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#101
11-21-2010, 04:57 AM
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The reason we can't consider it 1:2 is that Monty choice on which door to open is not always random. He will have this problem two thirds of the time, since you have two chances of picking the goat. This means that, two thirds of the time, he is telling you which door contains the prize. This means you are also not choosing randomly.
So basically, it goes back to Nemo's post above. To put it a different way: 2/3 chance to pick goat. Switch and win. 1/3 chance to pick prize. Switch and lose. 
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#102
11-21-2010, 05:19 AM
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Perhaps in response you can argue that the entire concept of a probability table is therefore flawed and perhaps as far as its ability to capture the subtleties of this hypothetical that might be true - or, it might not. IDK. What I do know is that when you look at the gross odds for every possible outcome at every stage of the process, the final odds turn out to be 50-50 no matter how you look at it. Quote:
I believe this addresses your objection.
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#103
11-21-2010, 06:05 AM
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The problem is you are ascribing a probability to switching and use a probability tree, while what I'm doing is setting up a table of possible outcomes, all of which are equally likely. The probability tree is the wrong approach to this problem as the 50-50 probability you introduce has nothing to do with the chance of picking the right door after one has been revealed, it's the probability of making either choice if you do so randomly.
Let's say instead of of flipping a coin you roll a six sided die and if it shows 1-4 you stay with the original door, that is a 1/3 chance you switch. Now the correct probability tree goes: 1) 1/3 chance you picked the right door 1a) 1/3 chance you switch = 1/9 chance you lose 1b) 2/3 chance you don't switch = 2/9 chance you win 2) 2/3 chance you picked the wrong door 2a) 1/3 chance you switch = 2/9 chance you win 2b) 2/3 chance you don't switch = 4/9 chance you lose Sums: Chance you lose 5/9 Chance you win 4/9 Now lets move on to the probability if you decide to always switch. 1) 1/3 chance you picked the right door 1a) 1/1 chance you switch = 1/3 chance you lose 1b) 0 chance you don't switch = 0 chance you win 2) 2/3 chance you picked the wrong door 2a) 1/1 chance you switch = 2/3 chance you win 2b) 0 chance you don't switch = 0 chance you lose Sums: Chance you lose 1/3 Chance you win 2/3 And to hammer the point in: the probability if you decide to never switch. 1) 1/3 chance you picked the right door 1a) 1/1 chance you don't switch = 1/3 chance you win 1b) 0 chance you switch = 0 chance you lose 2) 2/3 chance you picked the wrong door 2a) 1/1 chance you don't switch = 2/3 chance you lose 2b) 0 chance you switch = 0 chance you win Sums: Chance you lose 2/3 Chance you win 1/3 That's a second rigorous mathematical proof that you're wrong. Last edited by naita; 11-21-2010 at 06:07 AM. Reason: was missing a win
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#104
11-21-2010, 06:05 AM
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Suppose we have a game where you purchase one out of 1,000,000 lottery tickets. Once you do that, the lottery commissioner, who knows which one is the winning ticket, destroys 999,998 tickets which he knows are all losers, leaving only two tickets left, including yours. Do you really think that both tickets have an equal chance of being winners? If you do, I'd like to propose a little game. Quote:
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#105
11-21-2010, 06:51 AM
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I don't want to seem like I'm ignoring the thoughtful replies, but i'm a bit tired and just want to try a different approach.
I understand that the chance after the first draw is 1/3. My point is that even before the first draw, you know that one losing choice will be eliminated. Now I was hesitant to embrace the following concept, but it seems to be the only way to reconcile the result. Specifically, the fact that you know that a losing choice will be eliminated means that your odds from the first draw are not 1/3 and 2/3 but 50-50. Here's why. From the first draw, there are only 2 possibilities 1. you pick the prize, monty shows you a goat, the remaining choice is a goat. 2. you pick a goat, monty shows you the other goat, the remaining choice is the prize. It doesn't matter which goat you pick in 2 since monty is obliged to show you the other goat. It doesn't matter which goat remains in 1, since monty has already shown you the other one. Therefore, your odds were never 1/3 vs 2/3. It only appeared that way because you are calculating the odds without taking into account the effect of having a losing choice revealed - always. In fact, this situation is no different from letting monty pick first. Is it? If monty goes first and always picks one of the goats, aren't you in precisely the same situation you would be after the first draw?
