I thought we deserved a little break from arguing about politics and the skeletons of the Mormon closet, so, in the interest of arguing with the esteemed but poorly credentialed Reverend Mr. Gladhand, I thought I would throw out a classic epidemiology/probability problem, one that stumps easily 90% of epidemiologists in training (though I’m sure many fewer engineers, lawyers, humanities scholars, and non-academically inclined participants in blog culture will miss it; I got it wrong by way of confession). I’m not going to name the source in hopes of avoiding cheating through googling, as I’m sure the solution is posted on the web.
There are three refrigerator boxes and a prize inside one refrigerator box. You make your choice, and tell the boxkeeper (who looks like a female Brad Pitt on Survivor: Morocco) your choice. The boxkeeper then removes one of the other boxes, showing it to be empty, and asks you, “do you want to change your mind?” So, do you? Should you stick by your guns, or should the melodramatic gesture of the boxkeeper (clearly designed to unsettle your conviction) make you doubt your initial choice?
PS: Yes/No answers are insufficient. The correct reasoning is what is required. First person to get it right gets a backstage pass to Survivor: BCC or have her/his picture posted on this thread or other mutually agreed upon prize. In the interests of fairness, people who have already heard the problem and its solution should probably refrain from posting.
February 22, 2007 at 7:56 am
people who have already heard the problem and its solution should probably refrain from posting.
Dang.
February 22, 2007 at 8:17 am
I would stick to my guns – the now reduced odds are 50-50, so I have an equal chance either way don’t I??
February 22, 2007 at 8:22 am
Alright Howie, I would say. “No Deal!” and slam the case over the big red button. Though I’m not qualified to explain my answer in epidemiological terms, I’ll take a stab. The way I see it, removing one empty box does nothing to alter my chances of getting the prize. The only thing that would change my mind is if she removed a box I didn’t chose and the prize was underneath. Once I have chosen a box, the prize is either there, or not. It doesn’t matter if there are 100 boxes or just 3. Correct me if I’m wrong, but isn’t the lottery the same way? People think that buying more tickets increases their probability of winning, when in fact, it doesn’t. I’d like some more insight on these things, because if I’m wrong about all this, I’m buying some Powerball tickets tonight!
February 22, 2007 at 8:37 am
Moderator: If this is giving away too much (I don’t think it will), there’ll be no hard feelings if you delete this message.
In solving the problem, you should make the assumptions that
a) The boxkeeper knows where the prize is, and
b) No matter which box you pick initially, the boxkeeper will show you an empty box after you make your choice.
February 22, 2007 at 8:38 am
Alright Howie, I would say, “No Deal!” and shut the case over the big red button. Though I’m not qualified to explain my answer in epidemiological terms, I’ll take a stab. The way I see it, her removing an empty box does nothing to the odds of my choice. Once I choose a box, the prize is either there, or not. It doesn’t matter if there are 100 boxes or just 3. The only thing that would hamper my choice is if she showed the prize under a box I didn’t choose. Isn’t the lottery the same way? People think buying more tickets increases their probability of winning when it really doesn’t. I’d like more insight on this, because if I’m wrong, I’m playing the Powerball tonight!
February 22, 2007 at 8:48 am
oh, commenter, your collision with the thread is very sneaky. I’ll let it stand, though I’m not sure it’s particularly illuminating. Neither a) nor b) could possibly relate to the odds you chose correctly, so I’ll let it stand on grounds of showy irrelevance.
Does anyone want to succumb to this Brad Pitt character and actually waffle? I know it violates certain important cultural norms, but she’s pretty persuasive (maybe it’s the Indiana Jones-style whip).
February 22, 2007 at 9:08 am
If the boxkeeper planned from the beginning to reveal an empty box after I made my first choice, then I have always had a 50/50 chance, haven’t I? The offer to change my mind is irrelevant, added only for drama, and creating the illusion that my odds of winning have now gone from 1 in 3 to 1 in 2. Whether I understood that would depend on whether I knew from the beginning that the boxkeeper would remove one of the empty boxes. Changing my mind at this point makes no difference — am I stubborn, or do I want to play along with the illusion of drama? Brad Pitt in drag is off-putting, so I don’t play along — I stick with my first choice.
