You should choose Door B, the very door the male Boomer came out of.
If you can’t recall the setup, please first review it. Here is only the bare bones of the trilemma:
There were three closed rooms, A, B, and C. One room had two female HR Nice Enforcers. One room had one female and one old male Boomer, from the old Personnel Department, and the third room had two old male Boomers. As you got to HR, one of the men walked out of Door B. You had to pick the door that maximized your chance of the best result.
Which is to say, of getting into a room with an old male Boomer.
The chance room B has another male Boomer, given you have seen one male Boomer exit that room, is 2/3. Thus, you should choose room B.
The Explanation
Right off, you know the man could not have emerged from the room with two women. The means one of the other rooms, A or C, has two women. The man had to come from the room with two men or the room with one man and one woman.
There are two ways a man could have emerged from the room with two men. There is only one way the man could have emerged from the room with one man and one woman. Thus, given you saw the man leave B, you know there were two ways a man could have left B, and only one way the man could have left one of the other rooms. That gives a 2/3 chance the second man is in B, too.
That’s the final answer. Which implies there’s a 1/3 chance the man is either in A or C, or 1/6 in A and 1/6 in C. Meaning the best chance to not be lectured about the importance of DIEing is to stick with B.
The error most make is to suppose the probability is 50-50, thinking one man is left in the two-man room, and none in the man-woman room. And so the chance is 50-50. But there are also no men in the woman-woman room, which most forget.
More
Exaggeration is a good technique to help solve these kinds of problems. Instead of three rooms, exaggerate and think of 200, or more rooms, with only one with two men, one with one man and one women, and all the rest standard HR ladies filling out crucial paperwork, invented just last week, to rate your performance in the company. It now seems more natural the man would have more likely come from the room with the other man. And that all those other rooms are a distraction. The only two that count are those with the men.
However, there is a better exaggeration. Still three rooms, one with two women, one with one man and a 999,999 women, and one with a million men. A man walks out of room B. It is now (I hope) obvious that there were a million ways a man could have walked out of the many-men room, and only one way in the one-man-one-woman room. The man almost surely came out of B. Therefore the best bet is to pick the room the man walked out of.
The exaggeration changes the probabilities, of course. I’ll let you discover how.
This is a version of an old “paradox” due to Joseph Bertrand, his “box paradox”. He envisioned three boxes, one with two silver coins, one with one silver and one gold, and one with two gold. You reach in box B and grab out a coin, which happens to be gold. The rest is the same.
Just like the Monty Hall, you can try it if you disbelieve it. Get three identical opaque bags and put in, say, poker chips. Anything that feels the same but has different colors or numbers. Have somebody mix up the bags. Reach in any of the bags and pull out a chip (or whatever). If the first draw is “gold”, then on about two thirds of the time you try it, the other chip will also be “gold”. If the first draw is “silver”, that trial doesn’t count. Keep track.
That’s a slow and tedious way to prove it, though. So try the exaggeration instead. Put in ten chips in each bag. One with all “silver” (the women), one with all “gold” (the men) and one with one “gold” and 9 “silver”. It will very quickly become obvious.
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There is a simpler, less math-intensive solution that gives the same answer.
Knowing that it doesn’t take two men to do a simple job, you enter room B.
Wow, Wow, Wow! Someone else used the exaggeration method! Years ago I tried to figure out the Monte Hall problem using math, but I kept getting confused. Then I tried to intuit the problem and it made sense. Imagine that there are 1 million doors, and only one of them has a new car behind the door. You get to pick one. The probability that you have picked the right door is 1 in a million. You know almost to a certainty that you’ve picked the wrong door. Then imagine that Monte Hall opens 999,998 doors and shows you that none of these other doors contain the car. Now there are only two doors left, the one you picked and the remaining one that has not been opened. Monte Hall asks you if you want to switch doors. Should you switch doors? Yes. The one you chose is almost to a certainty not the door with the new car. Therefore the other door is most likely the one with the new car because Monte Hall has ruled out 999,998 possibilities. This doesn’t prove the mathematical probability calculation, but it gets to the same answer in a very convincing way.