# Outside the Beltway

## Math Puzzle: One Boy, Born on Tuesday . . .

The gang at The Browser tease Alex BellosNew Scientist essay “Magic Numbers, Mathemagical Tricksters” thusly:

If the Monty Hall puzzle drove you mad, here’s another one sure to do the same: “I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?”

The piece doesn’t disappoint in this regard.

Gary Foshee, a collector and designer of puzzles from Issaquah near Seattle walked to the lectern to present his talk. It consisted of the following three sentences: “I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?”

The event was the Gathering for Gardner earlier this year, a convention held every two years in Atlanta, Georgia, uniting mathematicians, magicians and puzzle enthusiasts. The audience was silent as they pondered the question.

“The first thing you think is ‘What has Tuesday got to do with it?'” said Foshee, deadpan. “Well, it has everything to do with it.” And then he stepped down from the stage.

The gathering is the world’s premier celebration of recreational mathematics. Foshee’s “boy born on a Tuesday” problem is a gem of the genre: easy to state, understandable to the layperson, yet with a completely counter-intuitive answer that can leave you with a smile on your face for days. If you have two children, and one is a boy, then the probability of having two boys is significantly different if you supply the extra information that the boy was born on a Tuesday. Don’t believe me?

Seems like nonsense to me, too.  I figured, it’s 50/50 and the Tuesday was a distractor.  After all, there are only two sexes and it’s 50/50 the second will be a boy.   Not so much.

After the gathering ended, Foshee’s Tuesday boy problem became a hotly discussed topic on blogs around the world. The main bone of contention was how to properly interpret the question. The way Foshee meant it is, of all the families with one boy and exactly one other child, what proportion of those families have two boys?

To answer the question you need to first look at all the equally likely combinations of two children it is possible to have: BG, GB, BB or GG. The question states that one child is a boy. So we can eliminate the GG, leaving us with just three options: BG, GB and BB. One out of these three scenarios is BB, so the probability of the two boys is 1/3.

Now we can repeat this technique for the original question. Let’s list the equally likely possibilities of children, together with the days of the week they are born in. Let’s call a boy born on a Tuesday a BTu. Our possible situations are:

• When the first child is a BTu and the second is a girl born on any day of the week: there are seven different possibilities.
• When the first child is a girl born on any day of the week and the second is a BTu: again, there are seven different possibilities.
• When the first child is a BTu and the second is a boy born on any day of the week: again there are seven different possibilities.
• Finally, there is the situation in which the first child is a boy born on any day of the week and the second child is a BTu — and this is where it gets interesting. There are seven different possibilities here too, but one of them — when both boys are born on a Tuesday — has already been counted when we considered the first to be a BTu and the second on any day of the week. So, since we are counting equally likely possibilities, we can only find an extra six possibilities here.

Summing up the totals, there are 7 + 7 + 7 + 6 = 27 different equally likely combinations of children with specified gender and birth day, and 13 of these combinations are two boys. So the answer is 13/27, which is very different from 1/3.

It seems remarkable that the probability of having two boys changes from 1/3 to 13/27 when the birth day of one boy is stated — yet it does, and it’s quite a generous difference at that. In fact, if you repeat the question but specify a trait rarer than 1/7 (the chance of being born on a Tuesday), the closer the probability will approach 1/2.

The bold emphasis is mine, indicating what I consider to be the key to the problem.

Of course, 12/27 is pretty damned close to 1/2.  But my reasoning was wrong.

UPDATE: This is actually a more sophisticated variant of Martin Gardner’s “Boy or Girl Paradox,” which dates to 1959.

FILED UNDER: General
James Joyner is a Security Studies professor at Marine Corps University's Command and Staff College and a nonresident senior fellow at the Brent Scowcroft Center for Strategy and Security at the Atlantic Council. He's a former Army officer and Desert Storm vet. He's a widower and father of two young daughers. He earned his PhD from The University of Alabama. Views expressed here are his own. Follow James on Twitter @DrJJoyner.

1. john personna says:

It seems remarkable that the probability of having two boys changes from 1/3 to 13/27 when the birth day of one boy is stated — yet it does, and it’s quite a generous difference at that.

That’s just evidence that this is all just numeric idiocy.