(We must be missing something, since we’re agreeing without coming up with some neat little trick…)
February 22, 2007 at 9:10 am
The answer is it really makes no difference either way.
February 22, 2007 at 9:11 am
But it’s so simple. All I have to do is divine from what I know of you. Are you the sort of man who would put the prize into his own box, or his enemy’s? Now, a clever man would put the prize into his own box, because he would know that only a great fool would reach for what he was given. I’m not a great fool, so I can clearly not choose the box in front of you. But you must have known I was not a great fool; you would have counted on it, so I can clearly not choose the box in front of me. Because a Female Brad Pit host comes from Morocco, as everyone knows. And Morocco is entirely peopled with criminals. And criminals are used to having people not trust them, as you are not trusted by me. So I can clearly not choose the box in front of you. Yes — Morocco, and you must have suspected I would have known the female Brad Pitt’s origin, so I can clearly not choose the box in front of me. You’ve beaten my giant, which means you’re exceptionally strong. So, you could have choose the prize in your own box, trusting on your strength to save you. So I can clearly not choose the box in front of you. But, you’ve also bested my Spaniard which means you must have studied. And in studying, you must have learned that man is mortal so you would have put the prize as far from yourself as possible, so I can clearly not choose the box in front of me.
It has worked — you’ve given everything away — I know where the prize is.
I choose –
– what in the world can that be?
February 22, 2007 at 9:13 am
Here is a very similar problem that may illuminate the one posed: Suppose there are 100 boxes, and you pick one of them. Then Mrs. Pitt (??) opens up 98 of the 99 boxes you didn’t pick and shows that they are empty. She offers you to keep the box you originally selected or trade it for the one box she could have opened but chose not to. Do you want to trade now?
February 22, 2007 at 9:14 am
Matt W – LOL! Love the Princess Bride reference!
February 22, 2007 at 9:15 am
Commenters assumptions reveal something very important about this particular problem: the solution will be different depending upon how the problem accounts for the behavior of the host. Actually commenters thoughts are extremely relevant to the solution to the problem, if one applies the formalism of Bayesian probability theory. According to that theory, and the math behind it based precisely on the assumptions stated by commenter, the right choice will always be to switch because switching will yield a prize 2/3 of the time. If the problem does not stipulate that the host will always show an empty box, and this version does not, then the correct solution is to stay with your choice because the odds remain 50/50.
February 22, 2007 at 9:25 am
It doesn’t matter whether you change or not. You don’t get any information as to where the prize is from the revelation of a wrong choice that you didn’t choose.
Unless I’m wrong. Given that the answer seems obvious to me, yet it stumps 90% of budding epidemiologists, I’m guessing that the answer is counterintuitive.
February 22, 2007 at 9:26 am
I stick to my guns, trust my first choice and not pay attention to what the host said. I go into the decision risking a 1 in 3 chance. Irrespective of what the host does, my chance is still 1 in 3 because there still are three boxes. I guess I’m an assertive guy who is confident in his decisions. As for the epidemiological stuff….well I’m a librarian/political scientist not a scientist, so I don’t even know where to begin discussing all that.
February 22, 2007 at 9:30 am
When you chose the first box you had a one in three chance of choosing the correct box. If you choose to change your pick after one box is eliminated, you have a one in two chance of picking the correct box–so the value of that choice has increased. The value of the first choice was worth .33 and the opportunity to choose again is worth .50, so you should take the lady up on the offer.
February 22, 2007 at 9:48 am
opportunity to choose again is worth .50, so you should take the lady up on the offer.
But you can choose again and stay on your original box.
What would change is your willingness to wager. If you place a bet before you know that you’ll have a 50/50 choice, you have to bet as if the odds are 1/3. But if you know beforehand that you will have the opportunity to make a 50/50 choice then you should bet as if the odds were 1/2. Your bet, or your willingness to bet, should increase when you find out that your odds are better.
However, if you don’t get the opportunity to make another choice after one wrong choice is revealed, then your bet shouldn’t change after the reveal—it should still be based on 1/3 odds.
