1. There is a couple with exactly two children. What are the odds that both are girls?

2. The same as #1, but now you are told that one child is a girl. What are the odds that both are girls?

3. The same as #2, but now you are told that one child is a girl named Idhitri. What are the odds that both are girls?

For ease of calculation, assume the boy:girl ratio is 50:50 in the population.

1. Consider Figure 1. There are four kinds of couples with two children where each child may be a boy (b) or a girl (g) : bb, bg, gb, gg. The area of each section is proportional to the number of couples with that combination. Notice that there are four equal possibilities, but only one, gg, of interest. Hence the odds are 1 in 4.

2. Consider the same diagram, but now that it is known that one child is a girl, the combination with no girls (bb) is ruled out. Hence the odds are now 1 in 3.

3. Now comes the interesting case: one child is a girl with a rare name. How do we incorporate this new information? Now we need a three-way classification- boy (b), girl-named-idhitri (G), girl-not-named-idhitri (g’). Since there are two children, we have 3×3 = 9 combinations, as in Figure 2 below. Notice that the areas of these 9 sets are not equal since Idhitri is a rare name. However, four of these are ruled out because at least one G is required. The case of both girls named Idhitri (GG) is doubly rare, hence can be ignored. Thus we are left with only 4 combinations, and these are approximately equal in area. Of these, two are the ones we are interested in. Hence the odds are 1 in 2 (approximately).

Notice that bG + bg’ = bg of Figure 1.

The inaccuracy of the calculation is equal to the probability of GG.There is a subtle point there, but I won’t try to explain it further.

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