A well-known probability riddle is “I have two children. [At least] one is a boy. What is the probability I have two boys?”; the answer is 1/3.

Richard shares a variation (imagined by Gary Foshee) which I find counter-intuitive: “I have two children. One is a boy *born on a Tuesday*. What is the probability I have two boys?” (Assume uniformity over genders and days of the week.) This is easy to solve, using the method of your choice (1, 2).

Spoiler alert: the answer is 13/27. The addition of a seemingly useless piece of information makes a big difference, which surprised me.

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This entry was posted on 03/06/2010 at 15:48 and is filed under Mathematical games. You can follow any responses to this entry through the RSS 2.0 feed.
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21/06/2010 at 04:09 |

I don’t believe that this is right, even the first formulation giving 1/3 as the odds of the man having 2 boys. As we know that he has at least one boy, then there MUST be a 50:50 chance that the remaining child is a boy (or a girl). If the answer was really only 1/3 for the 2nd child being a boy, then there is a 2 out of 3 chance that that child is a girl. If someone selected a group of 100 parents who have 2 and only 2 children of whom at least one is a boy, would you be happy to say that 66% of those parents had a girl as the second child? I don’t think so.

The logical impossibility of this idea can be clearly seen if we someone were to then select a group of 100 parents who have 2 and only 2 children of whom at least one is a GIRL. According to the idea presented, 66% of this group would have boys as the 2nd child. But this is not in accordance with the first result.

As for the Tuesday nonsense, if there was anything real to this at all, the whole of logic in this universe would be vastly different. We would be able to make intelligent bets on whether unknown children, siblings etc were male or female on the basis of knowing something on an unrelated subject. If you could do this, there would be massive use of this principle in every type of investigation, intelligence reporting, acturial and detective work. Has anyone ever seen a single exploitation of this principle used in the real world?

24/06/2010 at 10:05 |

I’m perfectly happy with the final statement of your first paragraph. Take 400 couples of parents with exactly 2 children each. On average, 100 have 2 boys, 100 have 2 girls and 200 have a boy and a girl (100 had a boy then a girl, and 100 had a girl then a boy). If you only keep those who have at least one boy, you are left with 300 couples of parents. 200 of those have a boy and a girl, which gives 66%.

I don’t understand your second paragraph.

If you want more details, you can read Wikipedia’s article on this. They have a section on “a boy named Jacob”, which is comparable to the “boy born on a Tuesday” issue.

06/02/2012 at 22:02 |

If you have two kids and one of them is a boy, what are the odds that you will have two boys? This is a well-known problem with an unexpected answer. To answer it, lay out the variables:

BB BG GB GG

The problem told you that one child was a boy, so you can eliminate the option of two girls, leaving you only three options:

BB BG GB

Of those, only one option gives you two boys. By this logic, the odds would be 1/3 that if you had two kids, and one was a boy, that both would be boys. Let’s boil out all the extra info though, and look at what the problem is really asking. At its core the question simply asks, if you have a child, how likely is it that it will be a boy. Lay out the variables now:

B G

Clearly, that is a 50% chance, so why is it that we get two disparate answers? Why do math aficionados get 1/3, while everyone else comes up with ½? It all comes down to a flawed assumption that has been taken for granted. The assumption is that we must know whether John was born before or after Sue, yet this assumption isn’t applied to whether John is born before or after Jacob, or whether Sue is born before or after Sally. There are two ways to fix this flaw. 1: apply an importance in birth order to all pairs:

B1B2 B2B1 BG GB G1G2 G2G1

Subtract variables with no boys at all:

B1B2 B2B1 BG GB

Solve: 2/4 = ½

The second possible solution is to remove the importance of birth order from all pairs. It doesn’t matter whether John was born before Sue, only that John and Sue are siblings:

BB BG GG

Eliminate all options with no boys:

BB BG

Solve: ½

By simply applying variables properly, we get a logically sound answer. This highlights the need to question everything, and explains why mathematicians frequently come up with many answers to the same questions.

Gary Froshee imagined a similar problem, also with the same flaw. I have two children. One is a boy born on a Tuesday. What is the probability I have two boys? Lay out the variables:

BtueBmon BtueBtue BtueBwed BtueBthu BtueBfri BtueBsat BtueBsun BtueGmon BtueGtue BtueGwed BtueGthu BtueGfri BtueGsat BtueGsun GmonBtue GtueBtue GwedBtue GthuBtue GfriBtue GsatBtue GsunBtue BmonBtue BwedBtue BthuBtue BfriBtue BsatBtue BsunBtue

There are 13 options that provide 2 boys, out of 27 given options. Gary used this as proof that the more variables you add, the closer an equation gets to 50%. After looking at the first problem in a corrected light, it would turn his notion on its head, but again, this has the same flawed assumption. It erases a repeated variable where both boys are born on Tuesday. It assumes that birth order matters unless two boys are born on the same day. Let’s correct this as we did with the previous problem:

BtueBmon BtueBtue BtueBwed BtueBthu BtueBfri BtueBsat BtueBsun BtueGmon BtueGtue BtueGwed BtueGthu BtueGfri BtueGsat BtueGsun GmonBtue GtueBtue GwedBtue GthuBtue GfriBtue GsatBtue GsunBtue BmonBtue BtueBtue BwedBtue BthuBtue BfriBtue BsatBtue BsunBtue

The first option is to replace the missing variable to account for birth order. The result gives us 14/28, or ½. The second option is to remove to variable of birth order entirely:

BtueBmon BtueBtue BtueBwed BtueBthu BtueBfri BtueBsat BtueBsun BtueGmon BtueGtue BtueGwed BtueGthu BtueGfri BtueGsat BtueGsun

We get 7/14 or ½. It turns out there is no magical value in adding unimportant information. It neither improves nor erodes the accuracy of the equation, it merely complicates it.