February 22, 2007 at 9:51 am
OK–after I answered it I googled it. I think I’m right. If not, Sam would you let me know so I can try again?
February 22, 2007 at 10:01 am
Costanza and commenter are proposing some kind of new stipulation on the problem, which I think has caused them to mislead themselves and perhaps you. The problem is correctly written, and neither of the commenter’s requirements (nor the costanza expansion) is actually required for the present problem. I’ll give you an obfuscatory hint: as the problem is originally and currently phrased, Brad Pitt need be no more clued in than you (and he likely isn’t).
February 22, 2007 at 10:04 am
Matthew, you have sinned against a CES-style object lesson based on this problem: remember that not choosing is in itself a choice. (these comments followed by a knowing Carradine nod).
February 22, 2007 at 10:04 am
I think it should be looked at from a different perspective (only slightly). In the first round you have a two in three chance of picking an empty box. Those are good odds. If I pick an empty box and the other empty box is taken away then my odds are 2/3 that the other box has the prize in it, the best odds that have been given yet.
It’s easier to see with the 100 boxes example. If the odds are 99/100 that I will pick an empty box and that after all the rest of the empty boxes are taken away the one with the prize in it is still there then really my chances are 99/100 to get the prize. You take the box offered because the odds will always be higher that it is there. Not 50/50 but 2/3.
February 22, 2007 at 10:08 am
I googled it and I know what the answer’s supposed to be. It is indeed counterintuitive; I would have gone the other way. However, I’ve seen lots of purported explanations, but none that really is clear. So I’m hoping that after enough people have beat their heads against this, you will provide us with a clear explanation of the solution.
February 22, 2007 at 10:29 am
Kevin,
When I googled it I found the answer rendered mathematically. Take a look–I think you’ll agree it is the most elegant rendering of the solution.
February 22, 2007 at 10:47 am
It all comes down to whether or not the host knows where the prize is. The original post is a little unclear — “The boxkeeper then removes one of the other boxes, showing it to be empty, and asks you, “do you want to change your mind?†Is the box she shows empty by chance, or by design? If she has prior information, then that can be used to your advantage — switch for a 2/3 chance of winning. Otherwise, your odds don’t change. (Though it would be embarrasing for the host to open the box with the prize by chance — then your choice would be easy!)
February 22, 2007 at 10:53 am
As is often the case in these brain-teasers, the trick is that the problem description here is generally underspecified. The actual answer depends on our model of the boxkeeper’s behavior. Specifically, if we believe the boxkeeper to have chosen a box to open at random — with a 1/3 probability of opening our box, and a 1/3 probability of opening either of the other two boxes — then the answer goes one way. If we instead believe the boxkeeper to have deliberately opened an empty box that we didn’t choose, the answer goes another way. Most versions of the problem description don’t make this point explicit, so finding the answer revolves around our decision to make a reasonable but not logically necessary leap regarding what the boxkeeper’s actually up to. Once you do that, it’s a simple application of probability.
February 22, 2007 at 10:53 am
Ah, right. Dan’s point exactly.
February 22, 2007 at 10:55 am
I think whether the box the hostess opens is empty by chance or because of foreknowledge, your odds still increase.
Looking at the 100 boxes example: there is a much greater chance (99/100) that the prize is not under your box than the chance (1/100) that it is. If the hostess opens 98 of the boxes that you didn’t choose and shows them to be empty, there is still a 99/100 chance that the prize is not under your box, which is the same as saying that there is a 99/100 chance that it is in the remaining box. You take it.
February 22, 2007 at 11:06 am
Tom, that’s actually not quite right. You’ve got to do this with Bayes’ rule, to avoid intuitive errors.
Let’s take the three-box case, because the math is simpler. Our prior belief is that we’ve selected the right box with 1/3 certainty. If the box shown to us is selected at random, then the event we’re observing is whether the box is empty or not.
What is the probability that the box is empty given that our selected box is nonempty? 1, obviously. What is the probability that the box is empty given that our selected box is also empty? 1/2.
