It's a Puzzlement

Loaded Die by John P. Robertson

This issue's puzzlement is todetermine how you can use a loaded die (which hasfaces that come up with probabilities that might not all be 1/6) to fairly select numbers from 1 to 6 at random. This puzzle is attributed to Al Zimmerman and appeared in Frank Morgan's Math Chat.

Parrondo's Paradox

The puzzlement was to consider three games, each of which involves flipping coins. You start with some amount of capital, say $1,000. Your capital goes up by $1 if the coin comes up heads, and goes down by $1 if the coin comes up tails. Game A uses a coin that comes up heads with probability 0.495. Game B uses two coins, one that comes up heads with probability 0.095, and another that comes up heads with probability 0.745. If your capital is a multiple of 3, use the first coin, otherwise use the second coin. Game C uses a fair coin to determine whether to play Game A or Game B (play one round of whichever game is selected, and then flip the fair coin again, etc.). Stuart Klugman provided the following solution.

1. Why is it a paradox?

Simulations can reveal that games A and B are losers and game C is a winner. This is already a paradox. How can randomly selecting between two losing propositions make one a winner? But there is an additional paradox and that concerns game B. Assuming one's fortune is a multiple of three one-third of the time, the probability of a win on a randomly observed turn is (1/3)(0.095) + (2/3)(0.745) = 0.52833 and so game B appears to be a winner, not a loser. Why is game B a loser?

Game B is a loser because more than one-third of the time is spent having a fortune that is a multiple of three. This seems reasonable. When one's fortune is a multiple of three, it is likely that the next play will be a loss, moving the fortune to being one less than a multiple of three. But then the most likely next move is a win, getting one back to a multiple of three sooner than expected.

2. Why is applied probability on the new Exam 3?

So all new actuaries can analyze game B. It is a Markov chain. There are three states, being one's fortune mod 3. The following table provides the matrix of transition probabilities.

Next State	Current State
	0	1	2
0	0	0.255	0.745
1	0.095	0	0.255
2	0.905	0.745	0

The steady-state probabilities are the eigenvector that goes with an eigenvalue of 1 and has elements that sum to 1. Equivalently, if A is the matrix above, solve the equation Ax = x constraining the elements of x to add to one. The solution is 0.383612, 0.154281, and 0.462108. The probability of winning on a randomly selected turn is 0.383612(0.095) + (0.616388)(0.745) = 0.495652 and we see that indeed, game B is a loser.

Game C can be analyzed the same way. The transition probabilities are different. For example, if in state 0, the probability of going to state 1 (winning) is 0.5(0.095) + 0.5(0.495) = 0.295. When in states 1 or 2 the probability of winning is 0.5(0.745) + 0.5(0.495) = 0.62. The steady-state probabilities are now 0.34507, 0.254108, and 0.400822. The probability of winning on a randomly selected turn is 0.507852.

This was also solved by Nolan Asch, Don Glick, John Herder, Glenn Meyers, Ira Kaplan, and David Skurnick.

In the last issue we failed to note that a correct solution was sent in by Amy Angell.