Nope. Your probability in that case is 2/3. 2/3 only. Not two different numbers simultaneously. Bet with either one of them in this situation and you’ve got two-to-one odds on your side.
Yep. I do. There are 100 families - 25 boy/boy, 50 boy/girl, 25 girl/girl.
You’ll win 2/3 of the time, because when you hear ‘I have an X,’ you’ve eliminated 25 families.
Frankly, I don’t know what language this is, but from reading the comments, I see what is going on. The if (all(s(ii) == 0)) is an unfair condition.
Find the 5 girls, then figure out how many times the 6th is a girl vs. a boy.
In other words, of all the sets where sum(s) <= 1, how many have sum 0 versus sum 1?
I’ll write a simulation in c# and try it.
Here’s my code:
And here’s the output:
2999189 total girls and 6000000 total kids - 0.499864833333333 are girls
15548 6 girl families and 108859 5 girl families - 0.142826959645045 are all girls
0.857173040354955 are 5 girls and one boy
Aw, crud, roark04, I completely missed the point yesterday. Here’s the problem:
[quote=roark04]
This is when I start getting greedy. I decide if I can rip off the parents, I can rip the kids off too. I know that if I see a girl on the street, she can’t belong to a family will all boys, so the odds of winning my bet should be the same. So I start to bet every girl I run into that I can guess whether she has a brother or a sister, and subsequently guess brother. To my dismay, I’m not making any money as it appears the odds are now 50/50.[/roark04]
False. The odds are still 2/3. You’ll make money off the kids. If you see a girl on the street, knowing she’s from a 2-child family, it’s more likely her sibling is a brother than a sister.
I took your conclusion on faith without really thinking it through, and started from that point.
[QUOTE=bup]
Aw, crud, roark04, I completely missed the point yesterday. Here’s the problem:
I’m sorry, but that’s not right. Like I said, you still have the chance of running into the girl’s sister and betting and losing with her. That evens out the odds to 50/50. If you still have any doubts read this article: http://64.233.161.104/search?q=cach...en&ie=UTF-8)
It’s just a fact that if you bet with the girls on the street you’re only going to win 50% of the time. Also read my last post. Do you really think in the card experiment I laid out in the last post that you’re going to win more than 50% of the time by guessing the opposite?
The only way you’re going to win in said card experiment is if:
Do you really think those 2 assumptions are implied in the original question of:
There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal? (Answer: 2/3.)
The whole flaw in this question is that it doesn’t take into account what will happen when the family you pick has two boys. How do you know the family has at least one girl? Because the parents told you? Were they randomly picking one of their kids and telling you his/her gender? If so, the odds are obviously 50/50, as demonstrated by the above card experiment. If they are restricted to only telling you if they have a girl, then what do they do in the case of them having two boys?
See, that’s the whole thing - the only way you’re increasing your odds is by eliminating the families with two boys, which in the real world YOU CAN’T DO.
Therefore, if you walk up to random family on the street, and they tell you “we have girl,” and assuming that there is just as much chance of them saying “we have a boy,” (if they are one of those families that has a boy and a girl) then your chances of winning a bet with them is 50/50 no matter if you bet they have a girl or a boy.
Can you come up with a real world situation where you only have to make bets with the families that have at least one girl? Put another way: Can you come up with a real world situation where the ONE family you walk up to (an isolated case) is guaranteed to have at least one girl?
Probability cannot ignore the chance that the family you walk up to will have two boys. Probability cannot ignore the fact that any family you walk up to has the choice of telling you the gender of either of their two kids.
I first read that Ask Marilyn column years ago, and I just KNEW she was wrong. I just couldn’t put my finger on what it was, and I kept trying to come up with the perfect rebuttal. Thinking of it here and there. Then, about a year later, I was thinking about it, talking it over with a friend at the bar, and then it hit me. I had that feeling you get when you had an insight, but hadn’t actually put it together yet, so I talked it out, and stopped mid-sentance when I realized that… she was right!
When you pick a door, you have a 66.7% chance of being wrong. Then one of the doors (obviously a wrong one) is opened up, and you’re asked to switch. You don’t only switch doors, but also whether you are right and wrong. As the door you are already holding has a 66.7% chance of being wrong, if you switch every time, you have, in the end, a 66.7% chance of being right.