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2. john personna says:

Seems like nonsense to me, too. I figured, it’s 50/50 and the Tuesday was a distractor. After all, there are only two sexes and it’s 50/50 the second will be a boy. Not so much.

You were right the first time, as a first approximation. For more assurance you wouldn’t look at BG, GB, or even BTu. You’d need actual population data on whether women who carry boys to term are more likely or not to carry another. Or if Tuesday births were not 50/50 with a margin of confidence. 😉

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3. Drew says:

I see it differently (I can be that way). The issue is understanding the question that was really asked.

If the questions was: I have one boy, he was born on Tuesday, now tell me the probabilities of having two boys your instincts (with two trivial technicalities) would be correct. The technicalities: a) if I remember correctly, the odds of girl vs boy are not 50/50, but more like 51/49. b) genetics matter; if you have a boy, you are more likely to have another boy. But your basic thrust would have been correct. Aside from my trivial adjustments, a dice is a dice. If fair, despite that 6 6’s in a row may have come up, the odds of another 6 on the next roll is 1/6th. Period.

The issue is the question that was posed, which was a slight of hand. It was not PREDICTIVE in nature, but after the fact. That is, potentially what are the probabilities of what could have happened? And so now you just get into raw numericy about what might have happened, as they did, with no notion of correlation or randomness, or prediction, which was your reaction.

Seems to me it was a cheesy, gotcha, trick question, and just numerical masturbation after that.

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4. rodney dill says:

All probabilities are 50% something either happens or it doesn’t.

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5. Franklin says:

Note to JJ: Second to last sentence should read 13/27 not 12/27.

That’s just evidence that this is all just numeric idiocy.

Wow … somebody must’ve flunked math and is still bitter about it.

Yeah, I suppose he could have stated, “assuming the chance of having a boy or girl is always 50/50 and the chance of having a child on any day of the week is equal …” but that would have been assumed at a math conference, particularly one for Martin Gardner. Speaking of whom, this guy was awesome and his type of writing made math enjoyable – look him up if you don’t know anything about him. Better yet, buy one of his books for your kid.

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6. john personna says:

That’s just evidence that this is all just numeric idiocy.

Wow … somebody must’ve flunked math and is still bitter about it.

Made it through the Cal State system’s junior level calculus, just as much as I needed for a Chem BS. But I’ll admit I didn’t enjoy all those matrix maths in 5 function calculator days.

Gardner initially gave the answers 1/2 and 1/3, respectively; but later acknowledged[1] that the second question was ambiguous. Its answer could be 1/2, depending on how you found out that one child was a boy.

Maybe my thinking is shaped by the physical sciences. “How you found out” is supposed to be about the accuracy and the precision of the measurement, not the phasing of the text.

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7. john personna says:

“phrasing”

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8. The Tuesday factoid is irrelevant. The order of the coin flips is irrelvant. This is basic probability stuff and frankly silly.

This is identical to saying I flipped a coin twice and when I flipped it on Tuesday it was heads. What are the odds that the coin came up heads both time I flipped it? The events are independent, so the probablity of both coming up heads given that one has come up heads is 50-50.

Now the only exception to this is that fact that strictly speaking genetically, if one child was a boy then your odds of the other child being a boy are not in fact 50-50, but we don’t have enough information to make a determination as to what the number should be. For instance, where this experiment conducted in China…

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9. Sorry, meant to say births rather than coin flips in the previous post. Got ahead of myself.

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10. Also, the 1/3 answer based upon options of BG, GB, or BB is sheer nonsense. BG and GB are the same option in this scenario unless you specify the boy born on Tuesday was born first or second, at which time one of these options is eliminated, leaving you with BB or the other one. Sorry but I forget the proper term for this.

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11. Grewgills says:

To answer the question you need to first look at all the equally likely combinations of two children it is possible to have: BG, GB, BB or GG. The question states that one child is a boy. So we can eliminate the GG, leaving us with just three options: BG, GB and BB. One out of these three scenarios is BB, so the probability of the two boys is 1/3.

If we are going to delve into the silliness then both GG and GB are out since the first child was a boy. That leaves us with the intuitively correct 50%.

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12. Grewgills says:

To be clear the options BG and GB indicate the order of births (or dice rolls etc), for GB to be a possible outcome the first child would have to be a girl.