So our posterior probability that our selected box is the right one is: (1/3*1)/(1/3*1 + 1/2*2/3) = 1/2. Which is identical to the posterior probability for the other non-revealed box.
February 22, 2007 at 11:07 am
The story is very, very different if the revealed box or boxes are chosen deliberately.
February 22, 2007 at 11:07 am
Unless, Tom, the hostess chose to open those 98 boxes only because she knows you have the prize and she is trying to persuade you to give it up, because she is a cruel temptress who likes to bestow favors but then snatch them away and convince you the loss is due to your own shortcomings.
February 22, 2007 at 11:07 am
interestingly 22 and 23 stipulate slightly different conditions. 22 claims that the boxkeeper must know the location of the box with treasure, while 23 claims that the boxkeeper must be restricted from choosing our box AND know which box contains the treasure. Is one claim more important than the other? How are they affected by the fact that the two choices are no longer theoretical at the time you must make the third choice?
and PS, I did make it a little murky to separate it from its original context, which is an old game show. If you’re setting up a game show, you’ll need to know where the treasure is and not choose the contestant’s box, or your boxkeeper is going to make you broke.
February 22, 2007 at 11:15 am
If the prize isn’t under your box and the hostess is opening boxes randomly her chances of opening the remaining 98 boxes without the prize are 1/99. Which is still better than your chances of having randomly picked the right box (1/100).
February 22, 2007 at 11:20 am
Tom, it doesn’t work that way. If the boxkeeper is going to open 98 boxes out of 100 at random, the box with the treasure is almost certainly going to be opened. But, conditional on the box not having been opened, the odds that your previously-chosen box was the right one are 1/2, as are the odds that the non-chosen box was the right one. You get there via Bayes’ rule, as above. So there’s no information gained if the opened boxes are selected at random.
February 22, 2007 at 11:25 am
Just for fun, here’s the math on the 100-boxes-with-98-opened problem.
The prior probability that you’ve chosen the right box is 1/100. Given that you’ve chosen the right box, the probability that the 98 boxes chosen to be opened are empty is 1. Given that you’ve chosen the wrong box, the probability that the 98 boxes chosen to be opened are empty is 1/99 (the problem is identical to choosing just one box not to open).
So, our posterior probability is: (1/100*1)/(1/100*1 + 1/99*99/100) = 1/2. No information gained about which of the remaining two boxes is correct.
By the way, because I’m picky about these things, let me note that the two analyses I’ve done have both assumed that the boxkeeper is selecting randomly among boxes other than the one you chose.
February 22, 2007 at 11:43 am
In the first choice I have 2/3 shot of choosing an empty box. That is to say that I most likely chose an empty box. The moderator removes one empty box. Consequently, when asked to switch boxes, I will do it as I most likely chose the other one.
February 22, 2007 at 12:27 pm
JNS and other caught me being sloppy. This is used as a classic example of an epidemiological collider, and when it’s fully specified it is indeed just that. Using an intuitive staged choice model on the fly this morning, I tried to generalize to allowing random choice by the boxkeeper, but I think JNS is right that Bayes won’t support it. I’ll attempt a Kevin-worth explanation when I get a chance.
February 22, 2007 at 1:35 pm
JNS,
So did I get it right? :)
February 22, 2007 at 1:42 pm
Wow, I understood what Stapley said. Amazing.
February 22, 2007 at 2:01 pm
Stapley has it right, I think. My wife teaches junior high math, and has talked about this problem over dinner several times. I never understood the answer, nor could I remember it. Switch your choice to the other box, and your odds go up.
Inconceivable!
February 22, 2007 at 2:05 pm
By the way, has anyone tried this with BYU students and prospective eternal companions? Do you switch boyfriends after your first choice when his roommate graduates?
February 22, 2007 at 2:49 pm
The randomly selected empty box problem that JNS describes is interesting to think about. It helps to keep in mind that under that scenario, the cases for which the randomly selected box holds the prize have been thrown out. Taking the 100 box problem, each time a random box is opened by the host and turns out to be empty, the contestant correctly feels luckier since he didn’t pick that empty box, and the box he did pick is now part of a smaller set. Most of the time, though, one of the 98 randomly selected boxes will contain the prize, and the contestant will have lost before reaching the stage of swapping with the final box.