She really did not explain it very well, which is where part of the debate really comes from. It also seems to me, from discussion here, that people are reading way too much into the problem as posed - for instance, does Monty have discression on whether to open a door or not. Not only was that not part of the question (it specifically states he makes the offer), but also a staple of the show (he didn’t just ask to switch, but often threw in a guaranteed $100 or something.
I don’t think anyone doubts the conclusion of the Monty Hall problem at this point. It’s been explained in enough ways that most people can understand the correct answer. What we’re discussing is the similar girl or boy problem.
Ack. You’re correct.
I think you only need the first condition. If he won’t say anything when he has a black card in his hand, you don’t bet. But that’s quibbling.
[quote]
Do you really think those 2 assumptions are implied in the original question of:
**There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal? (Answer: 2/3.)**Yeah, I think it is. “The universe begins,” if you will, at the point you’re told they have at least one daughter. It’s a conditional probability.
To my way of thinking, the question precludes it.
But you did because you already heard they had a girl.
Correct. The families that have two girls will tell you they have a girl, and you’ll lose. The families that have two boys will tell you they have a boy, and you’ll lose. The families that have one of each will tell you either one, and you’ll win. 50/50
What does the question have to do with the real world? What does math have to do with the real world?
Is this “real world”? Go up to a family. Ask if they have two kids. Ask if at least one is a girl. Guess that the other is a boy.
It can if you phrase the question correctly.
Back to your hypothetical situation, in the land of two-childrenville, where you’re in a house, ready to ask the woman, and a girl walks in. I think I had the wrong answer before. If you see a girl, the odds are fifty/fifty. Why? Because that’s more likely to happen (twice as likely) in a house with two girls as it is with a girl and a boy.
What’s the difference? Well, in a computer simulation, I’d code your hypothetical situation as ‘look in one cell of the family - is it a girl’?
In the other way, where you ask the woman if she has at least one girl, I’d code it as ‘is the sum of girls in the family >= 1’?
That’s different information.
(I think. I’m pretty damned sure the simulations would crank out different answers. I’ll try it.)
I tried it. The results:
**Created 1000320 boys and 999680 girls
Method 1 - ask random girls the gender of their sibling
made 100000 bets
50143 siblings were girls
49857 siblings were boys
Method 2 - ask random families if they have at least one girl - if so, bet they have a boy
made 100000 bets
33077 siblings were girls
66923 siblings were boys**
This is what happens when a question is poorly formed. The discussion here reveals many underlying assumptions, and the fact that the vague language of the problem conceals contradictory interpretations. It may be best to examine how the daughters question differs from the accepted solution to the more familiar Monty Hall problem.
The Monty Hall problem–which I think everyone here agrees upon–is carefully worded to give a clear, unambiguous procedure leading to the exact decision for probability analysis. Key to that process is the fact the Monty knows exactly where the prize is and will not reveal it when he opens his first door. If, instead, he were to open an unchosen door at random, then in the 2/3 of times where my first selection is incorrect (I have a 1/3 chance of choosing the right door first time), he will reveal the prize exactly 1/2 the time, and there is no point to the decision should I stick with my first choice or switch? Thus, in 2/3 * 1/2 = 1/3 of the games, my decision will be irrelevant; switching or sticking will produce the same (losing) result.
The remaining 2/3 is split evenly between the stick and switch strategies. Therefore, if Monty opens doors randomly, choosing one strategy over the other is a 50:50 bet. To recap, in the original problem, the fact that Monty is* intelligently selecting the information he reveals* is key to understanding why probability is 2:1 in favor of the switch strategy.
Applying this reasoning to the daughters problem “There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal?” The real question, as roark04 points out, is what assumption can be made similar to the intelligent selectivity Monty Hall demonstrates in his eponymous problem?
