# Cheryl’s Birthday

This is going to bug me until I get it out.

The riddle of Cheryl’s birthday has been going around the internets, and mathematical professors have made videos of the solution to help illustrate for the peeps who still don’t get it. I’m here to say they got it wrong. Here’s a quick rundown:

http://www.cnn.com/2015/04/15/living/feat-cheryl-birthday-math-problem-goes-viral/

Now, other than the obvious conclusion that Cheryl is infuriating to be around, the solution appears sound. The crux of the logic is, Albert is able to say he knows Bernard does not know, because he knows the month cannot contain a unique date. And, mathematically, this first clue allows you to solve the entire puzzle by process of elimination. As a mathematical problem, it completely works. However, because the problem is a logic problem, and not a mathematical one, it fails utterly at the presumption you can rule out May.

-In a mathematical problem, you solve by accepting and working within the givens.

-In a logical problem, you must first confirm that the givens are valid.

Now let’s run down the givens. You may want for pen and paper.

Given 1: Cheryl’s birthday lies within the dates of:

5/15, 5/16, 5/19

6/17, 6/18

7/14, 7/16

8/14, 8/15, 8/17

Given 2: Albert is given the Month, and Bernard is given the Day, of the correct date.

Given 3: Albert does not know the birthday, but is able to figure out Bernard does not know.

Given 4: Bernard, who does not know, hearing Given 3, is able to deduce the correct day.

Now let’s validate the givens. Is it possible for all of them to be correct? The first impulse is of course to start with Given 3, and by doing so you can follow the math professor’s process and get to the answer. HOWEVER. Given 4 must also be considered, because they all must be true for this solution to work. Is it possible that Bernard, having only the date, cannot figure out the birthday?

Step 1: In order for Given 4 to be true, the dates that do not repeat cannot be the birthday. That eliminates 5/19 and 6/18, leaving:

5/15, 5/16, 5/19

6/17, 6/18

7/14, 7/16

8/14, 8/15, 8/17

Albert can also come to this conclusion, because as we see, Bernard does not know the answer beforehand. Cheryl would also not tell Bernard the date straight out, because that would invalidate the point of having the puzzle.

Step 2: Now, for A still not to know, 6/17 must not be the birthday. If Albert knows the month, and it is June, this is the only remaining date. Given 3 would instead read: “A already knows the answer.”  Therefore, B can safely rule out June. Remember, the claim in Given 3 is A does NOT know the answer. The list now reads:

5/15, 5/16

7/14, 7/16

8/14, 8/15, 8/17

Step 3: For B to STILL not know the answer, 8/17 cannot possibly be the birthday, because it is now the only 17 on the board. Either this invalidates Given 4 and the answer is 8/17, or we remove 8/17. Which leaves:

5/15, 5/16

7/14, 7/16

8/14, 8/15

Final Step: Now we step into Albert’s shoes. A’s claim in Given 3 is he is absolutely certain B does not know the answer, but how can this be of any use to B? Each date now has a twin on the board. This means either Bernard is able to figure out the birthday from the information given, or Albert pointing out Bernard doesn’t know is of no use to Bernard.

Here we’re guided into thinking the only possible course of events is if A was told July and B was told 16. But actually, in any scenario neither man would actually know the date and would whittle down the options as shown. They would not be able to make the claims that eventually lead to the solution. The two givens contradict each other, and the problem is actually unsolvable.

Incidentally, it’s blindingly obvious Cheryl intended the two men to pool their resources and derive the date. I’m willing to bet this actually happened and some clever maths student thought it would be an excellent solution.