Suppose with the three box problem, that the host opens one of the two unselected boxes randomly, but add that the contestant can switch his box with either of the two boxes he didn’t select the first time. If the random box opened by the host contains the prize, then the contestant will trade his box for the opened box, and he wins. If the opened box is empty, we’re back to JNS’s problem, and from that point on the box selected in the first stage and the unopened, unselected third box each have 50% chance of containing the prize. There is 1/3 chance that the box opened by the host contains the prize, so the chance of winning is (1/3)*1 + (2/3)*(1/2) = 2/3, the same as the odds for the scenario where the host knowingly opens an empty box. Thus a random box selection that doesn’t abort the game and start it over does give the contestant information and changes the odds of winning compared to what they would be if no box were opened.
February 22, 2007 at 3:09 pm
I was thinking about this is light of Howie Mandel and Deal or No Deal.
So 20+ beautiful models walk out on stage with their numbered cases. Various dollar amounts have been placed in the cases randomly. Neither the models nor Howie himself knows the contents of the cases. The contestant picks one, without opening it. Then as the game progresses, the contestant picks cases, and the models open them, revealing the amounts of potential prize money inside.
Occasionally Howie gets a call from the “banker,” who makes an offer for the contestant’s case, based on the then odds of how much it might contain. Usually the contestant succumbs to audience pressure and rejects the deal and keeps playing.
So let’s say you get all the way to the end, where there are just two cases: the contestant’s case, and one left held by a model. At that point, Howie offers to let you swap cases. Do you swap?
It seems to me that, at that point in the game, the odds are simply 50/50 whether your case or the other one holds the higher amount. So it doesn’t matter whether you switch or not; you are neither increasing nor decreasing your odds.
If I am wrong about this and the right answer is that you increase your odds by switching, then I really don’t understand it, and I need someone to walk me through it really slowly to explain it to me. Because I just don’t get it.
February 22, 2007 at 3:24 pm
Kevin,
I hear ya. I’m a stubborn Eastern European and will stick with the case I picked from the beginning through thick and thin.
February 22, 2007 at 3:30 pm
FWIW, the “Dan” in #22 (me) is not the same Dan is in all the other posts (except this one, obviously)… I still say change if the hostess knows where the prize is!
February 22, 2007 at 3:50 pm
I think I’m allowed to make this a simple probability problem by listing the possible outcomes of me randomly picking a box followed by the hostess randomly picking one of the remaining boxes. The six equally likely possible outcomes are:
1) I pick A, she opens B
2) I pick A, she opens C
3) I pick B, she opens A
4) I pick B, she opens C
5) I pick C, she opens A
6) I pick C, she opens B
In choosing whether or not to change my pick, what I want to know is this: what are the odds of my choosing the prize box and the hostess opening an empty box relative to the odds of my choosing an empty box and the hostess also opening an empty box. If the former is less likely than the latter, I should switch. Say the prize box is box A. If you look at the six possibilities, the odds of my choosing the prize box and the hostess opening an empty box (possibilities 1 and 2) are 2/6. The odds of my choosing an empty box and the hostess opening an empty box (possibilities 4 and 6) are also 2/6. The two scenarios are equally likely. The fact that, by chance, the hostess opens an empty box tells you nothing more than that possibilites 3 and 5 are off the table. So you can bet more because your odds have increased to 2/4, but switching doesn’t increase your odds over staying.
February 22, 2007 at 3:52 pm
Kevin, in the case of deal or no deal, you are doing all the choosing without a priori knowledge. That is different than this game.
February 22, 2007 at 3:58 pm
Kevin Barney,
Howie Mandel logic is dangerous. I’ve rethought what my wife the math teacehr said to me when she explained this one. You make your first choice, and you have a 1 in 3 chance of being right. When the beautiful Sicilian, er, Moroccan, opens one of the remaining two boxes, and shows that it is empty, and then offers you the chance to switch, you then have a 1 in 2, or 50/50 chance.