The lynchpin lies in the passive form “You have been told…”; did the “teller” in this case:
[ul]
[li]take the first two-child family he/she found and report the sex of one of the children randomly (e.g. could he have been equally likely to report the family has a son?), or [/li][li]did he specifically weed out any two-boy families, looking specifically for one with at least one girl before reporting this family had at least one girl (i.e. did he create the information intelligently)?[/ul] [/li]I don’t think you can favor one or the other from the sketchy language itself, and if we rely on Occam’s razor the first is the simplest–and therefore the more likely–interpretation. roark04’s rephrase of the problem–“every mother I run into I ask if she has a daughter. If she says yes, I bet her a $100 that I can guess her other kid’s gender.”–is better because it is explicit about the intelligent selection in advance of the decision.
On Preview, bup’s computer analysis bears out this difference; nice work…hope this post isn’t beating a dead horse at this point…
Well stated. I get it now. I’ve always seen the ‘obvious’ interpretation as ‘we know’ (magically) the family has at least one girl. No likelihood they might have said, ‘we have a boy.’ I see it’s a valid interpretation.
bup wrote:
To put it back into terms of girls and boys, bup asks “of all the families which have at least five girls, how many have six versus five and a boy?” But that is not what Bryan Ekers asked for. He asked for the additional step, choose 5 of the kids at random, and then, if you get five girls, what is the chance that the sixth is a girl. That is what I programmed in the simulation (in Matlab, by the way).
If you look at my data, you can answer his question as well: I’ve got 108819 families with at least five girls, 15268 of which have six girls (14 percent), leaving 108819 - 15268 = 93551 families with five girls and one boy (86 percent). Those numbers agree with bup’s simulation.
To be clear, that last bit is answering bup’s question.
Actually, Bryan Ekers seems to have stated the question two different ways:
What you answered.
What I answered.
Thank you, CJJ
One last thing that I’d like to point out now that we have this all cleared up:
The original question is EXTREMELY misleading because it implies that if I meet a mother and father (in real life), and they tell me they have two kids, and one of them is a girl, then the odds are 67% that their other child is a boy. Obviously this isn’t true, but I’m sure that this is what many, many people have been led to believe. Just look at some of the previous posts, especially in the other thread that discusses this problem.
This doesn’t sound like the straight dope to me, and honestly, I think that Cecil needs to issue a retraction.
Just to firmly beat this into the ground, although the examples given so far have been puzzles, the Monty Hall problem actually has a real-world application in the card game Bridge, where it is called the Principle of Restricted Choice.
Suppose you have nine cards in the trump suit such that dummy holds A 10 x x x and declarer holds K x x x. You need to take five tricks in this suit (no losers) with the lead in declarer’s hand, how do you proceed?
Well, the opponents hold four cards in this suit, including the queen and jack. If these are split evenly between them, the right choice is to play the king, then lead to the ace. But suppose when you play the king, the opponent on your right–the last to play–drops either the queen or jack under the king. What do you do now?
The odds favor leading to the ace, but covering only with the 10 should the left-hand opponent (LHO) play low on this lead. In essence, you’re betting the right-hand opponent (RHO) is out of trump on this second lead, and by the principle of restricted choice you’ll be right about 66% of the time.
The reason for this is that, via the play up to the moment RHO plays to the first trick, you’ve restricted the possible distributions of the missing trump to 3 (approximately) equal possibilities:
LHO - Qxx, RHO - J
LHO - Jxx, RHO - Q
LHO - xx, RHO - QJ
Thus, whenever the right hand opponent plays the queen or jack, it is 2:1 he/she did this because it is the only card he/she had. This is why the principle is called “restricted choice”.
Note that this analysis assumes RHO never “falsecards”, i.e. plays the Q from Qx or J from Jx in an effort to trick declarer. This is equivalent to assuming Monty always knows where the prize is in the Monty Hall problem.
I wrote a lengthy amount of posts on this puzzle in the other thread and I won’t bore you with rehashing all of stuff again. One thing I didn’t get to do in the other thread was to comment on the evolution of this puzzle. This puzzle appears to have been around since at least the 1940s according to what I gather from Scientific America. Scientific America shot this puzzle down in 1959, as did Martin Garner, the logic puzzle guru. Anyway, before I get to the heart of my post please let be provide some reference material.
