# The Birthday Problem

Several people have queried me on this and thus here’s a detailed explanation.

This is actually an ancient Math logic problem (with many versions).

The idea to solve it “logically” is like this:

(1) A person only “doesn’t know” if there are at least “two possible answers” remaining from his “point of view”.

(2) A person “knows” the “answer” if there’s exactly only “one – unique” possible answer remaining.

In our case here, the “answer” is a “birth-date”.

Also, Albert only “knows” the “month” while Bernard “knows” the “date” (because Cheryl told this to them separately).

Let’s begin to solve it now.

Step #1: Albert first says that he doesn’t know the birthday.

So, knowing the month, two avoid situation (2) above (which will make Albert know the answer instead), the answer must be a “month with no unique dates” since “Albert also says that Bernard does not know too from the initial point of view.”

From our list,

Months with unique dates are only May or June since the numbers 18 & 19 only occur “once” (unique). The months they occur in, namely “June” (18) and “May” (19) are thus “rejected” to satisfy Albert/Bernard’s not knowing.

Hence, the two remaining months “July” or “August” are the possible months now —> (3).

Step #2: Bernard says that he knows now!

Bernard, by the turn of possibilities at the end of step #1 above, realizes the answer now.

Bernard knows (3) being a logical thinker.

Look at the possible dates in July and August.

The problem states that at this stage, he “knows” the “answer”. Hence situation (2) “must” happen and thus “any possibility” which amounts to situation (1) must thus be “rejected”.

From (3), we know that the “month” is either “July” or “August” (the latest information to date).

We “reject” the date “14” because situation (1) arises since there are “two” possible answers for this date, namely, “July 14” and “August 14”.

At this stage, the “remaining” possibilities are:

July 16, August 15 or August 17.

To “know” the answer, situation (2) must arise. Since the “dates” are all “different” here, we now look at the “months”.

From the list of the three remaining possibilities above, we find that the month “August” still gives rise to situation (1) since there are “two” dates here (15 or 17) with this “same month” (August).

Hence, only “July 16” yields situation (2) of a “unique-one” answer at this last step to cause both Bernard & Albert to know Cheryl’s birthday.

Thus, Answer: July 16.

P/S:

A common remark, “solving a problem is the science of Mathematics but presenting your solution is the art of Mathematics”.

P/S 2:

Here’s a little clarification which may help edify the argument above.