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Mathematical Probability Puzzle



Your new mathematics-obsessed friend says to you, "I have two children.
One is a boy born on Tuesday. What is the probability I have two boys?"

The first thing you think to ask him is, "What the heck does Tuesday have to do with it?"
"Everything!", he replies...

So what is the probability? :D

Tim Folkerts

Super Moderator
Just to clarify, do you mean "one and only one is a boy born on Tuesday"?

And we will assume that boys are born 50% of the time (when in reality it is typically more like 53% I believe).

Tim Folkerts

Super Moderator
I'm getting 6/13 for a quick answer (if exactly one is a boy). If both could be boys born on Tuesday, then I get close to 64% chance of two boys. But I need to think about that some more before I will commit to those answers.


The probability of two boys is 1/3. Please don't ask me how. I don't have time right now to go through the process again.

Tim Folkerts

Super Moderator
I changed my mind ...

For either child, that child could be
G) a girl (1/2)
BT) a boy born on Tuesday (1/2 * 1/7 = 1/14)
BN) a boy born NOT on Tuesday (1/2 * 6/7 = 6/14)

If we reguire exactly one child is (BT), the options are
(BT)*(BN) = 6/196
(BN)*(BT) = 6/196
(BT)*(G) = 1/14 = 14/196
(G)(*BT) = 1/14 = 14/196
where the order listed is the birth order.

These are all the families that have exactly one BT. So if we looked at 196 families, there would be 6+6+14+14 = 40 with one BT. Of these 40 families, 12 have 2 boys = 12/40 = 3/10 = 30%

If you include (BT)(BT) with two boys born on Tuesday, there is one more out of 196. So there are 41 possible families, of which 13 have at least 1 BT. Which is 13/41 = 31.7%


My reasoning is as follows:

The "Tuesday Boy" can be either the first or second child.

So the other child can be either a boy or a girl born on any day of the week. So the availabel combinations are(bTu = Boy Tuesday, gMo = Girl Monday etc):

1st child	2nd Child
bTu	          gMo
bTu	          gTu
bTu	          gWe
bTu	          gTh
bTu	          gFr
bTu	          gSa
bTu	          gSu
bTu	          bMo
bTu	          bTu
bTu	          bWe
bTu	          bTh
bTu	          bFr
bTu	          bSa
bTu	          bSu
gMo	          bTu
gTu	          bTu
gWe	          bTu
gTh	          bTu
gFr	          bTu
gSa	          bTu
gSu	          bTu
bMo	          bTu
bTu	          bTu
bWe	          bTu
bTh	          bTu
bFr	          bTu
bSa	          bTu
bSu	          bTu
So there are 28 combinations above which result in 14 boys and 14 girls.

However the combination bTu is repeated so should be removed resulting in 27 total combinations 13 of which are boys and 14 girls, i.e. 13/27
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