Thread: Stick or swap?

If the empty door that is opened after you make your choice was opened by chance then it doesn't make any difference if you swap.

However, if it is opened intentionally by someone who knows for sure there is nothing behind it then you should swap your choice to double your chance of winning from 33% to 66%.

Originally Posted by Looper

If the empty door that is opened after you make your choice was opened by chance then it doesn't make any difference if you swap.

However, if it is opened intentionally by someone who knows for sure there is nothing behind it then you should swap your choice to double your chance of winning from 33% to 66%.

I know now, but I didnt know the answer beforehand.
But I am glad to read that I wasnt the only one who initially didnt know, as can be read on the Wikipedia link for 'Monty Hall problem'.

Originally Posted by withnallstoke

Originally Posted by rawlins

Have you tried my other game yet?.. The PRISM tag game.

Link?

http://www.ilovecuttingoffheadsforal...ngnailbomb.com

That should see you right fella.

Mathematically:



Simply:

^If you are more interested in the goat then it does not matter if you stick or change since your chance of winning will increase from 66% to 100% after Monty's clue!

Originally Posted by quimbian corholla

Originally Posted by rawlins

^ Wrong on both counts there. 100 boxes.... You choose a box.... 98 are opened leaving your box and one other...

How is it wrong? I'm not trying to pick a TD squabble but I'm having trouble braining this, late on a Sunday night.

100 boxes minus my box = 99 boxes. If the other 98 remaining boxes are opened only my box is left unopened. Where does the other box come from?

100-98=2

2 closed boxes remain.... Those 2 consist of the box you chose and one other.

If you still don't get it I would have to be in front of you showing you with lego bricks or something....


Getting back to the original problem.... I should have stated that the box not containg the gold that is opened is opened by a person who knows which box the prize is in. It is not a random act. This is important in setting the odds.

If it was a random act and the prize was not revealed then the odds on the remaining 2 boxes would be 50/50 and it wouldn't matter if you swapped or not....

I only just found this out because it was explained more in the next chapter of the book I am reading.

Edit... And is exactly how Looper explained it at the top of this page.

Originally Posted by Looper

^If you are more interested in the goat then it does not matter if you stick or change since your chance of winning will increase from 66% to 100% after Monty's clue!

That's a fair point.

Originally Posted by Looper

^If you are more interested in the goat then it does not matter if you stick or change since your chance of winning will increase from 66% to 100% after Monty's clue!

But only if he takes the car away?

In which case why bother playing?... Monty might as well just say "Have a fukin goat"

Originally Posted by rawlins

Monty might as well just say "Have a fukin goat"

And that would be a much better game.

Originally Posted by rawlins

If you still don't get it I would have to be in front of you showing you with lego bricks or something..

Originally Posted by quimbian corholla

Last edited by quimbian corholla : Yesterday at 11:33 PM. Reason: I am stoopid

Thanks, I got it last night, that's why I went back to edit my post but didn't because I thought somebody might have replied to it. Don't know why I had such a brain fart, alcohol may have been involved..

Wheter you sick or swap you still have a 50% chance, and you're already on a roll.
I say stick.

^ Doh!

^^No, before Monty opens the empty door there is a 33% you are right and a 66% chance that the goat is behind one of the other doors.

After Monty opens the empty door for a clue those probabilities are still true but there is now only one door to choose for the 66% option so you should go for that one.

This was a scene in the film '21' with Kevin Spacey. I didn't know the theory had a name. Thanks.

Originally Posted by rawlins

If it was a random act and the prize was not revealed then the odds on the remaining 2 boxes would be 50/50 and it wouldn't matter if you swapped or not....

no
if one unopenened choice was randomly taken away from the 2 then the remaining two would have a 33 % chance each.
and the removed box 33%.

^ no...

what you are saying is true and what I said is true... We said different things. Read again please...

I think the way Iain McEwan uses the Monty Hall problem in the novel Sweet Tooth is flawed.

It is a while since I read it but I think his writer protagonist writes a short story trying to use the problem in a dramatic fiction. In the sub-story a husband is trying to guess which hotel room door out of 3 his wife and her lover are behind. He chooses a door and is about to bang on the door when an Indian couple emerges from one of the other two doors. In the story the husband decides to change his choice using Monty Hall probability.

The novelist's GF (a mathematician) then informs him that this is not a valid use of Monty Hall probability since the event is random and not predetermined. She advises changing the story so that a maid opens one of the doors announcing that the room is empty, so it becomes a Monty Hall based predetermined revelation.

This correction would be valid if it was not known which of the 3 doors the Indian couple would emerge from and they could emerge from the door the husband had chosen but in the short story the Indian couple emerges from one of the 2 doors the husband did not choose so the Monty Hall probability holds and the mathematician girlfriend's analysis is wrong and her correction is not valid.

Were the Indians married?

They had a child with a hare-lip.... Whats the chances of that?

Looper... In the correction to the story the Indian couple are taken out of it and the maid pre-announces that she is going to clean one of the 2 empty rooms.... She then enters the empty room which is not the one that the husband is standing in front of ready to kick the door in.

So.... It is the same Monty scenario isn't it?

I don't think the correction is required. The novelist character is effectively Monty Hall, he decides that the Indian couple emerges from one of the 2 unselected rooms so the husband is given a probability decision advantage which he takes by switching selection. What difference does it make if the novelist instead uses a maid as the dramatic device? The result is the same.

The Indian Couple and the Maid announcing the empty room both give the husband the same advantage as long as the room in question is not the one he has already chosen to enter. If that is the case then Iain McEwan's mathematician is wrong.

I can't cut and paste the paragraph where It is explained but it is because the Indian couples emergence being random.... The maid is not choosing one of the other two rooms at random.

Ok... The novelist decides so it isnt random.... But the scenario of the story to make it believable as a story would have the couple emerging being random.