Risk-Neutral Dice Probabilities

Article Quant Q&A · Author: noob-mathematician

Summary

The discussion explains why a fair die’s physical outcome probabilities do not, by themselves, determine the risk-neutral probabilities. Risk-neutral probabilities are inferred from traded prices and payoffs, so a price process and the available securities matter. In a one-period example where a ticket costs one unit and pays six only on a six, with zero interest, no-arbitrage pricing gives the six outcome a risk-neutral probability of one sixth, matching the physical probability.

A different setup, where a ticket costs the die’s expected value and pays the face shown, does not uniquely pin down all six risk-neutral probabilities. The market is incomplete, so multiple measures can price the traded payoff consistently. A further example shows physical probabilities need not be risk-neutral when the stated price process fails the martingale condition under those probabilities. The examples are simplified; the lesson depends on specified market prices, payoffs, and assumptions, rather than fairness alone.

Key ideas

Risk-neutral probabilities are inferred from asset prices and payoffs, not from the die’s fairness alone.
In a specified binary payoff example, no-arbitrage pricing can make physical and risk-neutral probabilities coincide.
If traded payoffs do not span every outcome, multiple risk-neutral measures may fit the same prices.
Physical probabilities are risk-neutral only when they are consistent with the relevant discounted price process.

Tags

Statistics Derivatives pricing Arbitrage

Full text

# 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?

## Answer by Cettt (score 10)

https://quant.stackexchange.com/a/51383

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.

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When Physical and Risk-Neutral Dice Probabilities Agree

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