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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.

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Pricing of this option is an interesting exercise. – Richard Sep 2 '14 at 14:43
up vote 4 down vote accepted

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 we can price the option as $1/2*3.5 = 1.75$ which is the probability that the option is worth anything after the first throw times the value of the option in the second throw.

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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? – RandomGuy Sep 2 '14 at 16:19
Your method seems fine, only note the expected payoff of the option theoretically also has to be discounted to today: $P=\frac{1.75}{(1+r)^T}$ – emcor Sep 2 '14 at 17:12
There is no discount as these kind of game are instantaneous. – Were_cat Sep 2 '14 at 18:56
@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. – Richard Sep 3 '14 at 6:27
@Richard: done. – RandomGuy Sep 3 '14 at 9:40

Suppose the dice is well balanced,

The game is fair if your overall expectation is 0.

What you gain: $E(n)$

What you loose: $Price$

So $Price = E(n) = \sum_n p(n) * n$

(You should really figure the latest sum by yourself)

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You are basically saying that the price of the option is just the expectation of the payoff of a single throw. This is what my first idea was. However, I am confused by the fact that apparently the option price is set after the first throw, not at the beginning of the game and I feel like this changes things inasmuch the price of the option is to be set after the knowledge of the first throw, so it should be different from an option bought on a single die roll. Does it make sense? – RandomGuy Sep 2 '14 at 13:29
It make sense. For the general case you have to take $p(n|n_1)$ taking into account prior information. But the fact here that the dice is balanced imply that $p(n|n_1) = p(n)$. – Were_cat Sep 2 '14 at 13:37
Please see Richard's answer, he subsequently included the conditional part. – emcor Sep 2 '14 at 17:14
ok, my answer is not relevant. I did not understood your question. – Were_cat Sep 2 '14 at 18:54

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?

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The dice is an ordinary one, with 6 faces with the same probability. However, I did not understand your argument, could you please elaborate? – RandomGuy Sep 2 '14 at 13:05
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? – Bob Jansen Sep 2 '14 at 15:06

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