The Monty Hall Problem and the Two Boys Riddle are such great riddles precisely because they exploit our tendency to make things *too simple.* That is, to (erroneously) “simplify” even to the point of discarding relevant information. But as Albert Einstein noted, “Everything should be as simple as it can be, but not simpler.”

With the Monty Hall Problem, it is easy to discard the fact that the initially chosen door has a 1/3 chance of being the winner. This fact does not change when a dud is later shown behind a different door. Still, we have a strong tendency to make the problem *too simple.* It becomes easy to assume that once we’re down to two doors, it must be a 50-50 chance.

With the Two Boys Riddle, it is easy to discard the day of the week knowledge because we know that has no impact on the gender of the other child. But again, doing so makes the problem *too simple.* This becomes clear if we add a fourth column for “Expected Number of Families with a Boy” (ignoring the day of the week information) to the table presented in the solution:

Family Type |
Number of Families |
Expected Number of Families with a Boy Born on Tuesday |
Expected Number of Families with a Boy |

BB | N | N * 13/49 | N |

BG | N | N * 1/7 | N |

GB | N | N * 1/7 | N |

GG | N | 0 | 0 |

Using these new column values gives us a different final probability calculation:

N / [N + N + N]

which equals

1/3

Clearly, “simplifying” the riddle by ignoring the day of the week information changes the problem; it makes it *too simple.*

## Seductively Too Simple

Back in 1985, my wife, Carol, was working at AT&T Bell Laboratories writing computer simulations for a new “object oriented” telephone switching system being built using a brand new programming language–C++. She was surrounded by Bell Lab’s best and brightest scientists and engineers. But even in that lofty setting, many of her colleagues made the Monty Hall problem too simple, arguing for a 50-50 outcome. They debated competing solutions for a number of days. Finally, Carol wrote a computer simulation to play the game 1 million times and analyze the outcomes. The “always switch” strategy delivered a 2/3 success rate. Only then, did all the Bell Lab-ians finally accept that strategy as correct.

I first heard the Two Boys Riddle on the Skeptics’ Guide to the Universe podcast (episode 393). Initially, I was so sure the correct answer was 1/2 that I wrote about it on Facebook. That post sparked a lively debate in the comment stream. Eventually, I wrote a computer simulation to demonstrate the correct solution. To my surprise, the simulation produced a 13/27 result, which forced me to step back and reconsider my assumptions and thought process. When I did, I realized I had made the problem too simple.

Next time you find yourself simplifying a problem, ask yourself, “Have I made this too simple?” You can often test your simplified solution by comparing it’s results to those of a non-simplified approach to the same problem. The results should agree. If not, perhaps you’ve made the problem too simple.

Let me paraphrase Andy’s response to the Two Boy Problem as a response to the Monty Hall Problem: “With the Two Boy Problem, it is easy to discard the fact that the initially chosen family has a 1/4 chance of having two girls. This fact does not change when it is later found to include a boy. Still, we have a strong tendency to make the problem too simple. It becomes easy to assume that once we’re down to three family types, each must have an equal chance.”

I can do the same sort of thing with the table Andy made for the TBP, by applying a similar approach to the MHP: Say the car is initially behind door the three doors in N games for each door. But when door #1 is chosen by the contestant and door #3 is revealed to have a goat, the expected number stays at N for door #1 and door #2, but goes to 0 for door #3. Using these values, the chances the car is behind door #2 are now N/(N+N)=1/2.

Both are examples of invalid reasoning. We must temper the initial chance for each family type in the TCP, and game in the MHP, by the chance that it would provide the information we have. In the MHP example, after the contestant chooses door #1, the4 chances Monty Hall will open door #3 are:

0 if the car is behind door #3,

1 if the car is behind door #2,

1/2 if the car is behind door #1.

So the chances the car is behind door #2 are not N/(N+N) as Andy’s logic would say if applied to the MHP, they are (N*1)/(N*0+N*1+N/2)=N/(3N/2)=2/3.

In the TCP, the chances a man would tell you he has a boy are:

1 if he has 2 boys (N cases)

1/2 if he has a boy and a girl (2N cases)

0 if he has 2 girls (N cases).

So the chances he has two boys are (N*1)/(N*1+2N/2+N*0)=N/(2N)=1/2.

I understand your analysis here, but it changes the problem. In particular, “1/2 if he has a boy and a girl (2N cases)” applies if the man has not yet answered and the man is free to flip a coin to select whether or not he reveals that he has a boy. However, in the case of the riddle, the man

hasanswered. There is no coin flip involved; he has elected to reveal he has a boy. So, we really have “1 if he has a boy and a girl (2N cases)” given he said he had a boy.