# Throwing a dice and risk neutral probability

Consider the game of throwing a "fair" dice. Not sure if the answer is obvious but is there any proof (e.g. replication argument) that under the risk neutral measure the probability of any outcome is 1/6 and hence the price of the game is 3.5? In other words why under the real probability measure the probability of 1/6 coincides with that under the risk neutral measure?

I guess, one could argue that if the risk neutral probability was not 1/6 then one could create arbitrage by entering the game infinitely many times. But is there any other way to answer the question above?

the information you provided is not sufficient to deduce risk neutral probabilities. You have to provide something like a price process from which risk neutral probabilities can be computed.

Here are some examples:

### Example1:

Consider a game where you pay 1 and you win 6 in case a six is thrown and 0 otherwise. So in financial mathematics terms we have a binomial model with the following parameters: $$S_0 = 1, \quad S_1(up) = 6, \quad S_1(down) = 0, \quad r = 0.$$ In particular, we get for the risk-neutral probabilities: $$\Bbb Q(up) = \frac 16$$, $$\Bbb Q(down) = \frac 56$$ such that the physical probabilities and risk neutral probabilities agree.

### Example2:

Consider a game where you pay 3.5 and you win whatever the dice shows. In this case the physical probabilities are risk neutral but there are much more risk neutral measures. In terms of financial mathematics this means that the market is not complete. Another example of risk neutral measure would be: $$q_1 = \frac 16, \quad q_2 = \frac 16, \quad q_3 = \frac 1{4}, \quad q_4 = \frac 1 {12}, \quad q_5 = \frac 1{12}, \quad q_6 = \frac 14$$

### Example3

Consider a game where you pay 1 and you win whatever the dice shows. In this case the price process is not a martingale under the physical measure and hence your physical probabilities are not risk neutral.

I hope this helps a little.