Double Crostic

Alan K. Putney devised this issue's double crostic. The first letters of the answers to the clues A to AD give the author and title of the work from which the quote is taken.

Capital Allocation
You had identified 11 companies, each with an 85 percent probability that each \$1 invested would grow to \$10, and a 15 percent probability that the investment would become worthless. The problem was to find a way to grow \$17 million to \$100 million with at least a 95 percent probability.

Frank Chang showed that dividing the original funds evenly among any of 5, 6, 8, 9, 10, or 11 of the possible investments gives the desired probability of reaching the goal.

David Uhland observes that dividing the funds among 10 of the possible investments gives:

• The maximum potential return with at least a 95 percent probability, at \$102 million (in addition to the original \$17 million), when the total funds are divided evenly;
• The greatest probability of ending up with at least \$100 million, also when the total funds are divided evenly;
• A probability of 95.00 percent, when rounded to the decimal places shown, of ending up within one dollar of \$100 million when \$1,383,333.33 is invested in each of the 10 companies, and the rest is not invested.

Jeff Carter, Kevin Cleary, Jon Evans, Shiwen Jiang, John Hinton, David Oakden, David L. Ruhm, Jon Tsou, Nathan Voorhis, Michael Wittmann, and Micah Woolstenhulme also solved the problem.

Highest Probability?
Several readers objected to the solution to the August 2005 puzzlement, so let me give a more detailed solution. The probability that the first ace is in position k for 1 ≤ k ≤ 49 is (52 - k)(51 - k)(50 - k)⁄(13⋅51⋅50⋅49). If the first ace is in position k, the probability that the ace of spades turns up next is 3⁄(4(52 - k)), and the probability that the deuce of clubs turns up next is (49 - k)⁄(48(52 - k)). These are easily verified. Multiplying and adding shows both of the required probabilities are 1⁄52, as given previously.