Riddle me this…
Your correspondent is headed today for reaches afar, where connectivity is uncertain… so these missives may well be more “roughly” these next ten days or so than “daily.” To keep readers amused in the meantime, a puzzle– or more accurately, a paradox…
Readers (and viewers of the film 21) will recall “The Monty Hall Problem”:
Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?
The answer, readers will recall, is that it is: Because there is no way for the player to know which of the two unopened doors is the winning door, one might assume that each door has an equal probability and conclude that switching does not matter. But the player should switch– doing so doubles the probability of winning the car from 1 in 3 to 2 in 3. That’s to say: switching is only not advantageous if the player initially chooses the winning door, which happens with probability of 1 in 3. With probability 2 in 3, the player initially chooses one of two losing doors– one of which is then eliminated (making the other a 1 in 1 chance… i.e., the winner). So the total probability of winning when switching is 2 in 3. (See the answer worked out here.)
Now to the new puzzle, “The Two Envelopes Problem” (Hint: it’s another Bayesian riddle):
The setup: The player is given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. The player may select one envelope and keep whatever amount it contains, but upon selection, is offered the possibility to take the other envelope instead.
The switching argument:
1. Denote by A the amount in the selected envelope
2. The probability that A is the smaller amount is 1/2, and that it’s the larger also 1/2
3. The other envelope may contain either 2A or A/2
4. If A is the smaller amount, the other envelope contains 2A
5. If A is the larger amount, the other envelope contains A/2
6. Thus, the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2
7. So the expected value of the money in the other envelope is:
8. This is greater than A, so swapping is favored
9. After the switch, reason in exactly the same manner as above, but denote the second envelope’s contents as B
10. It follows that the most rational thing to do is to swap back again
11. This line of reasoning dictates that envelopes be swapped indefinitely
12. As it seems more rational to open just any envelope than to swap indefinitely, the player is left with a paradox
The puzzle: The puzzle is to find the flaw, the erroneous step, in the switching argument above. This includes determining exactly why and under what conditions that step is not correct, in order to be sure not to make this mistake in a more complicated situation where the misstep may not be so obvious. In short, the problem is to solve the paradox.
More background– and the answer(s)– here.
Extra Credit: The 23 Toughest Math Problems
For those who finish early, Plan 9 from Outer Space, the epic Ed Wood, Jr. epic (aka, “the worst movie ever made”) is being hosted in its 78-minutes-of-black-and-white-incoherence for free here on Google Video. (Hint: there’s nothing Bayesian about it.) Watch, too, for the remake— 9.9.09… just in time for the 50th anniversary of the original, and including Conrad Brooks, the sole surviving member of the original cast/crew.
As we count our change, we might say the magic word and collect $100, as it’s the birthday of Groucho Marx, born this date in 1890.
From the moment I picked your book up until I laid it down I was convulsed with laughter. Someday I intend reading it.
– To S.J. PerelmanPLEASE ACCEPT MY RESIGNATION. I DON’T WANT TO BELONG TO ANY CLUB THAT WILL ACCEPT PEOPLE LIKE ME AS A MEMBER
– telegram to the Friar’s Club