It’s hard for my liberal arts major mind to get around it, but it is true. Your first chance is 1 in 3. If you don’t switch, you still are living with the 1 in 3 odds. By switching, you increase your odds to 50/50.
Howie’s deal really isn’t the same, because along the way you have eliminated many other cases, some with more, some with less. When you get down to the end, that final choice is 50/50, but your original choice was 1 in 25. And if you notice, the $1,000,000 case usually is gone way before they get to the end, which skews the odds even more. Which odds would you rather have?
February 22, 2007 at 3:59 pm
Here’s the theory behind it. Your choice affects her possible choices, and her choice affects thus can give you an edge over random choice. This is not just two people choosing at random simultaneously, and the fact that your choice affects her choice is the critical insight.
Here’s another way of looking at it: what are the odds that her choice will give you valuable information, assuming she won’t pick your box and she knows which one holds the treasure?
If you chose correctly, her choice will be entirely uninformative because she is choosing among two identically empty boxes.
If you chose incorrectly at first, her choice will be completely informative as she is choosing the empty box of the two that remain.
So then what were the chances that you were incorrect at the beginning? 2/3. The chance that you were correct at the beginning? 1/3.
As far as New Deal or whatever, I have no idea what that is, so I can’t comment. If it obeys the same rules, you would benefit from switching.
February 22, 2007 at 4:02 pm
And as far as my slipup, for which I apologize:
In my quick response to the expanded case (she can’t choose your box but doesn’t know where the treasure is), I was using intuition rather than actual calculations and assuming that she would be informative whether she chose the treasure (you would thus choose her box) or not (the standard case), but the key insight here is that her choice isn’t constrained by the history of your choices, and the chance that her choice will be informative without displaying the treasure is lower than in the non-random case.
February 22, 2007 at 4:06 pm
Sam MB, only if she knows where the prize is, right?. If she knows where the box is, you should switch. If she doesn’t know and, by chance, opens an empty box, it doesn’t matter if you switch.
February 22, 2007 at 4:30 pm
While the math can be confusing, the logic should be straight forward. As has been said, when you pick the box, you’ll be wrong 66% of the time. You are “probably” wrong. Think of the other two boxes as grouped in one choice. Would you rather choose the other two boxes or just your one? You’re twice as likely to win with the other two. But you can’t pick two. Wait, yes you can. The host opens one of the two. By opening that one, he is providing you new information. Of the two you didn’t pick (one of which “most likely” contains the prize), it’s not this one. Then it’s the other.
I’ve never heard the part about does the host randomly open a box. This means 33% of the time, he opens the prize and the games over. The real way to complicate this is the host doesn’t always ask you to switch. Then you wonder if he tries to get you to switch more often when you’re right (33% of the time) then when you’re wrong.
Here’s an even more illogical question using the same type of math: I have two children. One is named Harry. Is the other one a girl or a boy? Is it 50/50?
Oh, I forgot to mention Harry is the older child. Does your answer change?
February 22, 2007 at 4:32 pm
You are correct, Tom. If she doesn’t know, then she could have chosen the actual treasure box, so the probability that she will choose an empty box (and hence be informative for our purposes) is less than certain, whereas if she knows, assuming your original guess was incorrect, you are certain that she has made an informative choice.
February 22, 2007 at 4:33 pm
In Deal or No Deal there is no host knowledge of what is in the cases; they are just opened. And I assumed no host knowledge in the original question here (probably influenced by the TV show), which is why I didn’t see where the increase in odds was coming from.
February 22, 2007 at 4:39 pm
Aha. I figured a boxkeeper would know what was in the boxes and didn’t realize the No Deal context. It is explicit in the standard formulation of the problem.
February 22, 2007 at 4:41 pm
Sam,
Thanks for writing that up, it was interesting to read. I’m pretty sure I had heard this one before at some point but reading through all the comments of how people approached it was still interesting.
February 23, 2007 at 6:58 am
Monte Carlo simulation reveals the answer is . . . Pick the other box–you’ll win 67% of the time. I’m assuming Vanna knows the answer and won’t reveal the box with the prize. Yup, I’m an engineer.