17/03/2012 at 13:17 |

While the two responses above that get 1/2 as the answer are wrong, that doen’t mean the answer itself is wrong.

First, why they are wrong. When they count families with at least one boy, they count four of them: two where the older child is a boy and the younger is either a boy or a girl, and two where the younger child is a boy and the older is either a boy or a girl. The problem is, that of the four family types BB, BG, GB, and GG, they only counted three of them but counted up to four. They did this by counting BB twice, and GG not at all.

But the original argument makes a mistake as well. When a family has two boys but you know about only one gender between the two, that gender has to be “boy.” But if they have a boy and a girl, it could be either. So while the chances a family is BG are 1 in 4, the chances a family is BG *AND* you know about the boy this way are only 1 in 8. That makes the answer (1/4)/(1/8+1/8)=1/2.

The exact same reasoning gives 1/2 as the answer to Gary Foshee’s version. One case in 196 has two Tuesday Boys, 12 have a Tuesday Boy and a different-day boy, and 14 have a Tuesday Boy and a girl. But on only the first is it required that the parent tell you about the Tuesday Boy; in all the others, the chances are 1/2. That makes the answer (1+12/2)/(1+12/2+14/2)=1/2. And the reason the 13/27 answer seems absurd, is because it is based on the un-implied assumption that Gary Foshee was required to tell you about a Tuesday Boy.

The principle behind this answer was first published in 1889 by a French Mathematician named Joseph Bertrand, and it is known as Bertrand’s Box Paradox. It is the reason, for example, that the answer to the famous Monty Hall Problem is not what people initially expect. It seems they usually apply the same method of counting cases that Robin Ryder and Gary Foshee did. In games of cards, it is known as the Principle of Restricted Choice, because the options where the information you learned was based on a choice have a lower probability than you would expect by merely counting cases.

07/07/2014 at 19:58 |

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now I am using net for articles or reviews, thanks to web.

10/12/2016 at 17:58 |

Say I have four boxes, each containing two coins. One contains only gold coins, one contains only bronze coins, and the other two contain one of each type.

Q1) You pick a box at random. What are the chances it contains only one type of coin? This supposed to be easy: 1/2.

Q2) Say I peek in the box, and tell you that there is (at least) one bronze coin inside. What is the probability that there are two? Note that this question is now identical to Robin Ryder’s “well-known probability riddle.” He says the answer changes, to 1/3.

Q3) What if I had said that there is (at least) one gold coin in the box. Shouldn’t the probability change to 1/3 in this case, also?

Q4) What if, instead of naming a type, I reach in, and pull a out a coin in my closed fist. Now what is the probability that the coin still in the box is the same type?That is, instead of naming “bronze” in Q2, I would grab a bronze coin, and instead of naming “gold” in Q3, I would grab a gold coin. So this question is the same as Q2 adn Q3, and Robin Ryder said the answer is 1/3.

It seems that my action in Q4 has to change the probability no matter what I picked. But such a change can only happen if useful information is gained, and it has not. This is the paradox that Bertrand referred to.

The resolution of the paradox is that naming a type of coin does not allow you to infer that I was forced to name that type. If the box did, indeed, have two of the same type, then I was forced. But if it had both types, I had to choose, and you can only assume I chose randomly. That makes the answer to Q1, and Robin Ryder’s riddle, 1/2.

23/01/2017 at 22:27 |

196 parents of two children are selected to present a form of Gary Foshee’s question at a puzzle convention in their home town. By sheer coincidence, every one of them has a different combination of (gender,day)x(gender,day) for their two children.

Q1) Obviously, not all can ask it about Tuesday Boys, as Gary Foshee did. In fact, only 27 can. But 26 of those could ask a different question. How many do you expect to?

Q2) Is your expected number different for any description of a child, such as “Girl born on a Thursday?” If so, why?

Q3) How many of those, whom you expect to ask about have Tuesday Boys, have two boys?

Q4) How many of those who ask the question any about any specific description have two of the same gender?

Q5) What is the best answer to Gary Foshee’s question?

Answers, in a slightly different order:

A2) You have no information that would allow you to expect the number of questions to be different for different descriptions. Since there are 14 descriptions, and 196 questions, you should expect 196/14=14 questions about each description.

A1) 14. Not 27.

A4) You have no information that would allow you to expect the fraction of same-gender families to be different for different questions. Since half of the entire group has same-gender families, you should expect 1/2 of each group to have same-gender families also.

A3) 7 of 14. Not 13 of 27.

A5) 1/2.

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