Here's a very nice probability puzzler -
For a scientific experiment, I walk upto a random person in a mall. I ask, "Do you have exactly two children, and one of them is a boy who was born in Tuesday?" The person answers "Yes". Assuming that he is telling the truth, what is the chance that he has two boys?
Assume all necessary clauses, eg. I don’t know the person, I can’t see his children, etc. Also note that exactly two "one of them" here means "atleast one of them", just as in normal English.
In so many ways the solution conflicts with intuition, that I can’t help feeling that this question has been designed to show that our intuitive ability fails when it comes to conditional probabilities.
If there was nothing mentioned about Tuesday
The first thought that comes to mind is that if one is a boy, surely we must find the probability of the remaining one being a boy - so the answer must be half. Well that’s simply not true, even forgetting the ‘Tuesday’ condition, the chance of both being boys is 1/3. It’s because we have 3 equally likely cases given there is atleast 1 boy –
BG, GB, BB
Hence P(2 boys | atleast 1 boy) = 1/3, simply by counting cases.
Note however, this would change, and intuition would be right, and the probability will be half, if the question was framed in slightly different manner – “Do you have exactly two children? If so, if you choose any one of them at random, is he a boy?”
The original problem with one boy on Tuesday
We have to find P(both boys | atleast one boy on Tuesday).
Again intuition says, being born on Tuesday doesn’t change the question – atleast one is still a boy, and given that, the chance of both being boys is still 1/3. Wrong again – Tuesday is very relevant to the problem in an interesting way. It makes the boy being asked about special, and almost identifies him. Solution is given below .
By enumerating all equally likely cases
There are in total 196 cases, which are all equally likely. The table below lists them all.
The answer can be simply counted off the table, it’s 13/27.
Using conditional probability
The solution above is fine, but the approach breaks down if the problem was complicated – by increasing the number of children to 3 or 4. We then wouldn’t have the luxury to visualize the sample space (i.e. all possible outcomes) easily. Fortunately, another way of solving it which does not depend on a table exists, which is in nature and so will still carry forward in more complicated cases with ease. For the mathematically inclined, the solution using laws of conditional probability is given below.
We will use some abbreviations to remove the clutter -
BT = Boy born on Tuesday
BnT = Boy not born on Tuesday
G = Girl
So for example, (BT,BnT) means first child is a boy born on Tuesday, second is a boy not born on Tuesday.
P(both are boys, and atleast one boy on Tuesday) = P(BnT,BT) + P(BT,BnT) + P(BT,BT) = 1/2*6/7*1/2*1/7 + 1/2*1/7*1/2*6/7 + 1/2*1/7*1/2*1/7 = 13/196
P(atleast one boy on Tuesday) = P(BnT,BT) + P(BT,BnT) + P(BT,BT) + P(BT,G) + P(G,BT) = 1/2*6/7*1/2*1/7 + 1/2*1/7*1/2*6/7 + 1/2*1/7*1/2*1/7 + 1/2*1/7*1/2 + 1/2*1/2*1/7 = 27/196
So P(both are boys | atleast one boy on Tuesday) = P(both are boys and atleast one boy on Tuesday) / P(atleast one boy on Tuesday) = 13/27.
Here the added condition was of being born in Tuesday. It can be seen that if we replace this with any other random condition (e.g. “Is his name Jacob?”), having probability p, the answer will turn out to be (2-p)/(4-p). This can be worked out easily if you replace all 1/7 with p, and all 6/7 with (1-p) in the solution given above.
Wikipedia has some more on this: