Game Show
You are on a game show and have the option of selecting 1 of 3 doors ... behind one of the doors is a fabulous prize ... behind the other 2 doors are gag gifts. The show host knows which door conceals the good prize.
You select a door. The host opens one of the other doors and shows you a gag gift. He now inquires if you would like to keep the first door you picked, or change to the other unopened door.
Should you keep the door you picked ... or change doors ... or does it matter?
(guys, I searched and did not find this puzzle ... If one of you recognizes it as having been a previous post, please advise and I will delete ... thanks)
You select a door. The host opens one of the other doors and shows you a gag gift. He now inquires if you would like to keep the first door you picked, or change to the other unopened door.
Should you keep the door you picked ... or change doors ... or does it matter?
(guys, I searched and did not find this puzzle ... If one of you recognizes it as having been a previous post, please advise and I will delete ... thanks)
posted by Zaux at
9:30 PM
37 Comments:
This might be a laugh ;)
It's been posted before, but in a different guise. Leave it up and see how it goes.
thanks Chris ... there's so many, it's difficult to know
This question is one of the best
many of great minds have been fooled by it
hi Knightmare ... really glad to see you can now post puzzles ... it's much better than submitting and waiting
eessh.... Monty Hall strikes again
Cam
Yes, our choice is likely to change. Earlier when there were 3 doors, the probability of getting the right door was 33.33%, but now when one wrong door has been revealed and only two doors are left, the probability of getting the right door increases to 50%, which is a good sign. Now all you are left with is to make right decision.
Cam knows Monty Hall! Must be in his late 100s! Just like me.
Check this page on The Straight Dope
http://www.straightdope.com/columns/read/916/on-lets-make-a-deal-you-pick-door-1-monty-opens-door-2-no-prize-do-you-stay-with-door-1-or-switch-to-3
I believe the theory is:
you have a 33% chance of the prize in the door you chose, thus a 66% chance of it being in the 2 doors you didn't choose.
if 1 of the doors you didn't choose is shown to be empty, the other now has 66% chance of being correct against 33% for your original choice.
So logically you change your choice.......but wouldn't you feel a fool if you'd moved away from the winning selection!!
I saw the theory to this on an episode of "Numbers" a US crime series.
regards, Curtis
I dunno...normally, people in show biz will want all money for themselves (cruel cruel people :P ) and they will try to dissuade you from choosing the correct in ways like that... but then, if you think about it too hard, they may think you will think that and use reverse psychology! and then you may think that they will think of using reverse psychology... etc. I hate it when that huge thought process (actually a loop)comes into my brain :P
Moving away from that insane logic:
I have seen it before too, the Indian version though.(meaning an indian game show was used :P )
Yeah there was an Indian game show on that, with 3 door's choice. However that doesnt gives the theory behind it. The english movie "21" has this thing.
I've seen this theory before, and I still don't have any faith in it.
Maths, like statistics, can be manipulated to be misleading.
At the point where you have only 2 doors left, nothing that has gone before has any bearing on your current position: 2 doors, only one is a winner = 50/50 chance of winning.
You cannot manipulate your chances of winning by identifying one of the losing doors prior to this point and then carrying forward the initial probablity ratios into the new scenario because that losing door is no longer relevant once it has been discounted.
You might as well claim that what you had for breakfast that morning can help you to win.
Think about it this way: in this scenario you only ever actually have a choice of 2 doors all along. Whichever one you pick first the host will open one of the other two that is not the winning door (assuming he knows which is which) so your choice is always, right from the start, between one winning door and one losing door (50% chance), and at no point do you ever actually have a choice of three. This means that the odds of the winning door being the one that neither you nor the host has chosen is always 50%.
If the host has no knowledge of which is the winning door then when he opens the first door (one of the two you didn't pick) then 33% of the time that door will be the winner (and presumably the game will end).
In the question we have already got past this point knowing that the (randomly) opened door was a losing door so we need to remove 50% of the occassions when the winning door was one of the two not chosen initially.
This means that instead of a 66% chance of the winning door being the other one it is now 33% (the same as the chosen door).
Therefore we have an equal chance of each of the two remaining doors being the winner (each have 33% of the original probability which equates to 50% probability each in the new scenario).
Ok, now I'm completely confused because I disagree with everything I've said above.
