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Think you understand Monty Hall? Try the Tuesday boy problem.
141–150 of 152 posts
Re: Think you understand Monty Hall? Try the Tuesday boy problem.
#142I found this post confusing and ambiguous, so I restated it in simpler terms, with pictures: http://mikeschiraldi.blogspot.com/2011/11/tuesday-boy-proble...
Thank you for this. I slugged through the whole original article and felt like I was being beat up with words. However, I still fail to comprehend how the "at least one is a boy" quirk maths out to a 1 in 3 chance that his second child is also a boy. Taken literally, it does. I understand that, in a set of data, GB is different from BG. But for the sake of our comparison, the order the children were born in doesn't m…
You could equally well distinguish between BG and GB based on birth weight. Age has nothing to do with it. If the order of the children doesn't matter, the odds of conceiving a BG combination are twice as large as conceiving either a BB or a GG combination. The distinction is made so all combinations carry the same weight, which makes it easier to illustrate the solution.
As another alternative, you could distinguish between BG and GB based on a non-accidental property, like the alphabetic ordering of their names (assuming each starting letter is equally likely and no two siblings have the same name, both of which are probably false in practice).
Re: Think you understand Monty Hall? Try the Tuesday boy problem.
#143Earlier quoted context omitted.
You're making the exact mistake the author is cautioning against, which is assuming the day doesn't matter. Write out all the possibilities (see my other comment in this thread), eliminate the dupe, and you get 13/27.
Did you read my post carefully? I do get 13/27, when presented with the information from a neutral third party - i.e. filter for all 2 child families with one son male/Tuesday, what is the chance the other is also male/Tuesday. But the fact that the father voluntarily offered up the information changes the probability distribution. We can assume he's selecting one of his children at random, and revealing their birthd…
The probability that he would make that statement given what his children are is a different question than the probability that the other child is also a boy given that he made that statement. Your conditional probabilities are correct, but your conclusion about what they mean for the original question is flawed.
N = Total number of ways to have 2 children over 7 days = 14^2 = 196
B = Two sons born on Tuesday.
O = Exactly one son born on Tuesday.
A = At least one son born on Tuesday.
T = Two sons.
S = The statement.
P(B) = 1/N = 1/196 = 0.005
P(O) = 26/N = 26/196 = 0.133
P(A) = P(B) + P(O) = 27/N = 27/196 = 0.138
P(S|O) = .5
P(S|B) = 1
P(S) = P(S|B)P(B) + P(S|O)P(O) = 1.0 * 0.005 + 0.5 * 0.133 = 0.0715
P(B|S) = 1/27 = 0.037
P(O|S) = 26/27 = 0.963
P(A|S) = 27/27 = 1.0
P(T|S) = 13/27 = .481
P(S|A) = P(S|B)P(B|A) + P(S|O)P(O|A) = 1.0 * 0.037 + 0.5 * 0.963 = 0.519
Re: Think you understand Monty Hall? Try the Tuesday boy problem.
#144P(two boys, at least one on tuesday) = 1/4 (1 - 6/7 6/7). The 1/4 is the probability of two boys, and the 1 - 6/7 6/7 is the probability at least one is born on tuesday.
P(one boy, one girl, boy on tuesday) = 1/2 * 1/7. The 1/2 is the probability of having one boy, one girl, and the 1/7 is the probability that the boy was born on tuesday.
The man can make his statement if either of the above is true, and they are disjoint, so their sum gives the probability that he has two children, one a boy born on tuesday.
The ratio of the first to the second is the probability that both are boys. Plug in the numbers and you indeed get the 13/27 the article claims.
Re: Think you understand Monty Hall? Try the Tuesday boy problem.
#145I found this post confusing and ambiguous, so I restated it in simpler terms, with pictures: http://mikeschiraldi.blogspot.com/2011/11/tuesday-boy-proble...
Thank you for this. I slugged through the whole original article and felt like I was being beat up with words. However, I still fail to comprehend how the "at least one is a boy" quirk maths out to a 1 in 3 chance that his second child is also a boy. Taken literally, it does. I understand that, in a set of data, GB is different from BG. But for the sake of our comparison, the order the children were born in doesn't m…
Probability is much less meaningful when you're not talking about large groups of repeated trials.
In the context of the article, imagine that you didn't do this just one time, but that you asked 100 fathers about the compostion of their children.
The question posed in the article is essentially, "Of those fathers who responded 'I have one son,' (which is likely 75 of the 100), how likely is it that they have another son, (which is likely 25 of that group of 75, or 1/3).
When the article talks about the father standing next to one of his children at random and the probability of another son being 1/2 at that point, it helps to imagine those same 100 fathers all standing next to their children. Of that group, you're not eliminating the fathers standing next to girls based on the way the situation is posed.
The English words used to describe each case make it much less clear which group of 100 people we're talking about.
