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3 boxes

You are playing a game. A person (game host) shows you 3 boxes. 1 box contains $100, 2 others are empty. You need to choose one of the boxes. If the box you chosen contains money you win.

Suppose you make a guess and point at some box. The game host instead of opening the box you just pointed at opens some other box and shows you that that box is empty. Then he asks "I'm going to give you a last chance to change your decision" So you can either stand by your original choise and point at the same box or choose the other one. What gives you better chances? (e.g. what are the chances you win if you choose the box you originally selected vs. the other box?)

Peter
Tuesday, July 15, 2003

That's an easy one! The probability that the first box is the $100 box is 1/3 (there are three boxes); just like the other two boxes.
After removing one box, the initial chosen box still has a 1/3 chance of being NEO (the one). Since there is a 1/1 chance that the $100 is among the boxes, the other box has a 2/3 chance.
Switching will be a good idea.

Bart Willems
Tuesday, July 15, 2003

This is the same os the Monty Hall question. I disagree with the previous poster's response (2/3) as will as most other that have respond to that problem.

Here is my response to the Monty Hall problem (3 doors instead of 3 boxes, but the same idea)...

The odds are 50/50. Period!.

I've seen a lot of answers on this problem all over the internet, but they all failed to observe one fact: The actual choice of the doors is only made when there are two doors from which to choose. All of the other stuff (i.e. starting with 3 doors, Monty opens a losing door, etc) is just glitz to make the show fun to watch.

In the end, when the final decision is made the player will always choose one and subsequently open only one of two remaining doors.

It doesn't matter how many doors you start with,
It doesn't matter if Monty opens the doors, closes the doors, or goes to commercial,
It doesn't matter if you got to pick one, two or all of the doors.

The actual selection isn't final until you settle on which door is to be opened. At that time, you have only two doors from which to open. You can select one, then the other, and back to the original one but until you say "open door 'x' ", you haven't made the final decision.

This is somewhat analagous to having a pocket full of coins. You could flip two coins or ten or a hundred, but in the end you decide to flip only one,  thus the odds of landing on heads (or tails) is 50%.

Or another way to look at it, at the toss of a coin, I could call "heads", then "tails", then back to "heads". This does not change my odds.

Or, I could flip two coins knowing that one coin would be removed after the flip (by Monty?), and my odds of landing on my choice remains at 50%.

Steve B.
Wednesday, July 16, 2003

Ho Hum.

This is indeed the same as the Monty Hall problem. I refer you tto the discussion

http://discuss.fogcreek.com/techinterview/default.asp?cmd=show&ixPost=1260&ixReplies=10

for why switching is better, giving you a 2/3 chance of winning. Bart, your analysis is good. Steve, I think you are falling into the trap of believing that because there are two choices, each must be as likely as the other. Not the case here.

David Clayworth
Wednesday, July 16, 2003

David, then explain to me why one random choice has different odds than another random choice when your pool of available choices are the same.

The player gets to pick one door, and only one door, for the entire game. The player only has a pool of 2 doors to choose from. Being given the opportunity to change their minds doesn't change the odds.

I would agree with the 2/3 response if, and only if, you had a chance to pick a door and then get another choice if you lost.

But the player only gets ONE choice, how can their odds be two out of anything?

Using your scenario, the player would get 100% odds if they switched the first time (2/3) then switched back (3/3) before the door was opened.

I'll even go this far...

The player doesn't even have to be present for the entire intro presentation (pick from three doors, Monty takes away one, etc). The player could arrive when there are only two doors left and then make the final decision. 1 out of 2 equals fifty percent.

Everything else is smoke and mirrors.

Call it ho-hum, but going into the game you know the following:

You start with 3 doors, but your final decision will be made from 2 doors. Switching doors has no bearing on the odds. You ultimatley get one choice, which is finalized when the door is open.

Suppose you started with 50 doors, and Monty keeps taking the losing doors away until the only thing left is 1 winning door and 1 losing door. Then you make a choice from those 2. The odds are still 50%. The player doesn't get to chose another door and keep going if they lost.

