Damn maths puzzles

[oktalist]

I might be playing with fire in opening this up again, but since the thread [http://www.escapistmagazine.com/forums/read/18.199528-Poll-Difficult-Maths-question-try-it-98-of-people-cannot-get-it] was locked only at OP's request, I don't see any fundamental problem in starting my own thread. God help me.

Which is okay, since the probability is only correct if you count it twice. The probability of having two boys on tuesday is 2/28, and not 1/27.

The fact that one boy is born on a Tuesday is extraneous to the probability of having two boys. I could say I have one boy that I decided to name Matthew, and the chance of having a boy and naming him Matthew has a different probability than just having a boy but it is still extraneous information and does not affect the real question of having two boys.

No. One of them still older than the other (even if ony 20 seconds or so). So these are two diferent possibilities. We use formula for Variations, not Combinatorics.

Kinda hoping this gets seen by some other people as well who explained it like you did. It's a good explanation, but it's counting B1Tues B2Tues only once, when it should in fact be counted twice.

That was exactly what I thought. Then I asked the computer, as I usually do in these situations, to see what it thought. It thought I was wrong:

http://codepad.org/SWukSH8W

Two males in 48.03% of 137656 independent experiments.

[small]Why 137656 experiments? Because it does 1000000 experiments with "independent" random numbers, and rejects any in which neither child is both male and born on Tuesday.[/small]

If you run it a few times over, it stays around 48.14%, or 13/27.

I completely fail to understand why, but it does.

If you rewrite [http://codepad.org/Pi0jIbXq] the code to reflect the interpretation of the question that one and only one child is male and born on Tuesday, the result is an even smaller percentage, around 46%.

If you rewrite [http://codepad.org/AyczuwPf] the code to reflect the interpretation of the question that one specific child is male and born on Tuesday, the result is around 50%, and is completely obvious and pointless.

If you rewrite [http://codepad.org/IcAJndaf] the code to ignore the day of birth completely, the result is around 33%, as expected.

If you have a good, contentious maths problem/puzzle, feel free to post it. We've already had this one and Monty Hall.

 

[RYjet911]

Your code still takes both boys as a single thing because of the or. If both boys are born on Tuesday, it only goes through the if statement once.

I think it would be fixed by performing two if statements, one for each of the possibilities of B1Tues and B2Tues, rather than putting them both in a single if statement.

 

[oktalist]

Your code still takes both boys as a single thing because of the or. If both boys are born on Tuesday, it only goes through the if statement once.

I think it would be fixed by performing two if statements, one for each of the possibilities of B1Tues and B2Tues, rather than putting them both in a single if statement.

That if statement just discards the cases where neither child is a boy born on Tuesday, as we have been told that at least one child is.

You can't count a case twice, it just doesn't make sense. If, as you suggest, having two boys both born on Tuesday is twice as likely as, say, one boy born on Tuesday and one boy born on Wednesday, then that would tend to occur twice as often, and counting it twice would just make it appear to occur twice as often again.

But, B1Tues B2Tues is equally as likely as B1Tues B2Wed, which is equally as likely as B1Wed B2Tues.

Think about it. Enumerate the possible combinations:

[small](The grey is those combinations where neither day is Tuesday.)
[del]Mon Mon[/del]
Mon Tue
[del]Mon Wed[/del]
[del]Mon Thu[/del]
[del]Mon Fri[/del]
[del]Mon Sat[/del]
[del]Mon Sun[/del]
Tue Mon
Tue Tue <-----
Tue Wed
Tue Thu
Tue Fri
Tue Sat
Tue Sun
[del]Wed Mon[/del]
Wed Tue
[del]Wed Wed[/del]
[del]Wed Thu[/del]
[del]Wed Fri[/del]
[del]Wed Sat[/del]
[del]Wed Sun[/del]
[del]Thu Mon[/del]
Thu Tue
[del]Thu Wed[/del]
[del]Thu Thu[/del]
[del]Thu Fri[/del]
[del]Thu Sat[/del]
[del]Thu Sun[/del]
[del]Fri Mon[/del]
Fri Tue
[del]Fri Wed[/del]
[del]Fri Thu[/del]
[del]Fri Fri[/del]
[del]Fri Sat[/del]
[del]Fri Sun[/del]
[del]Sat Mon[/del]
Sat Tue
[del]Sat Wed[/del]
[del]Sat Thu[/del]
[del]Sat Fri[/del]
[del]Sat Sat[/del]
[del]Sat Sun[/del]
[del]Sun Mon[/del]
Sun Tue
[del]Sun Wed[/del]
[del]Sun Thu[/del]
[del]Sun Fri[/del]
[del]Sun Sat[/del]
[del]Sun Sun[/del][/small]