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#106
11-21-2010, 07:31 AM
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Plus there's just the simple fact that I actually wrote a program, and switching always gives better results, at a ratio of 2:1. And, honestly, that wasn't what I was expecting. After I saw it actually happen, all of the sudden I could make sense of all the explanations that seemed so wrong before. There's just no way around the experimental evidence. Heck, the experiment is so well-known, it's actually used to test animal behavior. Some birds will actually figure it out faster than humans because it doesn't let false logic get into the way. If the bird and the human are put together, the bird will win more often, after it finally gets it, while a human will win more only early on when he doesn't.
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#107
11-21-2010, 07:45 AM
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Everyone is quick to emphasize that Monty's choice gives you additional information. Well, if that's true (and my claim is that it is not), then it shouldn't matter whether you get that information before or after you make your own choice. In fact, if it really is valuable, it should be even more so if you have it ahead of time. The argument really revolves around the idea that there are 3 choices. Initially, that's true, and if we left things as they were after the contestant chooses a door, there is absolutely no doubt that he would only win 1/3 of the time. But you know before the first choice that whatever he picks, one of the 2 that remain will be eliminated. Therefore, there are only 2 choices in actual fact - the one the contestant makes and the one left after monty shows you a goat. I think this is about as obvious as it could possibly be. The 3rd choice is not only an illusion but a transient one at that.
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#108
11-21-2010, 07:47 AM
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Why do you think that "there are only 2 possibilities" leads inexorably to the odds being 50-50? Quote:
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Again, please refer to my previous post, where I point out that according to your own analysis, when Monty picks second, your odds of winning are 2/3 if you switch doors. Is your own analysis incorrect?
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#109
11-21-2010, 08:35 AM
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Monty's choice should tell you even more if he makes his first - shouldn't it? Monty has to show you a goat. This true whether he goes before you or after. If it's monty's turn, you get a goat. So if getting a goat from monty is so informative, why is it less so if he shows you the goat first? It makes no difference if this is the basis of your argument. i've already shown with the probability tables the correspondence between choices and odds. In fact, table doesn't care about how many possibilities there are, but only their weight.
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#110
11-21-2010, 08:40 AM
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Here's an easy example: You are asked to select a card at random from a fair deck and place it face down to your left. A team of 10 people then each select a card, look at all 10, agree which is the highest, and place it face down to your right. You now get to choose one of the two cards - if it proves to be the higher of the two, you win $100. You have two choices, and you knew in advance that you would - but I'd hope it's obvious that the probabilities of each card being a winner are not equal.
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#111
11-21-2010, 08:43 AM
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#112
11-21-2010, 08:53 AM
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I'd like to outline what happens if Monty picks first - given that he still always has to pick a door with a goat
Monty Player Result picks picks goat 1 goat 2 loss goat 1 prize win goat 2 goat 1 loss goat 2 prize win Each scenario is equally likely. The result is that the player wins 50% of the time. I submit that this sequence is statistically the same as the one where the player picks first. The only difference is that in this case, the illusory nature of the 3rd choice is highlighted.
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#113
11-21-2010, 08:53 AM
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prize - goat - prize = 1/3 * 1/2 * 1/2 = 1/12 = prize STAY prize - goat - prize = 1/3 * 1/2 * 1/2 = 1/12 = prize STAY goat - goat - goat = 2/3 * 2/2 * 1/2 = 4/12 = goat STAY prize - goat - goat = 1/3 * 1/2 * 1/2 = 1/12 = goat SWITCH prize - goat - goat = 1/3 * 1/2 * 1/2 = 1/12 = goat SWITCH goat - goat - prize = 2/3 * 2/2 * 1/2 = 4/12 = prize SWITCH So, when you STAY with the original door, you win the prize 1/3 of the time and get the goat 2/3 of the time. When you SWITCH to the other door, you win the prize 2/3 of the time and get the goat 1/3 of the time. So according to your own analysis, it's better to switch. If you don't believe that, then what, in your own analysis is incorrect? Quote:
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#114
11-21-2010, 08:55 AM
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And is this not very similar to the MH problem?