Steve B.
Wednesday, July 16, 2003

Another way to look at all this...

Monty says "I'm going to start with 1000 doors and keep removing the losing doors until you only have 2 doors left, one winning and one losing"

He goes on to say "You can keep picking doors while I keep removing doors. However, your selection won't count until there are only two doors left, at that time you must tell me your choice"

What would the odds be? Answer... 50%.

Same scenario as mentioned, except for the number of doors/boxes.

Steve B.
Wednesday, July 16, 2003

Did you read the previous discussion, Steve?

Once Monty has opened a door, then switching will get you the prize whenever you chose the wrong door first time (2/3 chance) and not switching will get you the prize only if you chose the right door first time (1/3).

The scenario where the chooser is not present until a door has been opened is equivalent only if the chooser knows which was the chosen door, in which case the above analysis applies.

David Clayworth
Wednesday, July 16, 2003

The real question here... have you read my posts at all?

I did read the previous discussion, including the link to the C+ program that "proved", the 2/3 solution. They are all flawed because they fail to account for one thing... the only selection that counts is the last one, when there is only 2 available choices.

One choice and 2 options equal 50% odds.

Giving someone the chance to change their mind does nothing to alter the odds. In fact, even in the beginning,the odds are NOT 1 in 3 because, even if the winning door is selected the player is not allowed to see the results, so nothing counts at this point. The rules of the game stipulate that another door will be removed and the player will then make their final decision. No final decision was made when there were 3 doors, only when there are 2.

One last time... changing your mind does not change the odds. You only have two doors to choose from. The third door will always be removed, thus not part of the problem.

Steve B.
Wednesday, July 16, 2003

David... one other thing I noticed in your post, thus furthering my cause...

You state that you get another choice if you pick a losing door the first time around. If that was teh case, you would be correct (2/3)

However, you get to change your mind and make a second choice even if the first selection was the winner.

Again, the only thing that counts is the final selection when there are 2 doors.

Steve B.
Wednesday, July 16, 2003

I read your posts in detail. You're not the first person to argue like this. Read the web sites I linked to.

I think your main problem is that you are assuming that because there are two doors, they are equally likely. Not true in this case.

David Clayworth
Wednesday, July 16, 2003

Steve, suppose that the game host did not give you the chance to switch--he just opens the box you chose. I think we would both agree that the probability of winning in this case is 1/3.

Then suppose that the host is going to show you one of the non-winning boxes you did not choose, but IS NOT going to let you switch. After your first choice he just opens you one of the non-winning boxes, then opens the box you chose. Again, I think you would agree that the probability of winning is 1/3.

Then suppose that you get to play the game WITH the choice to switch. But before you go out to play it you resolve to stick with your first choice, no matter what. So you make your first choice, and the host opens a non-winning box, and you announce that you will stick with your choice.

What is the essential difference between the second and third scenarios? In the second scenario you pick a box and do not switch, while in the third scenario you pick a box and... do not switch.

I think that the same thing happens in either case.

Do you claim that the act of choosing itself affects hte location of the prize? The boxes were packaged before you ever saw them or made and choices.

If you agree that the third scenario actually has a probability of 1/3, what do you think is the significance of making your decision before hte game as opposed to on the spot?

Do you claim that the presence of a choice can affect your winning, even if the choice itself leads to no action? Philosophers and theologians have been working unsiccessfully for milennia trying to prove the existence of free will. Imagine if simply running this experiment could prove that WILL affects the outcome while ACTIONS are unchanged!

If you doubt us, go ahead and run the game yourself a couple dozen times. It's easy and more productive than Internet debate. Who knows? You could put thousands of philosophers out of work!

Peter Meilstrup
Thursday, July 17, 2003

Steve, I think it becomes more obvious when there are more doors, as you said.

