Valuing a Reroll Option in a Fair Dice Game

# Option on a dice game


# Option on a dice game












I am sligtly confused by this problem, although it should not be difficult.

Let us roll a sigle dice. If the dice shows $n$, I receive $n$ dollars. I can buy an option to roll the die again. What is the price for the option?

My idea is that the price should be the expected payoff of the game, conditioned over the result of the first game, but I am not sure as to how write this down precisely.

## Answer by Richi Wa (score 7, accepted)

https://quant.stackexchange.com/a/14603

I would use the following arguments:

If the option were on the first throw of the dice, then we would price it using the expectation, which is $3.5$ (= $(1+2+\cdots+6)/6$.

Now we have a 2 stage game:





So the value of the game with the option is $1/2*5 + 1/2*3.5=4.25$.

The value of the game without the option is $3.5$.

The option value, which is the fair price of the option before the game starts, is then the difference $4.25-3.5=0.75$.

## Answer by dm63 (score 4)

https://quant.stackexchange.com/a/36688

Hang on a second. The value of the game assuming you have an option to roll a second time is 4.25, as established above. But the value of the game without the option to roll again is 3.5. Therefore the value of the option is 0.75.

## Answer by Rui Wang (score 2)

https://quant.stackexchange.com/a/36687

This is a dynamic programming problem.

For a roll, the expectation will be 3.5.

For two rolls, if your first roll gets 1 or 2 or 3, you will roll it again. In other words, you have 1/2 chance to do the second roll and get the expectation as 3.5. If 4 or 5 or 6, you will stop here since it is good enough above the expectation (3.5). For such outcome as 4 or 5 or 6, your average is 5 with another 1/2 chance. So the total expectation will be 1/2*3.5 + 1/2*5 = 4.25.

For three rolls, you will consider the case differently after the first rolls. If I obtain 5 or 6, I will stop here for just one roll since it achieve above my expectation. Otherwise, I will roll twice and obtain the expectation of two rolls as 4.25. In total, the expectation will be 2/3*4.25 + 1/3*5.5 = 14/3 = 4.67.

Suppose that we have k rolls and obtain the expectation $E$ above 5, we have to change the strategy again after the first roll. You can stop rolling if you get 6, otherwise keep going to obtain $E$ values. The new expectation will be like that: $$E_{new} = 1/6*6 + 5/6*E_{old} $$

For n rolls, write a program in Python as the following:

` def die_fair_value(rolls): cnt=1 val=3.5 while cnt < rolls: if val < 4: val = 1/2.0*5 + 1/2.0*val elif val<5: val = 1/3.0*5.5 + 2/3.0*val else: val = 1/6.0*6 + 5/6.0*val cnt += 1 return val `

## Answer by emcor (score 0)

https://quant.stackexchange.com/a/14601

The next throw is independent of the previous throws, so you only calculate the value of the future expected payoffs from the option to continue.

How many "$n$"s does the dice have, and what is their probability?

## Answer by Hunaphu (score 0)

https://quant.stackexchange.com/a/77787

Let $\pi_1$ denote the price of the die roll, $\pi_2$ denote the price of the option, $\Phi$ denote the payoff and $A$ the event that the option is used so that $$ \pi_1 + \pi_2 = E[\Phi] = E[X\mid A^*]P(A^*) + \pi_1 P(A) $$ The option will be exercised when $X\leq\pi_1$ so $$ \pi_2 = -\pi_1 + E[X\mid X>\pi_1]P(X>\pi_1) + \pi_1P(X\leq \pi_1)\\ =(E[X\mid X>\pi_1] - \pi_1)P(X>\pi_1). $$ In this case, $\pi_1 = 3.5$ so $\pi_2 = (\frac{15}{3} - 3.5)\frac{1}{2} = 0.75$

Intuition: The value is the average excess in the good case multiplied by the probability of realizing the good case.