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13. Franklin says:

BG and GB are the same option in this scenario unless you specify the boy born on Tuesday was born first or second

Incorrect. This has been demonstrated many times, but I’ll try my hand at explaining it. Or you could try it yourself – do a thousand sets of two coin flips. You will get roughly 25% HH, 25% HT, 25% TH, and 25% TT. HT and TH are different cases, but if you want to consider them as one case, then you have to admit the probability of getting one head and one tail is 50%. And it’s easy enough to show this: flip a coin once. What’s the chance that a second coin flip will be different than the first? 50%, obviously.

So BB, BG, GB, and GG are all equally probably, at 25%. If you consider BG and GB as “the same”, then you must admit the probability of that case is 50%.

Now you’re told that one is a boy. If it helps, it could also be stated as “at least one is a boy.” Either way, this only eliminates one case: GG. Period. The remaining cases are BB, BG, and GB, all still of equal probability. And only 1/3 of them include a second boy.

Or if you choose to consider BG and GB the “same case”, they were originally 50% possible while BB was only 25% possible, so again, BB is only 25/(50+25) = 1/3 possible.

If you had, as in your example, been told that the *first* child was a boy, then you’ve just eliminated two cases: GB and GG. In that case, the chance of another boy is in fact exactly 1/2. But that’s not what was stated.

Hope this helps. If not, I can’t help ya.

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14. Franklin says:

If we are going to delve into the silliness then both GG and GB are out since the first child was a boy.

The puzzle does NOT state that the first child was a boy. It states one of the children is a boy. It’s an important distinction – words matter.

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15. Franklin, well, no.

Either one of two cases is possible. Either the boy specified was born first in which case the only options are then BB or BG, or the boy specified was born second, in which case the only options are GB or BB. In either case, the odds of having two boys are 50-50. There are no other options.

BG and GB are invariant (that’s the word I think I was looking for) in this problem statement unless and until you specify the boy was born first or second. To say otherwise means that you must also have a B’B and a BB’ as your options to go with BG and GB, where B’ is the unspecified child, leaving you with, wait for it, a 50-50 chance once again.

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16. Drew says:

We seem to be going in circles, and I come back to my original point. JP, Charles and I seem to be on the same page. (Wow. What a Motley crew there?!)

Franklin, you can assert that at a math conference it would be assumed that the issue was to assess not predictive probability, but possible probabilities after the fact. But I think that devolves into a very simple and mechanical exercise. If that’s all this guy has, I’m really bored. Any dope can do that math.

I maintain, it was an intentional trick question and the essence of the debate they wanted to provoke is one that capitalizes on interpretation of the question’s intent, not the just raw math.

Did I just come to the defense of odo? Heh.

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17. john personna says:

So I’ve read the wikipeida page, and Franklin, I’m going to say this isn’t a math problem at all. It hinges on expectation, human nature, and psychology:

Thus, if it is assumed that both children were considered, the answer to question 2 is 1/3. In this case the critical assumption is how Mr. Smith’s family was selected and how the statement was formed. One possibility is that families with two girls were excluded in which case the answer is 1/3. The other possibility is that the family was selected randomly and THEN a true statement was made about the family and IF there HAD BEEN two girls in the Smith family, the statement would have been made that “at least one is a girl”. If the Smith family were selected as in the latter case, the answer to question 2 is 1/2.

My expectation, based not on math, but on framing of the Boy/Tuesday problem, that it was a random selection of a family, and that Boy/Tuesday were observations about that random sample.

What I’ve learned here isn’t about math, it’s about mathematicians ;-), for some reason they assume that someone has stacked the deck, and given them a non-random sample. They require that, to get the 1/3 answer.

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18. john personna says:

(Yes, this is one of those things Drew and I agree on. We normally, my mutual silent agreement, tend to keep quiet about such things.)

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19. Drew says:

“Yes, this is one of those things Drew and I agree on. We normally, [by] mutual silent agreement, tend to keep quiet about such things.”

In the upcoming reality TV show we scream and bitch slap each other.

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20. TangoMan says:

If you want to be a real stickler the problem doesn’t tell you to assume a fair coin and if you substitute real birth data then you need to contend with the fact that the coin is slightly biased towards males, in that 104-105 males are born for every 100 females born.

