Bob and Jane have three children. Given that one child is their daughter Mary, what is the probability that Bob and Jane have at least two daughters?

(I am also interested in wordings of this problem where additional given information such as "name is Mary" or "born on a Tuesday" are shown to alter the probability.)

Method 1:

Evidently there are eight equally probable families (here lower numbers are older children e.g.). There are 24 children who we could meet. But we met Mary, who is a girl. There are 12 girls.

(G1, G2, G3)
(G4, G5, B6)
(G7, B8, G9)
(B10, G11, G12)
(G13, B14, B15)
(B16, G17, B18)
(B19, B20, G21)
(B22, B23, B24)

Of the 12 girls we could have met (each equally probable), 9 of them are in a family with 2+ daughters. 3 are alone.

So the probability is 9/12 = 3/4.

Method 2A:

Evidently there are eight equally probable families. We met one of the seven families that contain a girl.

(G, G, G)
(G, G, B)
(G, B, G)
(B, G, G)
(G, B, B)
(B, G, B)
(B, B, G)
(B, B, B)

Of the 7 families we could have met (each equally probable), 4 of them have 2+ daughters. 3 are single-daughter families.

So the probability is 4/7.

Method 2B:

"Given a daughter" in a 3-child family is logically equivalent to "not three boys." "Bob and Jane have 3 children; the children are not all boys." There are seven equally possible families:


Four of these have 2+ girls; hence, 4/7.

Since this can be easily verified experimentally by using three coins and tallying P(2+ Heads, given 1+ Heads), answers that claim "3/4" may be up against a wall.

  • $\begingroup$ Any ideas? It will be better if you show where you actually stuck in this problem. $\endgroup$ Oct 6, 2015 at 20:29

3 Answers 3


I might miss something because it seems too easy to me. In case, please let me know.

The name of the girl we know, Mary, is irrelevant. So we have the equivalent question: "probability that Mary has at least one sister, in the two children we haven't met". As they are two, they can be (G,G), (G,B), (B,G), (B,B). In 3 out of 4 cases Mary has "at least one sister".

PS. In all the above, we assume that probability of Bob and Jane generating a male or female is equal and therefore 50% each. It might not be like that...


I don't agree with either of your methods, although the answer agrees with the first method. In the first answer, you assume that the girl you met equally likely to be any of G1 through G12; but that's not true -- each of the 8 scenarios is equally likely, so, for instance, G1 has only 1/3rd the likelihood as G13.

In the second answer, you don't use the information that you met Mary; you only use the information that there's at least one girl, which is less information.

The final answer should be 3/4, as it is equivalent to at least one of the two non-Mary children being female.

  • $\begingroup$ The text of the problem does not say that you met Mary, it simply says "given a daughter Mary." I edited my original writeup to include the parallel problem: if I flip three coins, and discard results that do not show at least one heads, then 4/7 of my experiments show two or more heads. ... not 3/4. $\endgroup$
    – user277577
    Oct 7, 2015 at 21:13
  • $\begingroup$ yes, but if you flip three named coins and then say "the coin named Larry flipped heads," the probability that you flipped at least two heads is 3/4ths, not 4/7ths. $\endgroup$
    – hunter
    Oct 7, 2015 at 22:54
  • $\begingroup$ What if the name Larry is only given after locating a heads coin, and could have just as easily been Curly or Moe? $\endgroup$
    – user277577
    Oct 7, 2015 at 23:18

This is a type of problem sometimes called a "probability paradox" where the answer is not intuitive. The information that the given daughter is named Mary seems irrelevant, but it is not.

One way to think of this is that the wording of the problem creates three types of children: 1. boys (B), 2. girls named Mary (M), and 3. girls not named Mary (G) So there are 3 * 3 * 3 = 27 ways to have 3 children; though they are not all equally probable.
Assume that the probability of having a boy is 0.5 and the probability of any girl being named Mary is 0.01, so:
P(B) = 0.5;
P(M) = (1-P(B)) * 0.01 = 0.005;
P(G) = (1-(P(B)+P(M))) = 0.495

To simplify, let's ask what is the probability of a family with only 2 children having 2 girls given that one child is a daughter named Mary. Of the 9 possible combinations of the three types of children (MM,MB,BM,MG,GM,BB,BG,GB,GG) only 5 include a daughter named Mary (MM,MB,BM,MG,GM):
P(MB) = P(BM) = P(B) * P(M) = 0.5 * 0.005 = 0.0025
P(MG) = P(GM) = P(G) * P(M) = 0.495 * 0.005 = 0.002475
P(MM) = P(M) * P(M) = 0.005 * 0.005 = 0.000025

P(2 daughters, given 1 named Mary) = (sum of combos with a Mary and also 2 daughters) / (sum of all combos with a Mary)
= (P(MM)+P(MG)+P(GM)) / (P(MM)+P(MG)+P(GM)+P(MB)+P(BM)) = (0.000025 + 0.002475 + 0.002475) / (0.000025 + 0.002475 + 0.002475 + 0.0025 + 0.0025 ) = 0.004975 / 0.009975 = 0.4987

Wow, approx 50% ! Note that as the probability of the name, "Mary", goes down, the probability of having 2 daughters approaches 0.5
As the name, "Mary", becomes more common, the probability approaches 0.333. In other words if every girl is named Mary then the answer is the same as if we did not know that one was named, "Mary".


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