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#115
11-21-2010, 09:22 AM
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#116
11-21-2010, 09:28 AM
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I can't really adapt that idea to your example in a way that is exactly comparable, but I'll try to come close. It would be as if, in your example, you knew that 9 of the 10 cards had to be 4 2's, 4 3's and 1 4 (assuming that ace is high and not low). In that case, the only real unknown is the value of the 10th card. Last edited by dzero; 11-21-2010 at 09:29 AM.
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#117
11-21-2010, 11:16 AM
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And, as a follow-up, perhaps you could explain why you would say Quote:
I honestly have no idea how you're approaching this problem, because it looks like you're saying contradictory things as well as completely missing the point of the whole question. Perhaps the best thing to do would be for you to restate in your own words what, exactly, you think this problem is and what question you're answering.
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#118
11-21-2010, 12:01 PM
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However I have changed my argument over the course of the discussion. I'm not trying to hide that fact but it may be contributing to the confusion. I started by arguing that the the initial odds were 1 in 3 to win and accepted the proposition that those odds were descriptive of the hypothetical until after Monty eliminated once of the choices by opening a losing door. I'm not sure whether or not that argument was valid but either way, I'm ditching it in favor of the argument that the odds were never 1 in 3 to begin with but 1 in 2. I'm not going to restate that argument here or subsequently since it's already been made in my most recent posts. BTW, I have no idea if this argument necessarily has any merit either. To be honest, I constantly flip-flop between thinking I might have some genuine insight here and thinking that I'm completely full of shit and my arguments are pure sophistry. Personally, I think the odds are 50-50, but I might just be saying that out of habit. 
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#119
11-21-2010, 12:38 PM
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I'm still not following exactly what question you think you're answering, and I can't help but notice you didn't bother answering any of my questions. Let me try this one more time. We're talking about Cecil's original column entitled, On "Let's Make a Deal," you pick Door #1. Monty opens Door #2--no prize. Do you STAY with Door #1 or SWITCH to #3?
1. Do you think that the concept of switching is irrelevant? If it's irrelevent, why is it explictly mentioned in the column title. If it's not irrelevant, why are you ignoring it? Quote:
2. Please state exactly how you think the Monty Hall problem is constructed and exactly what question you're answering.
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#120
11-21-2010, 01:02 PM
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#121
11-21-2010, 10:21 PM
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I hope the columns remain aligned. Code:
You pick Car Behind Monty Picks You Switch 1 1 2 or 3 Lose 1 2 3 Win 1 3 2 Win 2 1 3 Win 2 2 1 or 3 Lose 2 3 1 Win 3 1 2 Win 3 2 1 Win 3 3 1 or 2 Lose
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#122
11-22-2010, 02:16 AM
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As for scholarly articles, it's so ubiquitous that I probably am best off just giving you the Google Scholar link. But I can also throw in the study I was talking about, both the full PDF form, and the abstract available on PubMed. As for the information--Monty's picking a door from two choices does not give you as much information as him being forced to pick a certain one. Unless you think Monty can pick the door you already have. Two not random choices and one random one is not the same as three random choices between two.
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#123
11-22-2010, 08:44 AM
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Here's a simple example that I hope will help. Take a well-shuffled deck of cards. Cut the deck into two unequal piles, where one pile has about five cards and the other has the remainder. Now guess which pile is likely to have the ace of spades. I assume you agree that the large pile has approximately a 90% chance of containing it. Now decide which pile to pick by flipping a coin. You'll win 50% of the time. But that doesn't tell you what the odds are if you decide with your knowledge! The coin flip is a red herring that's throwing you off. One other point - you've said that you know from the outset that Monty will reveal a door, and that door will not have the prize, so when he does this, there's no information that you didn't have from the beginning. But you're mistaken here - at the beginning, you didn't know which door he would be revealing. That makes all the difference. Last edited by CurtC; 11-22-2010 at 08:45 AM.
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#124
11-22-2010, 05:55 PM
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dzero, it really isn't that hard. Get a deck of cards and a friend to play Monty. Take the Queen of Diamonds as the prize (the chick with the bling), and two jokers as the goats (or pick whatever set you want to designate).