Suppose there are 1,000,000 doors.
You pick one door.
After you pick a door, Monty closes 999,998 other doors, all the other doors except for the right one (unless your first pick is the right one)

Clearly, it's better to switch.  Unless you got the first pick right (1 in a million), you'll win by switching.

luv2puz
Thursday, July 17, 2003

well, that seems to be very confusing..I have only one point to say..
When u take the probability of the first door pick as fixed and compare it to the probability of the remaining door ..
what do u assume???
u assume that the sum of the first door probability and the remaining dorr probability is 1.
but in doing so u lose out a very important point..
what about the doors which were opened for u??????

shouldn't u divide the probability of these doors amogst the two remaining???

the whole catch lies here..i think

U just can not say that probability of the first door which was very less initially ..will remain as it is when u open all doors except one u picked plus one u can switch to..

look at this...
think u have only two doors.one leads to prize and other to nowhere ..u chose one..now if i ask u to switch ..woould u switch??
the answer can be yes or no..since its 50-50 probability..

the same is happenening here..just the point everyone is missing out is to divide the probability of the opened doors amongst the remaining..
anyway i can not explain more than this....!

amit
Sunday, July 20, 2003

Sampling without replacement.

You have a jar and 3 marbles in it. 2 black and 1 white. What is the probability of picking white? -> 1/3

A crow comes and takes one black marble away. What is the probability of picking white? -> 1/2

Lenko Donchev
Monday, July 21, 2003

This is a fascinating problem, isn't it? Something about it seems to challenge all our preconceptions about how probability works.

I'm just going to repost some websites that discuss this in more detail...
http://mathforum.org/dr.math/faq/faq.monty.hall.html
http://www.sjsu.edu/faculty/watkins/mhall.htm
http://www.jimloy.com/puzz/monty.htm

Lenko:

Your analysis is not equivalent, because it assumes I choose my marble after the crow has taken one. Suppose I pick a marble. Then a crow comes and takes a black marble. There are two marbles left, one in my hand and one in the jar, one back and one white. Whatever the crow has done there will still be a white marble in my hand 1/3 of the time.  The marble left in the jar is always the opposite colour, so I can get a white marble 2/3 of the time by switching with the marble in the jar. If you have three marbles I suggest you try it.

Remember, just because there are two choices does not mean they are equally likely.

amit, the probability that you chose the right door first time does not change because a door is opened later. Your first guess will always be right 1/3 of the time, and wrong 2/3. Monty opening the door is a way of allowing you to make a second bet, betting whether your first guess was right or wrong.

David Clayworth
Monday, July 21, 2003

let me try to take a stab at why i think 2/3 (and hence switching) is the right answer.

say you have a regular pack of cards (52). you win if you pick the ace of spades. you choose 1 card (your set, prob = 1/52 of it being the ace of spades).

i put the other 51 cards (my set) in 2 sets: 50 and 1.

now, the probability that the ace of spades lies in my set is obviously 51/52. the fact that i have kept them in sets of 50 and 1 is irrelevant. agreed?

however, i have not mentioned how i have placed the 50 cards in my set. i could have placed them face up...meaning you can clearly see what they are. this does not take away from the fact that the probability of the ace of spades lying in these 51 cards is 51/52. thus, if 50 cards are face up, and you can see no ace of spades in them, it doesnt alter the fact that the probability of finding the ace of spades in my set is 51/52.
hence, the only card left face down among the 51 must have a probability of 51/52 of being the ace of spades.

Amit
Monday, July 21, 2003

David,

you are rigth. I just wanted to fuel the fire :)

Lenko Donchev
Tuesday, July 22, 2003

The problem with most of these posts is the commets that sez "what if this, or what if that".

The rules do not chage during the game. Start with three doors, one is taken away, you then have 2. You can change your mind, but won't know the winning door until the very end.

So what if a door is taken away (and "supposedly" the odds change to 2/3), do the odds revert back to 1/3 if the door is closed again? Of course not!!!

And considering the scenario of 1,000,000 doors and 999,998 are closed. Using the other posters' theory, the the odds would be 999,999 out of 1,000,000, which obviously they aren't.