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21. I believe Franklin is correct. The stuff about psychology, a 51-49 split, etc. is off base in this example.

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22. Grewgills says:

Franklin et al
I realized that I had read into the the problem that the boy was child one shortly after I posted. Doh.

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23. Franklin says:

Either one of two cases is possible. Either the boy specified was born first in which case the only options are then BB or BG, or the boy specified was born second, in which case the only options are GB or BB.

I think personna’s Wikipedia excerpt (I haven’t read the whole page yet) may be identifying the assumption that I am making and that you aren’t (or perhaps the opposite – you are assuming that the person is describing the children in some order). Going back to one of my previous posts, would it change anything for you if the wording was, “at least one is a boy”? I think that more clearly eliminates only the GG case.

I would still classify this as a math problem; we used to call these “story problems” when I was a lad. But you were forced to interpret words to identify the underlying mathematical equations. That’s what this problem is doing.

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24. “I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?â€

Robert, that is the complete statement of the problem. Put into mathematical or logical terms (my college background, FWIW) the problem and the solution are exactly as I have stated. Everything else being argued depends upon some additional, fanciful information that is beyond the statement of the problem, as well as changing the ground rules midstream — which is the primary mistake made in the “solution” offered as well as by Franklin.

If you look carefully at the coin toss analogy I made you will understand why. The bottom line here is you can only get to a 1/3 answer by assuming that the births are sometimes independent events and at other times dependent events. Needless to say, or so I thought, it doesn’t work that way. I’ll admit it has been many years since I hung out at Altgeld Hall worrying about such things, so please don’t make me pull out my Probability and Statistical Inference by Hogg and Tanis to provide a more detailed exposition on conditional probablity, independent events and dependent events.

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25. And further….

The 7+7+7+6 formulation is wrong, because the first three situations resulting in 7 outcomes each treat the two births as an ordered pair, while in the last situation the ordered pair is conceptually maintained but the BB pair is discarded because a BB pair appeared in the third situation as well. Now if you label the births with a subscript of 1 and 2 respectively it isn’t hard to see that B1B2 is not the same ordered pair as B2B1, so you cannot throw it out to end up with only 6 outcomes after all.

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26. To get any other answer requires an ambiguity to be exploited in the statement of the problem, as well as confusing dependence and independence of the two births as discrete events.

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27. And Bayesians can bite me.

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28. Actually, the only thing to be learned from this is that poor problem statements can lead to ambiguous results. Well, that and a little knowledge can be a dangerous thing.

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29. Franklin says:

You are incorrect about Marilyn Vos Savant agreeing with you. She did not and does not. I was reading her column in Parade at the time.

In fact the link you give includes her first version, where it uses the “at least one of” phrase that I’ve mentioned twice and that you haven’t responded to. This is a completely unambiguous question, and clearly leads to the 1/3 answer.

By the way, I really dislike the 7+7+7+6 formulation as well, although it does work out to the correct answer. A more brute force way would be to label the possibilities of each child, using the gender and day-of-birth, BSu, BMo, BTu, … GSu, GMo, GTu, etc. There are 14×14 combinations, of which 27 include a boy born on Tuesday, blah blah blah.

I’m honestly not sure where you’re getting the “sometimes independent and sometimes dependent” argument from. This is more straightforward than that, and you’re hardly the only one around here who has taken classes in logic.

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30. No doubt. I think the application of Bayes theorem is incorrect because I don’t think the two events satisfy the definition of being independent, but I’ll have to check the textbook tomorrow to be sure.

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31. One last analogy before bed. Imagine there were two NFL games today and you don’t know how they ended, but that I tell you that the first game was won by the team that won the coin toss. Are the odds that the next game will be won by the team that wins the coin toss less than 50-50?

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32. Sorry let me rephrase that….

Imagine there were two NFL games today and you don’t know how they ended, but that I tell you that one of the games was won by the team that won the coin toss. Are the odds that the other game will be won by the team that wins the coin toss less than 50-50?

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33. Wayne says:

It is a little smoke and mirrors.

Their numbers are correct but it leads people to think it says something it doesn’t.
The probability of a having two boys where one is a boy being born on a non specific day of the week is still 1/3. It is only after actually specifying a day that it changes to 13/27. Once you specify a date you get a subset of instances of the original 1/3.