Now you look away and have Monty shuffle the three cards, check which is which, and lay them out as 1 2 3. Then you pick the first card. Now have Monty reveal a goat. Now pick a second time - switch or stay. Record the results of whether you switched or stayed and whether you won the prize or the goat. Now repeat this for a couple dozen times, and then tally the results. What, you don't have a friend? Okay, version 2: Take your three cards, shuffle them, and then lay them out. Pick your card. Now tally whether the prize was in the card you selected or the other 2 cards. Those are equivalent scenarios. And I think that is the root of your problem. I suspect you will argue that is not equivalent. Understanding why that is equivalent should clarify why the odds are 1/3:2/3.
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#125
11-22-2010, 06:55 PM
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#126
11-22-2010, 07:33 PM
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It's clear that acting without foreknowledge makes the situation similar to a coin flip, which means 50-50 odds. It's (slightly) interesting to note that this is entirely independent of what the actual distribution up to the coin flip happened to be. This seems to be at the root of dzero's failure to understand - he thinks that the situation pretty much requires a coin flip, and (correctly) concludes that a coin flip means 50-50 odds.
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#127
11-23-2010, 09:26 AM
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Powers &8^]
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#128
11-23-2010, 01:30 PM
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The simplest explanation I have ever heard is this:
what if Monty didn't revela but gave you the opportunity to switch to both doors versus your one door, i.e. he sais "Ok now you can keep what is behinnd Door 1 or you can have what is behind Doors 2 and 3" clearly the odds are favorable for switching. How does this relate to the actual problem at hand? Becasue Monty won't reveal the good door, he has effectively only shown that one of the doors is a dud, which wew already know. And we know at least one of the doors is a dud. So by revealing one of the doors as a dud, NO NEW RELEVANT INFORMATION has been learned, therefore no change to the odds.
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#129
11-23-2010, 02:25 PM
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The odds started 1/3 for each door. 1/3 - 1/3 - 1/3 Monty now reveals a dud door. He always reveals a dud door. What happens to that 1/3 possibility? Intuition suggests that 1/3 chance splits between the remaining two doors, and that you make a second guess at 50:50. The Monty Hall problem was devised to demonstrate that that intuition is false. The odds on the first door were selected at 1/3 and do not change. The odds on the other two doors that are Monty's doors sum to 2/3. Monty shows you either of his doors that is a dud, which means you get 2/3 possibility if you swap to the remaining Monty door.
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#130
11-29-2010, 11:46 PM
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Certainly, the person who had the two children knows whether they're boys or not. But we can still talk about the odds. What are the odds that out of thirty people in a room, two will share a birthday? Everybody knows their birthday, and I might even know the other people who shares my birthday. But, I can still talk about the odds. Quote:
My youngest son explained it this way: Monty's choices of which door to choose are limited. He can't choose the door we picked, and he can't choose the door with the car. He is constrained to pick one and only one door. Thus, revealing where the car is located. If you look at the truth table I published in post 121, you can see he's right. When Monty only has a single door to choose from, switching is a winning strategy. When Monty can choose multiple doors, winning is a losing strategy. If Monty doesn't know which door has the car, his choice of doors isn't constrained: He could pick either one. If he doesn't reveal which door has the car, switching in this case is only a 1:2 odds of winning.
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#131
11-30-2010, 05:02 AM
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If you initially chose the right door, only winning strategy is not to switch.
If you initially chose the wrong door, only winning strategy is to switch. 2/3rds of the time, you choose wrong, which makes switching the better strategy. (I tried to write the shortest explanation in the thread.)
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#132
11-30-2010, 11:33 AM
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I don't knwo much about mathematical probability, but this one is simple.
You have a choice of taking what is behind 1 door, or taking what is behind BOTH of the other doors. It's NOT BOTH you say. No, but one of the 2 doors you didn't pick is empty, and you are shown which one, so switching is essentially the same as taking what is behind BOTH of the other doors. 1 in 3 times your initial pick wins. 2 out of 3 times, the winner is one of the other doors, and you have been shown which one of those is not the winner. I don't know how it's modeled mathematically, but people seem to be confused about a key point. When Monty shows you what is behind one of the doors you didn't pick, it isn't random. At least one of those two doors is not the big prize, and he has to show you that one, never the one with the big prize. So the second part of the situation, where Monty picks a door to show you, is not random.
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#133
01-02-2011, 06:28 AM
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Hasn't Marilyn Vos Savant already definitively answered this question?
