I'll try and explain (my logic) again.

If I choose at random to start with 1 out of 3 times I would choose the car (so 33% of the time), but 66% of the time I would choose the goat. Now my the host opens up one of the boxes, revealing a goat. If I have previously selected the goat (which I will have done 66% of the time) and swap doors now, I will have won the car. If I don't swap doors now, then I would win the car 33% of the time.

Of course, you could argue that your first choice is simply irrelevant, and does not affect the second decision. It then becomes a 50:50 chance of winning the car, but only if you choose randomly to swap or not.

[selmer]

And therefore the odds that each door has the car is 50%, so it makes no difference if you stick or switch.

[anhamgrimmar]

but that means theres half a car between both of them! sod that, i'm off with the goat

[zattlebone]

If you don't switch doors you have a one in three chance of winning.
If you do switch doors you have a two in three chance of winning.

It's a well known problem:
www.sover.net/~nichael/puzzles/monty/problem.html

With a less known solution:
www.sover.net/~nichael/puzzles/monty/solution.html

[deeps]

But there is 66.6% chance that you will pick a goat - fact.

You're then left with two doors, the one you picked has a goat at chance 66.6%.

So it's best to change doors if you get the chance.

[selmer]

Plus the host was given a chance to open the other door, but didn't (assuming of course that he knows where the car is).
So I'd probab;y switch.

[moleamol]

zattlebone said:

If you don't switch doors you have a one in three chance of winning. If you do switch doors you have a two in three chance of winning. It's a well known problem: www.sover.net/~nichael/puzzles/monty/problem.html With a less known solution: www.sover.net/~nichael/puzzles/monty/solution.html

That only takes into account the host not having a random choice of which doors he opened. If he did have, there are more outcomes of the game. For example, he never opens door one first, if he did, that little solution would fall flat on its bumhole. As such, the choice between two doors will always be 50:50, one being chosen already cannot possibly negate that.

However, as it does explain that the host knows where the prize is and avoids a door then I shall say yes it does hold and next time I will read the problem properly before coming up with the solution. Damn whisky. *cough*

[joust]

Old old old....

The best option statistically is to change your choice.

If you want to know why, type "Monty Hall Problem" into google and read, and read, and read, and read, and read.

You'll then realise that it makes sod all difference in practice.

If everyone followed the statistical view, then everyone would stick 1,2,3,4,5,6 on their lottery card.

Statistics only describe perfect worlds. The world isn't perfect. If you are ever in that position, go with whatever you feel at the time, and pray you are lucky!

J

[joust]

moleamol said:

With a less known solution: www.sover.net/~nichael/puzzles/monty/solution.html

Ah - the wonders of assuming that what the stats say therefore proves it.

You are falling into the trick that if the numbers 10, 12, 27, 33, 48 and 49 come up on the lottery, should you buy a ticket the next weekend with the same numbers on it? Logic says no, status say "yes" (or, to put it in LB terms - "computer says no...."

Your "solution" that you quote assumes that you would choose the same thing more than once (that's how the stats work in that example), that's how it proves its better to switch.

It is indeed statistically correct, as it assumes the person sticking the prize in the boxes in the first place is a perfect random machine. However, people are not perfect random machines, and so the answer is, for normal human beings, go with the one you feel "lucky" on.

J

>> Edited by joust on Friday 9th December 00:16

[JonRB]

Interesting. Logic would say that you had a 33% chance of a right answer the first time. After one door has been removed, you then have a 50% chance that your choice was correct. The odds of your choice being correct have not actually changed - you have merely eliminated one answer.

It's like the "flipping the coin" problem. If you flip a coin, the odds of getting heads is 50%. If you get heads, what is the odds of flipping heads again? It's 50%. If you get heads, what are the odds of flipping heads a third time? 50% still - the probability is independant of previous actions. However, the odds of flipping 3 heads in a row is not 50%, of course.

So returning to the original problem, changing your mind and choosing the other door is an independant probability. You have a 50% chance of getting the right door regardless of your previous choice, so it makes little difference whether you stay with your original choice or choose the other as you are either right or wrong - the probability of success is independant of the previous choice.

>> Edited by JonRB on Friday 9th December 08:40