[Misc] Probability problem

[Coldeanseagull]

I don't care, it's not me who would be(or not) getting a prize. Stupid game anyways, I'm not playing anymore, I want to go to work, I'm bored, my hands are cold from going into the shady garden, so meh

 

[Diablo]

If you switch I believe probability of getting prize to be 2/3rds (you get it if you pick either of non prize doors) + 1/2 of 1/3rd (if you picked door with prize in first place).

Seems high but is chance of winning 5/6

 

[dazzer6666]

Ok, so using three permutations - prize behind 1, 2 & 3 and choosing Door 1 first, options work out as follows............

Door 1 Door 2 Door 3 Door 1 Switch
Empty Empty Prize Lose Win
Empty Prize Empty Lose Win
Prize Empty Empty Win Lose

So by sticking to door 1, would win 1/3 attempts and by switching would win 2/3 attempts, so 66.6% probability of winning by switching vs 50% of winning by not switching ?

 

[beorhthelm]

i dont see why the change to 50/50 means you should change your mind.

 

[PILTDOWN MAN]

This is the classic error.

The way I explain this to people is to ask whether they would rather take a choice about a one in three chance or a one in two. Obviously people would rather a 50/50 shot but this puzzle is interesting because it kids you into thinking the odds are 50/50 for both boxes left (which is true) BUT when you picked it you had a 33% chance. So you switch.

The removal of one box and the relatively small change in implied odds makes it tricky to change your mind. No law imagine there were 100 boxes and you picked one. 98 incorrect boxes were removed. Would you swap? Of course you would because it is obviously now a 50/50 shot whereas on first guess you had 1% chance of being right. It is all about the information available when you make the decision.

I now get it ha ha silly me. Here's me who studied high level analytical maths when studying aeronautics , but that was 36 years ago

 

[sparkie]

In actual fact it doesn't matter whether Monty knows or not. We are where we are, with one door having been revealed not to have the prize. Monty may have opened the door with the prize and the game would have been over, but he didn't, and here we are with two closed doors, one of which hides the prize

Oh, so Monty wasn't deliberately picking an empty door to prolong the game after the ad break... he was making an arbitrary choice , so might as well have done it randomly...?

 

[Berty23]

i dont see why the change to 50/50 means you should change your mind.

Because when you made your initial choice it was a 33% chance of winning. Given that there is always going to be two wrong answers with the initial choice then one box can always be removed but it can’t be the box you chose. So the chance of it being the box you chose was always going to be 50/50 once one other was removed but when you picked it it was 33% chance. It messes with your mind but when you assume you had 100 boxes to choose from in the first place and 98 incorrect boxes are removed then it is more obvious why you should switch.

 

[Diablo]

i dont see why the change to 50/50 means you should change your mind.

Monty new where prize was so had to pick empty one. If you also had picked empty one means other one def has prize so swap. If you had picked one with prize in it would make no difference whether you changed or not.

 

[Berty23]

I guess another way of looking at it would be to switch it.

So if you were asked to pick a box but then told you could swap it to the other two boxes then would you? Surely the answer is yes, so that is effectively what you are doing when one of the others is removed.

 

[beorhthelm]

Because when you made your initial choice it was a 33% chance of winning. Given that there is always going to be two wrong answers with the initial choice then one box can always be removed but it can’t be the box you chose. So the chance of it being the box you chose was always going to be 50/50 once one other was removed but when you picked it it was 33% chance. It messes with your mind but when you assume you had 100 boxes to choose from in the first place and 98 incorrect boxes are removed then it is more obvious why you should switch.

still not obvious. it implies switching would always reveal the prize. two seperate events, first 1/3 (or 1/100), the second 1/2.

 

[father_and_son]

In actual fact it doesn't matter whether Monty knows or not. We are where we are, with one door having been revealed not to have the prize. Monty may have opened the door with the prize and the game would have been over, but he didn't, and here we are with two closed doors, one of which hides the prize

The gameshow gets cancelled if Monty starts revealing the prizes, so the floor manager obviously tells him which door to reveal.

But yes. The maths doesn't require Monty to know, just that if an incorrect door is shown, you are better swapping.

If X is the prize and Y is the choice made...

1. 2. 3.
X
Y

(Monty reveals 2 or 3, lose if swap)

1. 2. 3.
X
....Y

(Monty reveals 3, win if swap)


1. 2. 3.
X
........Y

(Monty reveals 2, win if swap)

This repeats if you put the X under 2 or under 3 (but that's a lot of typing, so won't bother).

If you choose the right one first time you are better sticking with that choice (obviously). If you didn't chose the right one you should always swap because then you will always win.

Your chances of choosing the correct one first time is 1/3 so the chances you were wrong first time is 2/3. If you always win when you were wrong and swap then swapping increases your chance of winning from 1in3 to 2in3.



If you can't see why this works, draw out every possible combination with 4 doors (there are 16 possibilities and your initial chances are 1in4) and Monty always showing 2 empty ones. Then the swing in probability is even higher.


If you are given more information, it is always better to reconsider your choices (true here and in all the political threads!)