Her answer seemed logical and made sense; as I thought the same when I read the question myself--because....by increasing your choice, you then increase your chances of a win. Simple math....see below for ENTIRE debate from Marilyns website: http://www.marilynvossavant.com/articles/gameshow.html [Material deleted by moderator] --------------------------------------------------------------------------------------------------------  Old School "Scholars"--and after all those people wrote in PROVING that Marilyn's theory was accurate over and over again, the men still need to shake their heads.What a shame so many "intellects" are still so close-minded...alas, the earth is still flat for them. Pity... Last edited by C K Dexter Haven; 01-02-2011 at 07:37 AM. Reason: Quoted too much material -- ckdh
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#134
01-02-2011, 07:44 AM
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Welcome to the Straight Dope Message Boards, ecco477, we're glad to have you with us.
You might want to take a look at our rules on quoting from copyrighted material: FAQ: What's the policy on copyrights?. Basically, you provided a link to the material (that's good!) and so there is no need to cut-and-paste the whole thing (which is bad, and a violation of our rules.) I understand you're new, so no problem, I've edited it out -- those who want to read it, go to the link -- and no worries. For future ref, though, you might want to read through our other rules, just to avoid awkwardness: Please read first: Rules, Guidelines, Etiquette, and Technical Issues
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#135
07-18-2012, 12:52 PM
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You're getting the 1/3 in the first example by taking this chart: G G G B B G B B And on hearing that there's one boy, you cross off GG and have: G B B G B B Ah, so there's a 1/3 chance they are both boys, and a 2/3 chance there's one boy and one girl. So there's a 2/3 chance the other child is a girl. This should mean you can, on hearing the gender of one random child, predict the gender of the other with 2/3 accuracy. You can't. Go ahead, get a friend and go test it. (There's a nifty iOS dice app that includes boy/girl dice in case you don't actually want to do lots of reproducing and wait 18 months or more for results). You'll find that if try to predict the gender of the other child by picking "opposite gender to the one I know about", you'll be right 1/2 of the time, instead of the 2/3 the probabilities tell you. Keep that in your head for a moment. Now you get to your second part. You are told "The older one is a boy". Ah ha, you say. Now you started with: O Y (O = Older, Y = Younger) G G B G G B B B and can cross off G G and G B, leaving you: B G B B A 50/50 chance. You can now predict with actual statistical accuracy. "Well, I got more information this way", you say. But actually, in the first case, you got the same amount of information. Because we can just make that chart "The one you first learned the gender of is a boy". 1 2 B G B B And we're back to our 50/50. There is always an order, even if it was not explicitly stated in the problem. If you really want to ignore order, you don't have four options. You have three. Two Girls. One of each. Two Boys. B G and G B are the same when ignoring order. You can't set up your sets as ordered and then ignore order when considering them. Back on page 1, a poster comments that "Sandy is a girl and Chris is a boy" is not the same as "Chris is a girl and Sandy is a boy" are not the same. But as far as probabalities are concerned, they are. Adding the names is a distraction that sounds logical, but isn't mathematically correct. It's like saying that using one sparkly die and 4 white dice in Yahtzee changes the odds. Last edited by BrianSK; 07-18-2012 at 12:54 PM.
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#136
07-18-2012, 04:23 PM
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Sorry, but nope.
When you list your "unordered" possibilities, you're assuming each set is equally likely. Saying the possible outcomes are Two Girls. One of Each. Two Boys. isn't wrong, but listing them that way doesn't mean the chance of each occurring is one-third each. It's 1:2:1. (This is a zombie - I just didn't have any good, new zombie jokes to use.) Last edited by W0X0F; 07-18-2012 at 04:24 PM.
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#137
07-18-2012, 04:52 PM
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Since my point was that saying that logic doesn't actually work, we're not really disagreeing here.
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#138
07-18-2012, 05:12 PM
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We start with four equally likely possible outcomes:
On the other hand, if you're only told that at least one child is a boy, that only rules out outcome 4, and you're left with three equally likely outcomes. The probability that both children are boys is 1/3. What's so goddamn hard about this?
