Maybe it's easier to think of coin tosses. Think that we are sampling from a huge population of two coin tosses. The distribution in order of toss: HH 1/4

Maybe it's easier to think of coin tosses. Think that we are sampling from a huge population of two coin tosses. The distribution in order of toss: HH 1/4 HT 1/4 TH 1/4 TT 1/4 Problem two: Two coins were tossed, one is heads. What is the probability the other is tails? Looking at the first three rows of the distribution, weight of rows where there's one tails is 2/4 while there's only one row with both heads, whose weight is 1/4. So (2/4)/(3/4) = 2/3 probability of tails. Problem three: Two coins were tossed, first was head. So we look at the first two rows of the distribution. HT and HH each have equal weight so it's (1/4)/(1/4+1/4)=1/2. First problem: Two coins were tossed, one was tossed on tuesday and is a heads. What is the probability that the other is tails? Let's assume all coins are always tossed on random days as well. We are now interested in a sub-population where one was tossed on a tuesday and was heads. If we tabulate all combinations in the population that we are interested, x axis is first coin toss week day and y axis is second coin toss weekday: tuesday.txt mon tue wed thu fri sat sun mon HH HT tue HH TH HH HT TH HH TH HH TH HH TH HH TH HH TH wed HH HT thu HH HT fri HH HT sat HH HT sun HH HT $ egrep -o "HH|TH|HT|TT" tuesday.txt | sort | uniq -c 13 HH 7 HT 7 TH 14/27 chance that the other toss is a tails. Edit: to put it in words, the slight asymmetry is that if the other child was also born on tuesday, from the population where both children were born on a tuesday there are more boy-girl + girl-boy samples than boy-boy samples.