Old challenge (George Murray)

You have a fair coin and a trick coin with two heads. You pick one at random, without noticing which one. The first time you toss it, it's heads. What is the probability it will come up heads the second time? (What if you don't actually see the first toss, but three of your friends, each of whom is right 2/3 of the time, all say it was heads?)

Answer

Gregory Sahagen got both answers right. The chance of choosing the fair coin and getting heads on the first toss is (1/2)(1/2) = 1/4, the chance of choosing the two-headed coin and getting heads on the first toss in (1/2)(1) = 1/2; the total chance of getting heads is 3/4. Hence the probability of the fair coin is (1/4)/(3/4) = 1/3 and the probability of the two-headed coin is (1/2)/(3/4) = 2/3. Therefore the probability of heads on the second toss is (1/3)(1/2) + (2/3)(1) = 5/6 (about 83 percent). The version with the three friends is harder. The chance that the first toss is heads and they all say it is heads is (3/4)(2/3)(2/3)(2/3) = 2/9, while the chance that it is tails and they all say it is heads is (1/4)(1/3)(1/3)(1/3) = 1/108. The first case is 24 times as likely. Therefore the probability of heads on the second toss is (24/25)(5/6) + (1/25)(1/2) = 82 percent (exactly), a bit less than the previous 83 percent because the evidence is less reliable.

More 4s (May 23 and June 13 columns)

Two more readers have found ways to bend the rules and express 73 with four 4s. If you start with 4 and keep pushing the square root key on your calculator, you get closer and closer to 1, so Carl Pomerance repeats that process infinitely many times and writes 73 = ...4 + 4! + 4! + 4!. Harry Stern uses special mathematical notation to write the number of ways to choose 2 of 24 objects, which happens to be 276, in terms of 2 = 4 and 24 = 4!, i.e. in terms of two 4s, and then notes that 73 = 276/4 + 4. Marc Abel's unsuccessful computer search for solutions did not exhaust all possibilities, so I still don't know if 73 can be expressed in terms of four 4s using just +, - , x, /, , powers, decimal point, factorials (!), and parentheses.

New challenge (Jonathan Kravis and Dave McMath)

A process involving infinitely many

Ping-Pong balls is carried out in one hour as follows. In the first half hour, two balls are placed into a huge empty barrel and then one is removed and discarded. In the next 15 minutes, two more balls are put in and then one of the three inside is removed. At each step, in half of the remaining time, two more balls are added and one is removed. How many balls are in the barrel at the end of the hour after infinitely many steps?

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