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#139
07-18-2012, 05:23 PM
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No, either way works to give two-thirds
 ne-third logically.I'm not sure how you're coming up with 50% as the correct answer using your iOS dice app. It sounds to me like you're saying that given a boy, the next child has a 50-50 chance of being a girl. That's absolutely correct (assuming the odds are 50-50 for any given birth, which isn't quite true in reality). But then you're the one who has imposed order by implicitly assuming the boy is born first. That is, you've actually eliminated two of of the initial four possibilities: GG and GB - not just the GG case. Am I misunderstanding what you were trying to do there? Last edited by W0X0F; 07-18-2012 at 05:24 PM. Reason: More or less ninja'd by ultrafilter
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#140
07-19-2012, 12:26 PM
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#141
07-19-2012, 12:37 PM
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That's just it though - you don't know whether the boy is older or younger than his sibling and that provides different information than if you do know whether he's older or younger. Look at ultrafilter's four cases. They're are equally likely, correct? If you know the boy is older (or younger), you eliminate two of the possibilities and you're left with a 50-50 chance as you say. But if you only know one is a boy and not whether he's older or younger, then you've only eliminated one of the possibilities - the girl-girl one - and you're left with two of your remaining three possibilities having a girl as the remaining sibling, i.e. two-thirds chance.
Last edited by W0X0F; 07-19-2012 at 12:39 PM.
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#142
07-19-2012, 12:46 PM
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#143
07-19-2012, 01:32 PM
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Yes, but as BrianSK was alluding to upthread, we always have some form of identifying information. What's the difference between the following two sets?
1. The oldest child is a boy and the youngest child is a boy. 2. The oldest child is a boy and the youngest child is a girl. 3. The oldest child is a girl and the youngest child is a boy. 4. The oldest child is a girl and the youngest child is a girl. ---------------------------------------------------------------------------- 1. The one I know is a boy, the one I don't know is a boy. 2. The one I know is a boy, the one I don't know is a girl. 3. The one I know is a girl, the one I don't know is a boy. 4. The one I know is a girl, the one I don't know is a girl.
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#144
07-19-2012, 01:36 PM
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-------------------------------------------------------------------- Here's an even stranger example of how conditional probabilities depend on exactly what information you're conditioning on. Suppose that we're looking at families with two children, and we want to know the probability that a family has two boys given that they have at least one boy born on a Tuesday. As is standard, we assume that everything is uniformly distributed over the appropriate category and independent of everything else. In this case, there are 196 different equally likely outcomes, so it's not all that much fun to write them all out, but that's what computers are for. Here's R code to compute the desired probability: Code:
days <- c("Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday")
sexes <- c("Boy", "Girl")
events <- data.frame(expand.grid(sexes, sexes, days, days))
names(events) <- c("Sex1", "Sex2", "Day1", "Day2")
index <- (events[, "Sex1"] == "Boy" & events[, "Day1"] == "Tuesday") | events[, "Sex2"] == "Boy" & events[, "Day2"] == "Tuesday"
events.good <- events[index, ]
index2 <- events.good[, "Sex1"] == "Boy" & events.good[, "Sex2"] == "Boy"
p <- nrow(events.good[index2, ]) / nrow(events.good)
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#145
07-19-2012, 01:43 PM
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#147
07-19-2012, 02:18 PM
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Since that's just a convoluted way of saying that at least one is a boy, I know the gender of the one that is a boy.
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#148
07-19-2012, 02:23 PM
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In particular, there is no "child that you know". The language of the original problem that we're talking about (if it's the one I think) is: You have been told this family has a daughter. It's tempting to imagine there's one particular child — the oldest one, the blond one, the one whose name begins with a vowel, something — being revealed as a girl. But in fact the sentence only means, "the number of girls is at least 1".
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#149
07-19-2012, 02:42 PM
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It's sometimes easier to think in terms of absolute sets rather than probablities, so try this one on for size: There are exactly 1000 families with two children in my hometown. I have indeed verified that in exactly 750 of them, the two children are not both girls. Of those 750 families, how many do you think contain two boys?
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#150
07-19-2012, 02:59 PM
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1. The child whose gender I know is a girl, the other child is a boy. 2. The child whose gender I know is a girl, the other child is a girl. 3. The child whose gender I know is a boy, the other child is a boy. 4. The child whose gender I know is a boy, the other child is a girl. And the probability of the other child being a boy is 0.5 Only if the child whose gender you reveal was NOT chosen randomly can you come up with a probability of 1/3 for the second child being a boy. The wiki article actually has an explanation of the assumptions necessary for conditional probability to apply to this problem.
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