Only one Tue-Tue pair. It's equally as likely as any other combination. But for example, Tue-Wed and Wed-Tue are two distinct and equally likely combinations. They sum together so the outcome "one child born Tuesday and one child born Wednesday, regardless of order" is twice as likely as Tue-Tue.

This program is more instructive:

http://codepad.org/kfx9SSWr (it doesn't use random numbers, it just enumerates all the (equally likely) possibilities)

I think I'm starting to understand.

 

[RYjet911]

snip

But again, your code produces a list of Male Male entries where Boy1Tues and Boy2Tues only appear once, despite there being two possibilities. Otherwise the entire list of opposites should not count either.

If Boy1Tues and Boy2Tues is the same as it occuring the other way (Technically of the order Boy2Tues Boy1Tues, although lack of identity makes it awkward), then Boy1Mon and Boy2Tues should also be the same as Boy1Tues and Boy2Mon. Someone in the actual thread made a brilliant explanation where he assigned the boys (And possible girl) names to help show why Boy1Tues Boy2Tues should be shown twice in the list of possibilities.

The question itself also implies the boys are to be given seperate identities, as it says one child is a boy born on Tuesday. It just doesn't say which of the two children (First or second). Could either be boy 1 or boy 2, but in both cases, the other child could also be a boy born on Tuesday too. This is why Boy1Tues and Boy2Tues has to be counted twice.

 

[Rakenar]

Yeah, as Ryjet is saying : if the ordering is important, then you must count tue-tue twice, like you did with all the other mon-tue / tue-mon. This should bring you to 50%.

 

[jerrrry]

If the question is phrased "What is the probability that I have two boys, at least one of whom was born on a Tuesday, given that I definitely have at least one boy born on a Tuesday," then the correct answer is 13/27. As said, two boys born on Tuesday is only one possibility. Think of it as dice rolls. You are rolling two 14-sided dice(14 possible sex/day outcomes). There is only one possible way for you to get two of the same side(Male/Tuesday) to show in two rolls. There are two possible ways for you to get any non-identical pair. It is the same idea as why you are 50% likely to have a boy and a girl, but only 25% likely to have two boys and 25% likely to have 2 girls(assuming boy/girl equally likely).

My original problem was not with his math, but with the wording of the question. He did not phrase the question properly for what he was looking for.

 

[oktalist]

If Boy1Tues and Boy2Tues is the same as it occuring the other way (Technically of the order Boy2Tues Boy1Tues, although lack of identity makes it awkward), then Boy1Mon and Boy2Tues should also be the same as Boy1Tues and Boy2Mon. Someone in the actual thread made a brilliant explanation where he assigned the boys (And possible girl) names to help show why Boy1Tues Boy2Tues should be shown twice in the list of possibilities.

If Boy1Tue-Boy2Tue and Boy2Tue-Boy1Tue are two different outcomes, then Boy1Mon-Boy2Tue, Boy1Tue-Moy2Mon, Boy2Mon-Boy1Tue and Boy2Tue-Boy1Mon are four different outcomes. All you've done is double the number of outcomes. 26/54 = 13/27.

The question itself also implies the boys are to be given seperate identities, as it says one child is a boy born on Tuesday. It just doesn't say which of the two children (First or second). Could either be boy 1 or boy 2, but in both cases, the other child could also be a boy born on Tuesday too.

That is taken into account.

They are given identities by being first or second.

There is only one outcome out of 27 in which they are both boys born on a Tuesday.

If we are told nothing about the sex or birthday of either child, then there is one outcome out of 196 in which they are both boys born on a Tuesday. There is also one outcome out of 196 in which they are both boys born on a Monday, and so on. But there are two outcomes out of 196 in which one is a boy born on a Monday and the other a boy born on a Tuesday.