 

[beorhthelm]

Monty new where prize was so had to pick empty one. If you also had picked empty one means other one def has prize so swap. If you had picked one with prize in it would make no difference whether you changed or not.

ah so we have to assume a rigged game.

 

[father_and_son]

ah so we have to assume a rigged game.

No. The prize cannot be moved, that would rig the game.

 

[Diablo]

The gameshow gets cancelled if Monty starts revealing the prizes, so the floor manager obviously tells him which door to reveal.

But yes. The maths doesn't require Monty to know, just that if an incorrect door is shown, you are better swapping.

If X is the prize and Y is the choice made...

1. 2. 3.
X
Y

(Monty reveals 2 or 3, lose if swap)

1. 2. 3.
X
Y

(Monty reveals 3, win if swap)


1. 2. 3.
X
Y

(Monty reveals 2, win if swap)

This repeats if you put the X under 2 or under 3 (but that's a lot of typing, so won't bother).

If you choose the right one first time you are better sticking with that choice (obviously). If you didn't chose the right one you should always swap because then you will always win.

Your chances of choosing the correct one first time is 1/3 so the chances you were wrong first time is 2/3. If you always win when you were wrong and swap then swapping increases your chance of winning from 1in3 to 2in3.



If you can't see why this works, draw out every possible combination with 4 doors (there are 16 possibilities and your initial chances are 1in4) and Monty always showing 2 empty ones. Then the swing in probability is even higher.


If you are given more information, it is always better to reconsider your choices (true here and in all the political threads!)

Nearly agree, you always win when you`re wrong so 2/3 but also if you`re right first time you have a 50 % chance of the 1/3 surely ?

 

[Mtoto]

i dont see why the change to 50/50 means you should change your mind.

Because your original choice is still 2-1 against.

It's a classic problem in probability, and I read many years ago that when it was first brought to wider attention - I think in the puzzles column of Scientific American, but could be wrong - it generated many weeks of heated debate in the letters column in which prominent mathematicians took lumps out of each other.

Personally think Berty23's explanation is best, esp if you crank it up from 100 doors to a million. If you picked one door out of a million and Monty opened 999,998 others to reveal nothing behind them, would you still stick with your first choice or jump to the one which he overlooked for some reason?

 

[Berty23]

still not obvious. it implies switching would always reveal the prize. two seperate events, first 1/3 (or 1/100), the second 1/2.

Switching will always give you a 50% chance. Not switching gives you a 1 in however boxes were there chance.

With the 100 box example.

You pick one so have a 1% chance of winning. If 98 incorrect were to be removed leaving either the correct one or another incorrect one (if you have the correct one) it is the same as you picking one and if you were playing against someone else them getting the other 99 boxes. The chance of your box being correct in the first place has not changed but it is now appears to be a 50/50 when in reality it isn’t now you have more information.

 

[beorhthelm]

...
If you choose the right one first time you are better sticking with that choice (obviously). If you didn't chose the right one you should always swap because then you will always win.

well obviously. but how do i know if i got the right one or not?

 

[beorhthelm]

Personally think Berty23's explanation is best, esp if you crank it up from 100 doors to a million. If you picked one door out of a million and Monty opened 999,998 others to reveal nothing behind them, would you still stick with your first choice or jump to the one which he overlooked for some reason?

in a fair game, if the other 999998 box had not revealed the prize, the final two are still 50/50. the only "some reason" one has been overlooked is a rigged game, so yeah, switch.ludic fallacy and that. i can see why this would exercise the Scientific American readers, a clash of probability versus logic.

 

[pb21]

Switching will always give you a 50% chance. Not switching gives you a 1 in however boxes were there chance.

With the 100 box example.

You pick one so have a 1% chance of winning. If 98 incorrect were to be removed leaving either the correct one or another incorrect one (if you have the correct one) it is the same as you picking one and if you were playing against someone else them getting the other 99 boxes. The chance of your box being correct in the first place has not changed but it is now appears to be a 50/50 when in reality it isn’t now you have more information.

In the 100 example, the original box you selected has a 1% chance of being correct, so there is a 99% chance it is in one of the other 99 boxes.

So if you remove 98 empty boxes from the 99, there is a 99% chance it is in the remaining box, isn't it?

Edit: assuming empty boxes were removed knowing they were empty, or regardless

 

[father_and_son]

Nearly agree you always win when you`re wrong 2/3 but also if you`re right first time you have a 50 % chance of the 1/3 surely ?

If you are right first time and you choose to stick you win.
If you are right first time and you choose to swap you lose.
If you are wrong first time and choose to stick you lose.
If you are wrong first time and you choose to swap you win.

These are the only 4 outcomes possible. Agree?

Compare the 'stick' and 'swap' strategies...
Stick
To win you MUST be right in the first place... 1in3 chance.
If you pick wrong first time and stick then you lose... 2in3 chance.


Swap
To win you must be WRONG in the first place... 2in3 chance.
If you pick right first time and swap then you lose... 1in3 chance.


Clearly the 'swap' strategy is best. It doesn't guarantee a win because you might have picked right in the first place. But chances are higher you picked wrong first time and with the removal of all the other wrong options, you can only be left with the right option... So swap.