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.

  • $\begingroup$ Pricing of this option is an interesting exercise. $\endgroup$
    – Richi Wa
    Sep 2, 2014 at 14:43
  • $\begingroup$ Do you buy an option after rolling the first time? Than the price should be 3.5. Are you buying an option to roll twice? Then the price should be 4.25. $\endgroup$
    – mfnx
    Nov 12, 2019 at 13:20
  • 1
    $\begingroup$ take the difference of the expected winnings with and without the option $\endgroup$
    – The One
    Aug 26, 2021 at 1:29

5 Answers 5


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:

  • First throw : if the player throws more than $3.5$ points, i.e. $4,5,6$, then there is no sense in throwing again. If he throws $1-3$ then it makes sense to throw again.
  • Second throw: given that the first throw was $1-3$ we expect a payoff of $3.5$.

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 $4.25-3.5=0.75$.

  • $\begingroup$ Nice argument. If I understand well, you are pricing the option "at time 0", right? I mean, with your technique, which looks ike basically the same as pricing on a 2-step binomial tree, you are pricing the option at the beginning of the game, i.e., before the first roll. Is it so? $\endgroup$
    – RandomGuy
    Sep 2, 2014 at 16:19
  • $\begingroup$ Your method seems fine, only note the expected payoff of the option theoretically also has to be discounted to today: $P=\frac{1.75}{(1+r)^T}$ $\endgroup$
    – emcor
    Sep 2, 2014 at 17:12
  • $\begingroup$ There is no discount as these kind of game are instantaneous. $\endgroup$ Sep 2, 2014 at 18:56
  • $\begingroup$ @RandomGuy yes, I think it should be priced similar to an American option in a Binomial tree. If you like the answer, then please accept it (klicking on the button). And yes - if we throw dice then we don't need any discounting. Anyways, it would not really change anything - if you discount, then use $2$ time periods and an appropriate interest rate. $\endgroup$
    – Richi Wa
    Sep 3, 2014 at 6:27
  • $\begingroup$ @Richard: done. $\endgroup$
    – RandomGuy
    Sep 3, 2014 at 9:40

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.

  • $\begingroup$ This should be the correct one; the accepted answer forgot to subtract with the expected winnings of the first roll. The answer, following his argument, should also be $\frac{1}{6}(0.5 + 1.5 + 3.5) = 0.75$ $\endgroup$
    – The One
    Aug 26, 2021 at 1:31

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


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?

  • $\begingroup$ The dice is an ordinary one, with 6 faces with the same probability. However, I did not understand your argument, could you please elaborate? $\endgroup$
    – RandomGuy
    Sep 2, 2014 at 13:05
  • $\begingroup$ I think the point is to see the first throw as a kind of strike. If I'm right the question is: What is the value for the possibility to get a better roll after the first? $\endgroup$
    – Bob Jansen
    Sep 2, 2014 at 15:06

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.


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