Also, when we talk about one father and not a group of fathers, the 1/3 or 1/2 number is much less meaningful. This is where insurance companies make their money (ideally). It's impossible to predict whether a single person will die in a car accident over their lifetime, and any number is essentially a guess. But it's very easy to predict that, say, 1 in 50,000 people will.
Re: Think you understand Monty Hall? Try the Tuesday boy problem.
#146I finally understand it.
Here are my events (taking 2 children as given):
B = I have a boy
BT = I have a boy and he was born on Tuesday
2B = I have 2 boys
B+G = I have a boy and a firl
Pr(2B|BT) = Pr(BT|2B)Pr(2B)/c
Pr(B+G|BT) = Pr(BT|B+G)Pr(B+G)/c
where c = Pr(2B|BT) + Pr(B+G|BT)
Pr(BT|2B) = 1/7 + 1/7 - 1/49 = 13/49
Pr(2B) = 1/4
Pr(BT|B+G) = 1/7
PR(B+G) = 1/2
c = 13/49 * 1/4 + 1/7 * 1/2 = 27/49
Pr(2B|BT and 2C) = (13/49 * 1/4) / (27/49) = 13/27
Pr(B+G| BT and 2C) = (1/7 * 1/2) / (27/49) = 14/27
It's much more likely you have a boy born on Tuesday given that you have 2 boys (rather than you have a girl and a boy) so it's more likely that you have 2 boys given that we know you have a boy born on Tuesday.
Re: Think you understand Monty Hall? Try the Tuesday boy problem.
#147I stopped caring about 'paradoxes' ever since I understood there's no isomorphism between pure mathematical concepts and human language.
The paradoxes are important in highlighting the ambiguities in our language.
The annoying aspect is when someone uses a paradox to show off how mathematically clever they are, instead of to show how ambiguous language is. As XKCD illustrates: https://www.xkcd.com/169/
Re: Think you understand Monty Hall? Try the Tuesday boy problem.
#148I've been thinking about this for an hour, and I'm now convinced that the author is wrong. The fact that we found out about one of the children from the father means that all probabilities are not equal, even though they're treated here like they are. The difference is between the information being offered, and determined independantly. I'll do this with the boy/girl problem, for simplicities sake. If we ask a man if…
The distinction you're making is in your first example both the man is arbitrary, as well as the day of the week.
In the second example, only the man is arbitrary.
With those assumptions what you say is correct.
However, the question is "You meet a man on the street and he says, “I have two children and one is a son born on a Tuesday.”" and not "You meet a man on the street and he tells you he has two children, that one is a son and he tells you what day of the week he was born". So the day is not arbitrary, it's specifically Tuesday.
Re: Think you understand Monty Hall? Try the Tuesday boy problem.
#149Earlier quoted context omitted.
Did you read my post carefully? I do get 13/27, when presented with the information from a neutral third party - i.e. filter for all 2 child families with one son male/Tuesday, what is the chance the other is also male/Tuesday. But the fact that the father voluntarily offered up the information changes the probability distribution. We can assume he's selecting one of his children at random, and revealing their birthd…
The probability that he would make that statement given what his children are is a different question than the probability that the other child is also a boy given that he made that statement. Your conditional probabilities are correct, but your conclusion about what they mean for the original question is flawed. N = Total number of ways to have 2 children over 7 days = 14^2 = 196 B = Two sons born on Tuesday. O = Ex…
The problem I have is that you're assuming that each of the 27 outcomes has equal probability, but the chance that we received the information in the way we did makes that a flawed assumption.
The best analogous problem is the German Tank Problem:
http://en.wikipedia.org/wiki/German_tank_problem
If we have destroyed a single German tank with a serial number 100, we can at least to begin to make an estimate on the size of the German force, by basically asking the question:
"If they have 200 tanks, what was the chance one we randomly killed was this serial number? 500? 1000?"
And then combining n=100->infinity to form a probability distribution. You can then say that there is an x% chance that Germany has 500 tanks, and a y% chance that Germany has 10,000 tanks.
However - if instead, we asked 'does there exist a German tank with a serial number 100', and the answer is yes, this does NOT tell us anything past the fact that their tanks are >= 100 in number.
We have the exact same information, but how it was determined changes the outcome drastically.
Does that make sense?
Re: Think you understand Monty Hall? Try the Tuesday boy problem.
#150Earlier quoted context omitted.
Yeah, what matters is the contents of the initial set of families over which we determine probability. "A man has two children, and one is a son born on a Tuesday. What is the probability that the other child is also a son?" If the man is randomly chosen from the set of all families the answer is 1/2. If the man is randomly chosen from the set of all families with a son born on a Tuesday then the answer is 13/27. The…
>If the man is randomly chosen from the set of all families the answer is 1/2. By assumption, the man has a son born on Tuesday, so this is hardly relevant. If A is a subset of B, then choosing x uniformly at random from A given that x is in B is the same as choosing uniformly at random from B.
I agree but set B is not a uniformly chosen subset of A in this case. That is the core of the trick. The rule for choosing B is intuitively uniform but actually slightly favours families with a girl and a boy over those with two boys.