If you add instances where you have two boys with at least one boys being born on Tuesday (13) to the instances where you would have two boys with neither being born on Tuesday (36) (total of BB (49) and compare it to all the instances where you have one girl and one boy (98) you get back to your 1/3 ratios.

A clearer wording would be “What is the probability I have two boys with one boy being born on a Tuesdayâ€. Yes it is the same information but IMO puts it in better context.

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34. Tomid says:

There’s another thing to consider. Twins. If you say the first child was born on a Tuesday, then the probability of the second also being born on a Tuesday are greater than 1/7. Even if they aren’t twins, one might suggest that the sexual activities of the parents may have led the two children to be conceived on the same day of the week, slightly increasing the probability that they are both born on the same day too.

Also, if they are twins, is the probability of them being the same gender increased?

This puzzle reminds me of the game where you have to find the ball hidden under one of three cups. You pick one and the game vendor eliminates one empty cup and asks if you would change your selection to the cup she didn’t remove. There are 2 cups left, but the probability remains that your first choice is 1/3. So the other cup is 2/3 and you should swap. This puzzle is a little bit like that one, maybe.

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35. Wayne says:

Tomid
Why would the probability of the second also being born on a Tuesday be greater than 1/7 ?

As with most probability exercises there is an assumption of all else being constant.

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36. Tomid says:

Because twins are usually born on the same day.

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37. Ben Collier says:

I found this site after listening to piece about Martin Gardener on “More or Less” on BBC Radio4

The way I thought of this is as follows:
Assuming equal probability of gender and birth day
1. There are 196 equally probably distinct families (2 genders*7 days)*(2 genders*7 days)
2. In 14 families the older child is a boy born on a Tuesday
2a. Of these 14 half, 7 will have a sister
3. In 14 families the younger child is a boy born on a Tuesday
3a. Of these 14 half, 7 will have a sister
4. In 1 family both children are boys born on a Tuesday
5. So there are 27 families including one boy born on a Tuesday (14+14 – 1 for the overlap).
6. Of these 27 families 14 include 1 girl (7 older sisters, 7 younger).
7. So 13/27 families consist of 2 boys where (at least) one was born on a Tuesday

So still (7+6)/(7+7+7+6)

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38. Ben Collier says:

Oh and using the same logic, but posing, “I have two children. One is a boy born in the first half of the day. What is the probability I have two boys?” I get 3/7 – Is this analogous to Charles Austin’s NFL games and coin tosses?

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39. Wayne says:

Tomid
I didn’t articulate myself well. Yes if they are twins the probability of both being born on the same day is much higher. However the facts that twins exist (3% of time) does not increase the probability of it being another boy significantly, especially considering that of that 3% some of them would be mixed sexes.

Regardless throwing in more criteria than the problem stated is going outside of the problem. Remember the “assumption of all else being constantâ€. That fact that the parents may be ones who abort all fetuses that are not male does not apply to this math problem since those facts are unknown an is not stated. It is a good idea to keep those circumstances and others in mind when trying to predict real life outcome but it doesn’t apply on how to properly do the math of the known situation.

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40. Wayne says:

Ben
As for your 3/7, once again you are taking a subset of the BB,GB,BG problem. You are taking a larger % subset of the 1/3 BB instances than you are the GB, BG instances. One should take in account what instances they are excluding as much as what they are including.
For example as in the above problem. 13 is a larger % of 49 than 14 is of 98.

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41. Wayne says:

Ba = Boy afternoon Ga = Girl afternoon

Bm= Boy morning Gm = Girl morning

In your example you are only excluding from original set (4) of instances Ba,Ba(1) while excluding from bg,gb set(8) Ba,Gm + Ba,Ga + Gm,Ba +Ga,Ba (4). So you are excluding taking 25% of first set and 50% of second set. Or inversely you are including 75%(3 out 4) of the first set and 50% (4 out 8) of second set. Naturally doing so changes the probability.

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42. Tony says:

It is a question of semantics, like most ‘clever’ maths problems. We misinterpret the original question as “What are the chances of me having another boy”.