Like jerrrry alluded to, if you roll 2d6, there is only one way to get two sixes, but two ways to get a six and a five.

Yeah, as Ryjet is saying : if the ordering is important, then you must count tue-tue twice, like you did with all the other mon-tue / tue-mon. This should bring you to 50%.

The order is important, but only because it was how the children were differentiated. Thus Boy2Tue-Boy1Tue is an invalid outcome because in that notation, Boy1 is always placed before Boy2. If we use names instead of position to identify:

Alex = Boy of Monday, Bobby = Boy of Tuesday
Alex = Boy of Tuesday, Bobby = Boy of Monday
^^- these are two different outcomes

Alex = Boy of Monday, Bobby = Boy of Tuesday
Bobby = Boy of Tuesday, Alex = Boy of Monday
^^- these are one outcome, specified in two equivalent ways

Alex = Boy of Tuesday, Bobby = Boy of Tuesday
Bobby = Boy of Tuesday, Alex = Boy of Tuesday
^^- these are one outcome, specified in two equivalent ways

Now order is not important because we are using names to identify the children, instead of their position in an ordered list. (Fairly androgynous names so they could be girls, too.)

The obligatory updated Perl program:
http://codepad.org/4it7sTRj
(Gives the children names, and counts every outcome twice by reversing the order.)

[small][EDIT]The answer would only be 1/2 if the question had stated which of the two children was the boy born on a Tuesday. But then it would be completely obvious and pointless. You can try to say well, if the question had said the firstborn was a boy of Tuesday then there would be a 1/2 probability that the second is a boy, and if the question had said the secondborn was a boy of Tuesday then there would be a 1/2 probability that the first is a boy, and the probability that the question would have said either of those is 1/2, therefore the probability of getting two boys is &frac12; &times; &frac12; + &frac12; &times; &frac12; = &frac12;, but you can't do that because they are not independent variables.[/EDIT]

I roll two six-sided dice. They are both the same colour. I only let you see one of them; it has landed on 6. What's the probability that the other one is 6? Answer: 1/11.

I roll two six-sided dice. One is red, the other blue. I only let you see one of them; it is red; it has landed on 6. What's the probability that the other one is 6? Answer: 1/6.[/small]

 

[Maze1125]

As some people have mentioned, the issue is a matter of ordering.

With the question "A woman has two children, one is a boy, what is the probability that both are boys?" we have absolutely no ordering information on the two children if they are both boys.

As such, we end up with three possibilities, BB BG GB, of which, only one is the one we want, so the probability is 1/3.

If, on the other hand the question was "A woman has two children, the first is a boy, what is the probability that both are boys?" Then we do have an ordering and so not only is the possibility of GG removed, but so is GB. So we are left with just BB and BG, which gives a probability of 1/2 either way.

And it doesn't matter how you order them, any information about the children can be used to give an ordering, provided we know that information. For example the question could be "A woman has two children, one is a boy call Michael, what is the probability that both are boys?" and we'd get a probability of 1/2 again, because we could safely assume the other one's name was not also Michael, and so we could order them by name.

And that is the issue when we know one is a boy born on a Tuesday.
Knowing that means that we can order the children in all cases except when they are both boys and both born on a Tuesday. In that one single case, no ordering can exist from the information we have, and as such the total probability is almost 1/2, but skewed very slightly by the single case without an ordering, giving a final result of 13/27.

 

[oktalist]

snip

Very good explanation.

With the question "A woman has two children, one is a boy, what is the probability that both are boys?" we have absolutely no ordering information on the two children if they are both boys. [small][My emphasis.][/small]

Where the controversy seems to occur is that some people interpret that situation as meaning "a woman has two children, one is a boy, [small]and it is known which one that is[/small]." They insert a clause which wasn't there before, but which they believe is implied. And no offence to them, but they simply fail at English. But perhaps if the situation was stated as "a woman has two children, at least one is a boy," then they would not have the opportunity to interpret it erroneously. Or maybe, "one or the other is a boy."

And then explaining the answer in the way that the OP of the original thread did, is a blatant troll move. It looks wrong, because it's sleight of hand; it obfuscates the true reason of why the answer is what it is. I was convinced that the real answer was wrong until I tried to prove it was wrong by programming and realised I couldn't.