Taking a sample of 99 contestants on this game show, 33 of them will pick the right door first time, 66 of them will pick a losing door, in which case they will win if they change after the host has revealed the other losing one.
I guess it all comes down to the knowledge that the host has of where the winning door is, but I still can't fully grasp the fact that you ultimately only get to choose between 2 doors, but the odss aren't 50/50.
I need more coffee!
I'm going to sit back and enjoy this thread aboard my ROFLCOPTER
ROFL:ROFL:LOL:ROFL:ROFL
______|_____
L / \
LOL=== []\
L \______________\
| |
-------------/
Cam
Is it an air/sea rescue ROFLCOPTER?
I'm in need of some assistance.
Can you see me waving? (that's not waving, it's splashing around trying to keep from drowning in a sea of confusion)
:-)
As Curtis states above, the math wizards say you should switch doors.
There is an excellent explanation on Wikipedia ... search "MOnty Hall Paradox"
Thanks Karl ....
The Straight Dope site suggested by Karl also has some interesting arguments concerning the door switch.
Ver interesting discussion - it's so hard to get my head round the logic of this. Just in case it's of interest, I set this up in Excel as a Monte Carlo simulation - ie setting up a load of random examples (I know they're not really 'random') of choices and decisions, and the percentage 'win' with sticking comes out pretty close to 33% and the percentage 'win' with changing your mind comes out pretty close to 66%. I know the idea of these challenges is to puzzle the problem out rather than to throw trial and error at the problem, but I thought it might be interesting since there seem to be schools of thought in here that the mathematicians' calculations might be flawed. (Mind you, it also depends on whether my hastily cobbled together Monte Carlo simulation is 'right'!).
Torsionwire ...
that's pretty cool ...your simulation verified the gurus theory that switching doors is the best choice.
Hi dual ...
your thought process is nearly identical to mine as I struggled to understand after first reading this conundrum.
I think it best to only consider the probability after getting the
logic right. If you choose the right box, then whichever door is
opened by the host, it will contain a gag prize - swapping would
be a mistake. If you initially choose a gag prize door, the the
host will open the other gag prize, but the remaining door will be
the star prize door - swapping would be the right thing to do.
There's two doors that you could have chosen for the gag prize.
Now for the probabilities. The probability of choosing any door
is 1/3. In particular, the probability of you initially choosing
the star prize door is 1/3. If you swapped then you'd win a gag
prize. The probability of initially choosing a gag prize door
is 2/3. If you swapped, then you'd win the star prize. So swap.
This assumes that the host knows where the prize is. If he
didn't, then there is a 1/3 probabilty that he'd reveal the star
prize - and then you've lost already. But if he happened to open
a gag prize door, what would you do? (I'd swap).
Chris ...
if you haven't already, go check "Monty Hall Paradox" on Wikipedia ... there is an outstanding analysis ...
I'm off shopping for a bit ...
Have a good day!
I know this problem from a long time (at least 15 years, I heard it based on an old popular portuguese tv contest), and it is quite obvious to me (without calculations) that one should switch (based on the assumption that the host always opens a door without the prize).
Why do I find it so intuitive? Well, I imagine the same thing with 100 doors: You pick one, and the host opens 98 without prize. Either you are very lucky and get the prize at the first try (1% chances), or the remaining unopened door has the prize with 99% probability.
(once again, sorry for my crude english; I hope I made my point clear enough)
This post has been removed by the author.
It's a great puzzle, it seems to defy all logic because we all know that a choice of two equally likely outcomes results in a 50/50 probability.
It took me a while to appreciate that, even though it appears to be, it is not in fact equally likely that each of the 2 doors you are left with contains the prize.
The fact that the host knows which is the winning door causes the probability to be skewed, and when he opens a door he is massively increasing the odds of the unopened door that you didn't originally choose being the one with the prize.
Having read the Wikepedia article I feel much better knowing that Nobel Prize Winners have been caught out by this. I'm in good company!! :-)
Pick the other door.
You have a 1/3 chance of picking the right door and a 2/3 chance of picking the wrong door. With the other door being revealed certain elements change. Knowing that the other door is not the prize door you know that there is a 2/3 chance that the other door is it and a 1/3 that the door you currently have is it. The diagram below explains everything in mathimatical terms.