In the end, the player has a choice of 1 door from an available pool of 2 doors. Changing your mind or opening and closing other doors don't have any bearing on the outcome, regardless of when the doors where opened, or when the person changed their minds. Two doors, one choice... 50/50.... 1/2.

Bottom line: the player makes his final (and thus,  the only) selection when there are two doors left.

Flip a coin while changing your mind and take other coins off the table do nothing to change one fact: you have a 50/50 chance of picking the correct side.

Steve B.
Wednesday, July 30, 2003

Let me ask this question...

What would the odds be if you were told not to make ANY selection until after Monty opened one of the losing doors?

Steve B.
Wednesday, July 30, 2003

Steve:

you're absolutely right that "2 doors, one choice" would make the odds fifty-fifty, presupposing of course, that whatever is behind the doors is randomly distributed. That's not the case though, because the box that's being displayed to you was not picked at random and just happened to be empty. The box that's being displayed to you was deliberately chosen to be one that was empty.

If the box to be displayed to you were chosen at random, your strategy would change: if the box doesn't contain the prize you'd have a 50-50 chance and switching or not switching wouldn't make a difference. On the other hand, the game would be boring, because you'd have no incentive to continue playing because no matter what you do, the prize is gone.

The fact that the box being displayed is not chosen at random changes the odds.

If you still don't agree, I'd like to propose the following bet:

Name a card at random -- let's make it an ace of spades for this example. I'll shuffle a deck of cards, you pick one. I'll be allowed to look through the deck of cards and sort out all cards except for the one you chose, unless of course you picked the proper card in which case I'll just pick a random one.

So now there's two cards on the table. According to your theory, the chance that either of the two cards is the ace of spades is 50-50. I'll give you 2 dollars for every one dollar you wager that the card you picked is the ace of spades, that should leave you with a handsome profit. Up for it?

a2800276
Thursday, July 31, 2003

Not sure if I understand the logic behind the card comparison.

But let me re-ask the following question:

What would the odds be if you were told not to make ANY selection until AFTER Monty opened one of the losing doors? 

The scenario would thus be: "Pick one door from a pool of two doors. One door is a winner, one is a loser". 50/50 odds.

That is, in reality, what the game is.

Keep this in mind...

1. Monty will ALWAYS open a losing door.
2. This will ALWAYS leave you with two doors.
3. The remaining doors will ALWAYS be a quantity of two: one winner and one loser, thus making the doors evenly distributed.
4. Your choice is ONLY made when there are two doors, thus 50/50. The first opportunity to choose is simply entertainment as it has no impact on the outcome. You are not actually making a decision until there are only two doors.

Changing your mind prior to the final selection does not alter the odds of ANY game. The doors, or the cards, or the coins can't read your mind.

The mistake that is being made by the 2/3 odds proponents is that you get 2 choices out of 3 doors. That is absolutely wrong. Whille you are given the opportunity to change your mind, you only get to select one door to open. When that opportunity is given, there are only two doors from which to choose.

Steve B.
Thursday, July 31, 2003

The card comparison is simple: you try to pick a prespecified card (the door with the prize behind it). Once you've made your pick, I (Monty) get to select which cards to show you (which doors to reveal), basically sorting out all non-winning cards (showing you the doors which don't conceal the prize). I keep one card, which is either (a) the winning card or (b) if you have picked correctly on your initial guess, a random card.

Another example: 1000 beads, one of which is black the others are white. You blindly pick one bead, and put it in a jar. I will reveal 999 remaining white beads to you. The last bead though, I keep concealed. Do you think there's an equal chance your or my bead is black? No, you've split the 100 beads into two sets on your first decision. One has 1 bead and a 1 to 1000 chance of it being black. The other has 999 beads and a 999/1000 of _one_ of those being black. I get to manipulate the second set and remove 998 beads that definately aren't black. There still a chance og 999 to 1000 that my set contains a black bead, but there's no longer 999 beads.

>3. The remaining doors will ALWAYS be a quantity of two: one
>winner and one loser, thus making the doors evenly distributed.