The four possible family combinations:
Column A ….Column B
1)……Girl………… Girl
2)……Girl………… Boy
3)……Boy………… Girl
4)……Boy………… Boy
Statements of Equality
*If the families are equal, then the children are not. If we are given a girl and the families are equal, then each families (#1, #2, #3) are each likely 33.3% of the time… making each girl in family #1 worth half the value of the girls in families #2 and #3.
*If the girls are equal, then the families are not. If we are given a girl and the girls are equal, then each girl has an equal chance… making family #1 50% likely to be the home of the girl we are talking about.
Yes, I totally agree Cecil’s two children puzzle was badly worded, but the fact is… it just wouldn’t be accurate if worded it correctly anyway. What makes this puzzle even worst is that at each retelling the puzzle gets more and more inaccurate.
In Marilyn’s question she stated that the Mother told of a daughter (in her real version she used a son). The answer 2/3 could only be true under the following conditions:
In Cecil’s version we are told of the family and a daughter. This not only has all the same assumptions of Marilyn’s puzzle but now we have to assume that the source knew of both children but told us about just one. If the source knew of only one of the children then that child is specific. A specific child can be either child in a same-sex family(#1, #4). This makes the children equal, the family unequal.
In **Bryan’s **example (post#2), we have parents that say as their daughter enters the room, “This is one of our two children.” In this example we can make no assumptions, the child is specific and is either the child from Column A or the child from Column B. If she is Child A then families #1 and #2 are possible; If she is Child B then families #1 and #3 are possible. She is either one or the other, not both. In either case she has an equal chance to have sister or a brother.
An example that keeps the families equal is a Census taker, who stops at a house and talks to the woman inside. He asks her if she has any kids, she says yes, “two.” He then asks if she has any daughters. If she says “no” then she must be of family #4. If she says, “yes”, then she must be from one of the first three families (#1, #2, or #3). These families would be equally possible, so the chance the other child is a boy is 66.6% because of families #2 and #3. In reality, the 2/3 rule applies only because we, 25% of the time, already know the sex of both children (family #4).
Now if before I get to ask her about having a daughter, her daughter walks to the door, the chance for family #1 is now doubled. Why? Because this event is twice as likely to happen to family #1 than either family #2 or #3. This is the same if a boy walks into the room, family #4 is twice a likely than either #2 or #3. Families #2 and #3 have only one possible child that can be a girl, and one possible child that can be a boy. Because families #1 and #4 have an equal chance to have either child enter the room and provide me with the “same” clue, the chance the other child is the same sex is 50% and the chance of it being of the opposite sex is 50%.
Anyway, my point here is that this puzzle doesn’t work in the first place and each retelling people keep “cutting corners” and make this puzzle even more unlikely.
Even if it was accurate at 2/3, it would not be necessarily true if you rewrite the puzzle. Logic puzzles are like jokes, if you mess with the setup too much the punch-line won’t work.
An real life example of this evolution of the facts is the Wright brothers at Kitty Hawk. The Wright brothers are officially credited with being the first to fly a heavier-than-air, controllable and sustainable airplane. They were not the first to fly, they missed that by over a hundred years. They were not the first to fly a plane, just one that was both controllable and sustainable. The Wright brothers record is only true if we refer to a controllable and sustainable heavier-than-air aircraft. This is the same for our puzzle, even if it were true all the variations are true only if this maintain the same conditions are the original puzzle.
Lastly, Jiffy1234 started this threat about the Two children question, but Monty Haul keeps popping up. Monty Haul gave is opinion on this in USA Today .
maybe some of you might be interested.
Fascinating… From that article, a quote by Marily vos Savant:
What’s fascinating about this is that one of her detractors, in fact the one to which she devoted the greatest amount of column space, did in fact raise that exact objection. So not only does she need lessons in humility and in critical thinking, as we already knew, she also needs lessons in honesty.
It occurred to me that an easier (for me) to get this was to realize the question, “Do you have any girls?” is equivalent to the question, “Do you have zero boys?”
Whereas talking about one child, whether ‘random’ or specified in some way, and finding out that the child is a daughter, is not equivalent.
And Monty Hall (the person) is cool. That’s an excellent point about the mind-fuck of offering more money working to convince people that they were right in the first place. It was great of him to offer his expertise on a question he understood pretty well after 28 years of offering it.