W means wrong R means right for this diagram below
If you stay
R->R: (1/3)(1/3)=1/9
W->W: (2/3)(1/3)=2/9
(1/9)+(2/9)=3/9= 1/3
1/3
If you choose the other door
R->W: (1/3)(2/3)=2/9
W->R: (2/3)(2/3)=4/9
(4/9)+(2/9)=6/9= 2/3
2/3
Message from the Godson
This is so interesting. It seems as if the only explanation worth having is the one that works for you personally. (Just like the best way to make an omelette is the way you make it yourself.) For me, I think I've found my own logic for it. Having chosen a door initially, the chances of it being the 'prize' door is one in three. The chances that either one of the other two doors hides the prize is thus two in three.
Let's imagine the host puts in an intermediate step - he says 'would you like to stick with your original choice, or change to have the keys to both the other two doors?'. This would be a no-brainer - you'd have a two in three chance of winning if you could have 'the other two door keys'. So you change. He then opens one of the two doors that contains a 'gag', and you open the other door.
There was a two in three chance that one of your two doors contained the prize. He only showed you that one of the two doors in your set didn't contain the prize, and so your job is simply to open the remaining door.
Nothing along the way changed the probabilities. You had a two in three chance that your two keys would lead to a prize. So having had one of these two doors opened for you, you still had a two in three chance that the remaining one of your two doors would contain the prize. The fact that he opened the door before asking you if you wanted to change altered nothing in the probabilities. You were still choosing 'my original door' or 'the best of the other two doors'.
Works for me, anyway!
Hi Torsionwire. That was a nice
argument in your second paragraph.
yeah Torsionwire...the way i like to think of it is:
once you pick one of the three doors,what the host is REALLY asking you is "now,do you want to bet that you are right(1 in 3),or do you want to bet that you are wrong(2 in 3)"
I found this hard to believe, so I programmed a test in VBA. You can run this in any MS office program.
The function picks a door... 1 to 3
W will be the winning door
The contestant chooses door A
Monty Hall will display door B, as a loser
Door C will be offered to be swapped for Door A
Door A or Door C will be the wining door.
If door A = W then winnernc (not change) will be incremented
If door C = W then winnerc (changed) will be incremented
I ran it for a million times in 3 seconds
Winnerc got 66.7%
Winnernc got 33.3%
If the contestant changes he will win 66% of the time
IF the contestant sticks to his original choice,
he only wins 33% of the time.
======================================================
Sub MontyHall()
Randomize
h = 3
time1 = Timer
For i = 1 To 1000000
W = pickone(h)
A = pickone(h) 'select a door = S1
B: B = pickone(h) 'open another door, not W and not A
If B = A Then GoTo B
If B = W Then GoTo B
c = 6 - A - B ' change to offer instead of A
'Debug.Print "Winner was "; W
'Debug.Print "First Choice was "; A
'Debug.Print "Show Door, no win"; B
'Debug.Print "Change to offer "; c
If c = W Then Winnerc = Winnerc + 1 'win if change
If A = W Then WinnerNC = WinnerNC + 1 'win if no change
Next i
Debug.Print
Debug.Print "Change doors wins"; Winnerc / i
Debug.Print "If no changes "; WinnerNC / i
Debug.Print i - 1; " Loops"; (Timer - time1); " seconds"
End Sub
===================================
Function pickone(h)
pickone = Int(Rnd() * h + 1)
End Function
===============================
Change doors wins 0.666955333044667
If no changes 0.333043666956333
1000000 Loops 3.359375 seconds
Ragknot ...
that's impressive ...
Shervan, the host always offers the choice of swapping. You really should examine the other posts, most have got it right and have given good explanations.
It is much easier to see that this makes sence if you add more doors. Say there are 9 doors and you pick door number 1. The host shows you door 9, 8, 7, 6, 5, 4, and 3 which have no prize. It would be perfectly logical to pick door number 2. If you pick door 1 you have a 1/9 chance. Pick number 2 and you have a 8/9 chance. It is the same thing for 3 doors as for 9 only it is more confusing.
Message from The GodSon
hi their is chances increase of to win the game. wen their is 3 doors their is 33% of each doors has probablity but now one door has 66% chance so the other door has jackpot
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