Why all of those financial books start with a very simple game of 50/50-chance bet on coin toasting. Even better, they describe roulette betting with one or double zeros. Moreover, the theory of minority games goes from one extreme to another. However, I think the closest-form game to financial markets is backgammon. A novice player will consider his chances to win as 40% because of the strategy and 60% luck, while a top professional player will consider his odds as 30% luck and 70% strategy. There are plenty of strategies in backgammon, even when you can double up your bet during the game and the opponent will accept or reject your offer. Hence, like in the market, when the odds are in your favor, the market may not take the other side of your bet.Maybe there is something here, as Ed Thorp discovered a formula while playing Blackjack five years earlier, of what Black and Scholes mathematically did.Any other closed-form games in mind?

I agree that backgammon captures important aspects of financial markets. But it is a complete information game, both players have the same information. Clearly that is not true of the financial markets. So you need to incorporate some aspects of Poker, where you need to pay to find out the other person's information. Liar's Poker (which is not a true Poker game) is also interesting, because (a) individuals are betting on the group's total information and (b) each player is playing against the neighboring players, only indirectly against the group.

I completely agree that the dispersion of information in the financial markets is unequal. Hence, there are more market manipulations than anyone could think off. The Poker game reflects the deception process of these manipulations and unequal information dispersion. However, in backgammon, there are two meaningful approaches that it possible to correlate to a financial model and potential strategies as well. For example, in two dice game, it possible to manifest between conditional expatiations (you only need a variables of 1-6 irrespective to the total dices outcome), and unconditional expectations, for example the sum of the two dice should be 12. Clearly, the outcome of two dices probability to get 11 or12 is 20-1. It seems there is a pure martingale process here.However, a strategy for an arbitrary time interval is a must. In backgammon competitions, the winner circle identifies is often the same at every important competition. In other words, a strategy that adopts a longer time interval may have a better chance to win. Another aspect is; when to double your bet in term that the other player will agree to take the bet. In my humble opinion, as in the financial markets, dynamic strategy for longer time interval have the best chance, relatively to shorter time interval strategies that tend to be pure randomness (martingales).

Clearly, the outcome of two dices probability to get 11 or 12 is 20-1. >>I think it's 1 chance in 12 (three rolls out of 36 possible). But I agree that backgammon captures important elements of strategy under martingale uncertainty. Like a good trader, a backgammon player does not expect to improve on every roll of the dice, the trick is to have a long-term strategy that does better-than-average over many rolls of the dice. And the doubling decision is a lot like certain option strategies.

Clearly, the outcome of two dices probability to get 11 or 12 is 20-1. >> I think it's 1 chance in 12 (three rolls out of 36 possible).You "think"? C'mon, Aaron

<< Clearly, the outcome of two dices probability to get 11 or 12 is 20-1. >><< I think it's 1 chance in 12 (three rolls out of 36 possible). >>The correct probability to get 6,6 is 36-1, as we double each probability of one dice 1/6*1/6. And for 11 is 24-1. However if we want a probability of two possible outcomes such as 11 and 12 the probability is 1/12 as Aaron said. It necessary to distinguish between conditional and unconditional, such as, in unconditional expectations (where the total sums are seven or above) the chance to get 12 is 22-1 in two fair dices games. Ooops..

"However if we want a probability of two possible outcomes such as 11 and 12 the probability is 1/12 as Aaron said.You mean or -- The point is that both are allowed, so you add up the probabilities: 1/36 (for [6, 6])+ 2/36 (1/36 for [5, 6] and another 1/36 for [6, 5])= 3/36 = 1/12.And, David, shouldn't you be studying for that exam?

The Let's Make a Deal Applet

http://www.stat.sc.edu/~west/javahtml/L ... aDeal.html