The most common wrong answer was this:
Flip the coin twice. If you get two heads, go to the beach. If you get two tails, go to the mountains. If you get one heads and one tails, in whatever order, go to the desert.This approach doesn’t work, because you can get one heads and one tails in two different ways (HT and TH), each one having 1/4 probability. So with this approach, you’d go to the desert 1/2 the time.
As I indicated in the first post, there are many acceptable answers. Here’s the simplest solution, first submitted by Mike Glover. This was also the most common solution, with about 15 people suggesting it (or something very similar):
Label the options A, B, and C. Flip two coins. If you get HH, choose option A. If you get HT, choose option B. If you get TH, choose option C. If you get TT, start over.Here’s the answer I had in mind, some variant of which was submitted by 7 respondents:
Flip a coin for each pair (A vs. B., B vs. C, and A vs. C). If any option wins both of its head-to-head matches, choose that option. If no option wins both its matches, then start over.The principle behind these two solutions is the same: attach A, B, and C to equally probable outcomes, and repeat the procedure if any other outcome comes up. Mike’s solution uses two coin flips to generate four outcomes, and then repeats on one of those outcomes. My solution uses three coin flips to produce eight outcomes, and then repeats on two of them.
But I was mistaken to think every solution would have to rely on this principle. Lauren Fisher submitted a very clever solution:
Flip a coin until heads comes up. There is a one-third chance that the first head will come up on an even-numbered flip.Doubt it? The math works. The probability of the first head occurring on an even-numbered flip is:
(1/2)^2 + (1/2)^4 + (1/2)^6 + (1/2)^8 + …This solution does not, like the prior solutions, rely on repeating the procedure for certain outcomes. If you’re concerned that this procedure only generates the 1/3 probability, without choosing among the three options, that’s easily remedied: if the first heads comes up on an even-numbered flip, choose A; if the first heads comes up on an odd-numbered flip, then flip the coin one more time to decide between B and C.
= (1/4) + (1/4)^2 + (1/4)^3 + (1/4)^4 + …
= (1/4)/(1 - 1/4) = 1/3
Lauren’s answer does share one thing with Mike’s and my answers: it could in principle require you to flip the coin an infinite number of times. Some respondents suggested procedures that would avoid that problem. For instance, two respondents suggested using the positioning of the coin when it falls – e.g., treating the word “LIBERTY” as one-third of the quarter’s circumference, and choosing option A when LIBERTY is closest to your body. Another couple of respondents suggested something like dividing a piece of paper into three sections, then setting it on the floor and flipping the coin onto it. Although these approaches could work as a practical matter, I think they run counter to the spirit of the puzzle. Glen Raphael defends these approaches, saying:
You might criticize my proposals to make use of rotational or positional info on the grounds that they don't produce exactly a 1/3rd probability. However, any proposal involving merely flipping the coin fails the same test - coins aren't perfect. Flip a quarter and you don't get an exactly 50% chance of heads; it'll be off by a few hundredths of a percent.Agreed. But in keeping with the conventions of puzzles like these, I wish to assume that we have a “perfect” quarter. Given a perfect quarter, the algorithms discussed earlier will generate a probability of exactly 1/3. (BTW, Glen R. came up with Mike’s algorithm in addition to his position/rotational proposals.)
So, of the three mathematical solutions – mine, Mike’s, and Lauren’s – which is best? All three produce the correct outcome, so the question is which one will get you to a decision fastest. Mike’s clearly dominates mine, because the probability of getting a decision on any given round is 3/4 for both, but his involves only two flips per round while mine requires three. So it’s down to Mike’s and Lauren’s solutions. For Mike’s solution, the average number of rounds it takes to reach a decision is 4/3 (which results from a guaranteed first round, a 1/4 chance of a second round, a 1/16 chance of a third round, etc.). Since each round requires two coin flips, the average number of coin flips is 8/3. For Lauren’s solution, the average number of rounds it takes to for the first head to come up is 2 (which results from a guaranteed first flip, a 1/2 chance of a second flip, a 1/4 chance of a third flip, etc.). But 2/3 of the time, it will be necessary to flip one more coin to decide between options B and C. This adds another 2/3 of a round to her average, for a total of 8/3, which exactly ties Mike’s algorithm. I’m sure this is no coincidence, but I’m not going to try explaining why.
Thank you all for playing!