And that's the mistake: Two doors, one a winner, one a looser, I always put the winner behind the left door: not an uniform distribution.

>4. Your choice is ONLY made when there are two doors, thus
>50/50.

Your choice is between one door that has a 1/3 chance of a prize being behind it, and one that was in a set that contained 2 doors, thus had a 2/3 chance of containing the prize. One definate looser was removed from the second set. Chances are still 2/3 that the winner is in that set. You have one choice between two doors, one that has a 2/3 chance of concealing the prize and another that has a 1/3 chance of concealing the prize. Which door do you pick?

By the way: you can "prove" this to yourself by just trying it out, or writing a little program tries it out for you. Have a a look at:

http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

a2800276
Thursday, July 31, 2003

Wow... amazing how these things get dragged out...

Are people getting emotionally invested in their arguments? ;-)

I remember this one at uni.. I just refused to believe that the odds on your decision can change because you are shown one of the wrong answers...

But once I understood the value of the knowledge that is required by the game show host to take away the wrong box / door, I realised it is better to switch.

The million doors example is a good one. Choose a door, and then all the other doors except one are closed. What are the odds you choose the correct door?

1 in a million.

The odds on the other door?

999,999 in a million.

Would you switch? :-)

Giles
Monday, August 11, 2003

This is one problem that really made steam come out my ears. At first I thought that switching made no difference. It would always be a 50/50 chance. Then after reading more and more of the replies, I feel that switching will give you a better chance. I conducted some tests from one of the links supplied by David.

Out of 120 attempts, 60 switching, 60 not switching, I got these results:

Switching - 35 wins
Not Switching - 15 wins

Clearly this shows that switching will give you a higher probability of winning, near 66%. However, I found something very interesting. When you add up the total number of wins (50), it approaches 50% of the total number of attempts. It appears that the overall chance of choosing the correct door is 50% given that the number of times a switch occurs is even to the number of times a switch does not occur. Thus, it is more probable that switching will give you a better chance since the first decision is based on 3 doors. Still, it is a 50% chance of winning, just that most of the 50% is made up of switching doors.

Tim Mok
Monday, August 18, 2003

The important point is, when Monty opens one of the boxes you didn't pick, he is specifically opening one of the empty ones. Therefore, Monty's choice does not involve any odds.

After that, since you have a choice between two remaining boxes, and one of them certainly has $100 in it, and the other one is certainly empty, and you still don't know which is which, the chance is 50/50.

James R. Manka
Friday, August 29, 2003

Its really an interesting discussion... After reading all the posts i feel that i agree with Steeve.

I quote a post from another link.

There are 4 possible "paths" this game can go - I am referring to the table in the explanation on the techinterview website - let's say it's case 1 ($ behind door 1):

1. you pick door 1, Monty opens door 2
2. you pick door 1, Monty opens door 3
3. you pick door 2, Monty opens door 3
4. you pick door 3, Monty opens door 2

I know paths 1 & 2 seem equivalent, but they are NOT the same path. So all those four paths are just as likely. If you change your mind path 3 & 4 wins. If you don't, path 1 & 2 wins. And it is fifty-fifty in the end.

This clearly prooves that the chances of winning if we switch is just 1/2. I think since this shows all the available paths, all the possibilities, its absolutely clear that the chances are always 1/2...

Doing a random expiriment and getting more results in favour of switiching doesnot show you that its chances are more. No. If you flip a coin 10 times and happened to get all heads, do you say that it has a more chance?? No Never. We cant proove probability by experiment.

Joseph
Monday, September 15, 2003

Well, after thinking a little further i feel that swithicng will be a better idea... I can give you a more clear argument in its favour as follows.

Think it in this way:
You first select one door. Now you are given the choice of either retaining the door you selected or to select "both" the other doors.  We clearly will swith and select the other two doors together, because selecting two doors has more probability of winning than selecting one.

This problem is almost similar... the difference is that one door among the two we selected is shown that it doesnot contain the treasure.

Joseph
Tuesday, September 16, 2003

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