The answer given is the answer to the question “What are the chances of having two boys, one of whom was born on a day other than a Tuesday, given that one of them is a boy who was born (irrelevantly) on a Tuesday”.

There are then 27 (rather than 28) allowable combinations of boy / girl / day, 13 of which (rather than 14) are boys.

I should know. Have two boys, one of whom was born on a Tuesday.

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43. Lloyd says:

It seems to me that if we truly believe this ‘numerical masturbation’, as Drew so eloquently put it, then we have to accept that even if the day of birth of the given boy is NOT stated, there is STILL a probability of 13/27 that there are two boys.

What I mean by that is this:
If the question were stated as “I have two children. One is a boy. What is the probability I have two boys?â€, then surely it would be a fair assumption that the given boy was indeed born on *a* day (be it monday or tuesday or wednesday…). From there on, we can use the same mathematical proof that gave us 13/27 (since regardless of which of the seven days it is, the fraction1/7 for each day still applies). Am I missing something major here?

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44. Tomid says:

I think you’re right, Lloyd. I’ve been coming back to the original conclusion that Tuesday doesn’t matter. Somewhere it says we are to discount one option of BTu-BTu, but I don’t think we should. Count it twice, giving 14/28 probability.

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45. Tomid says:

Of the 196 permutations Sex1Day1:Sex2Day2, how many include at least one BTu? How many of those have a second boy?

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46. Tomid says:

27 and 13. But, of the 27 permutations we had a double chance of picking the BTu:Btu one. So, 14/28.

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47. leo says:

the people who are confused and can’t see why it s not just 50/50 – you are assuming that the FIRST child is a boy born on a tuesday, but the question says that ONE OF THE TWO (you don’t know which) is a boy born on a tuesday.

This is the major difference that changes the answer from 50/50 to 13/27

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48. Lloyd says:

Yeah you may be right leo… but i still agree with Tomid. I just think that in this case, BTu-Btu is different to Btu-BTu (if you get what I mean). And that even though you have no timeline to refer to the children as different people, they still can be counted as distinct entities

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49. Geoff says:

Once you know one child is a boy, then there are only four equally likely possibilities for the second child–a younger boy, an older boy, a younger girl or an older girl. In two of these cases we get two boys, so the probability of having two boys is 2/4, ie 0.5. forget about twins–one is always a little older than the other. Tuesday is a distractor. The essential info is that one child is known to be a boy.

The 13/27 answer is the right answer to a different question, which is:
Given that I have two children and that they are not both girls, what is the probability that I have two boys with at least one boy born on a Tuesday?

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50. Rob M says:

I have randomly chosen a parent who has a boy and just one other child.
What is the chance they have two boys.

I have randomly chosen a parent who has a boy born on Tuesday and just one other child. What is the chance they have two boys.

I have randomly chosen a parent who has two children.
They have been asked to describe the sex of one of their children, and have replied “boy”.
What is the chance they have two boys.

I have randomly chosen a parent who has two children.
They have been asked to describe the sex and birth date of one of their children, and have replied “boy Tuesday”.
What is the chance they have two boys.

Do we think that Monty Hall was specially chosen because he happened to have two children one a boy born on Tuesday, or was Monty Hall always going to be the speaker and is just describing his particular circumstances.

The answer to the above question will answer the 1/3 13/27 1/2 part.

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51. Wayne says:

Geoff
You have four “possibilitiesâ€ but only three instances\events. Older Boy\younger Boy or Older Girl\younger Boy or Older Boy\younger Girl. Therefore the “probabilityâ€ of having BB is still 1/3. Sort of like saying there are two possibilities of rolling number 5 on a dice three times in a role. You either you do or you don’t. That is true but the probability of doing so is not 50%.

Rob M
How do you figure Â½ for the last two examples?

If you would ask them to describe specifically their first or second child that would be true but to have them choose either child at random than no.

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52. xavier says:

I don’t agree with this analysis.
The Tuesday/Tuesday scenario should be double counted – each boy could be be born on either “this” or “that” Tuesday (ie a bit like there are two ways of getting heads and tails with 2 coins) – therefore probablitiy is 14/28 + 50%. The 1/3 hypothesis is clearly wrong – this is the Monty Hall trap.
I am not a mathematician so cannot express this in mathematical terms, but I am still sure it is right!

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