February 23, 2007 at 8:59 am
It doesn’t matter whether the host knows or not (other than it’s kind of pointless if they reveal the correct box). And the odds don’t change to 50/50, otherwise what would be the point of switching?
It is to your advantage to switch, and you will win 2/3 of the time. The easiest way to see this, as some have said above, is to work through it with the odds that the prize is not in your box. If you think of the two that you did not choose as a unit (one of which will then be eliminated), there is a 67% chance the prize is here (and you will win by switching). There is a 33% chance that you chose right originally and will then lose by switching. it is to your advantage to switch every time.
If you don’t believe it, sit down and try it a bunch of times. it is fun.
February 23, 2007 at 9:32 am
Re #50:
Assuming Harry is a boy’s name…
In the first case, there is a 67% chance that your other child is a daughter.
If, however, you tell me that Harry is older, that chance changes to 50% boy and 50% girl.
Strictly speaking, this is not entirely correct, as human males have a significant birthrate advantage compared to females
February 23, 2007 at 9:38 am
Furthermore, in Deal or no Deal, the banker always offers you significantly less than your expected return, so a completely rational player would be correct to reject all banker offers and continue to the end.
February 23, 2007 at 10:06 am
woodboy, i think the problem with the uninformed case is that the fact that it was NOT the correct box slightly increases the probability that your box is the correct one.
jose–could you run a MCS with the random assumption–>does it prove the fastidious Bayes approach?
February 23, 2007 at 10:13 am
yeah, you are right. I changed my mind shortly after i posted my comment but didn’t see a way to edit. My error in thinking was that I had assumed the host exposes the correct prize (and the game ends) 1/3 of the time. but it is not an unbiased 1/3 of the time, it is 1/2 of the 2/3 of the time you chose the wrong box (and would have won by switching). So this negates your switching advantage (because half of the time you lose before you get the chance to switch)
February 23, 2007 at 10:46 am
Dan (#2)
(#43)
Sorry, I had seen your post after I made my comment and realized the error of my ways. Good work on getting it right though. :)
February 23, 2007 at 12:31 pm
So 20+ beautiful models walk out on stage with their numbered cases.
What? The other six are dogs?
;-)
February 23, 2007 at 12:32 pm
woods, i made a similar error–we were both eager to ignore the times when the host inadvertently showed you the actual prize as irrelevant.
February 23, 2007 at 4:34 pm
what if they empty box she revealed IS the prize? What if the other two boxes are full of rotten eggs?
February 23, 2007 at 5:26 pm
Why is everyone wasting time discussing the situation in which the host is selecting at random and doesn’t know what is in the boxes? In practice that isn’t going to happen or else the host would reveal the prize 1/3rd of the time which would be exceptionally lame. Of course the host knows and of course you should switch.
February 24, 2007 at 1:49 am
Alright, I know I am late, but the puzzle is OK, and I am game.
If you picked the right box there are two ways the boxkeeper can choose the empty box. If you chose the wrong box there is only one way the boxkeeper can choose an empty box. There are three possible outcomes, of which two favor you. So the chance of finding the prize is 2/3 if you do not change you choice.
However, if the boxkeeper can choose the prize box, there are also three outcomes: one you fail immediately, one you win on the second round, one you fail on the second round. So of the three, there is only one chance to win out of three.
I think if you switch, your probability falls to 1/2.
So, in this life, how much information is gained by being initially informed about wrong choices and right choices? If I have chosen something, and someone shows me that I have not chosen wrongly, then I have increased my chances of being right.
So, of the many choices in life which God does not confirm positively, but not negatively, indicates that we are on the right track and our probability of success increases. So, no answer to prayer is also information, not as good as a burning in the bosom or your mind going numb.
Lead us not into temptation by tempting us with boxes and prizes and chances and wagers. Next it will be a non-mathematical seduction. Maybe by physics?
March 15, 2007 at 5:57 pm
This is a simple question just don’t over analize it. if there are two boxes left, the probubility is always going to be 50/50. It dosn’t matter wich one you started out with it’s simple fith grade math
March 15, 2007 at 6:05 pm
LOL, Donpunkraka.