Game theory: dice game to aim for sum no greater than 9

[Master1022]

Homework Statement

You have been chosen to play a game involving a 6-sided die. You get to roll the die once, see the result, and then may choose to either stop or roll again. Your payoff is the sum of your rolls, unless this sum is greater than 9, in which case you "bust" and get nothing. What is your strategy for this game? That is, for each possible outcome of the first roll will you choose stop or to roll again?

Relevant Equations

Probability

Hi,

I was attempting the following question and would appreciate any insight on how others would approach this game theory/probability-type question.

Question: You have been chosen to play a game involving a 6-sided die. You get to roll the die once, see the result, and then may choose to either stop or roll again. Your payoff is the sum of your rolls, unless this sum is greater than 9, in which case you "bust" and get nothing. What is your strategy for this game? That is, for each possible outcome of the first roll will you choose stop or to roll again?

Attempt:
So if we get a sum greater than 9, then our pay-out is 0. Some things that I am keeping in mind are:
- if we are on a sum of 1, 2, or 3, then we should re-roll as there is no chance of overshooting 9 from one more roll
- For the higher sums, the general principle I was planning to use when deciding whether or not to roll again was the 'marginal expected gain of rolling again'

For example, if we are on a sum of 6, then the marginal expected gain is given by the expression below (I hope this is correct?). We can lose our current pay-off with 3/6 of the outcomes of the roll.
$$ E[\text{marginal gain at sum of 6}] = \frac{1}{6} (+3) + \frac{1}{6} (+2) + \frac{1}{6} (+1) + \frac{1}{2} (-6) = -2 $$
and thus we shouldn't roll as this 'marginal expected gain from rolling' is negative. I suppose this makes sense if EV[dice roll] = 3.5 because $6 + 3.5 > 9$

Following similar logic for a sum of 5:
$$ E[\text{marginal gain at sum of 5}] = \frac{1}{6} (+4) + \frac{1}{6} (+3) + \frac{1}{6} (+2) + \frac{1}{6} (+1) + \frac{1}{3} (-5) = 0 $$
and thus we could roll if marginal EV = 0

For sum of 4:
$$ E[\text{marginal gain at sum of 4}] = \frac{1}{6} (+5) + \frac{1}{6} (+4) + \frac{1}{6} (+3) + \frac{1}{6} (+2) + \frac{1}{6} (+1) + \frac{1}{1} (-4) = \frac{11}{6} $$
and thus we could roll if marginal EV is positive.

By this logic, the strategy would be to act as follows:
- For sums of 1, 2, 3, 4: re-roll
- For sum of 5: re-roll
- For sum $\geq 6$: don't re-roll

Any help would be greatly appreciated!

 

[FactChecker]

I think that you got the correct answer and that your method is valid
EDIT: I think that the following is wrong.
but more complicated than it needs to be.
I think this is simpler. Calculate the expected result of a second roll. Subtract that from 9 and you should get the value of the first roll that tells you if you should try a second roll or not.

 

[FactChecker]

I think this is simpler. Calculate the expected result of a second roll. Subtract that from 9 and you should get the value of the first roll that tells you if you should try a second roll or not.

Sorry, I think I was wrong. When the second roll puts the total over 9, the entire amount is lost. So the expected value of the second roll is much less.
I do not see an error in your approach and think that it is as simple as it can be.

 

[PeroK]

By this logic, the strategy would be to act as follows:
- For sums of 1, 2, 3, 4: re-roll
- For sum of 5: re-roll
- For sum $\geq 6$: don't re-roll

Any help would be greatly appreciated!

I like your logical approach. What if the score not to be exceeded is $8$? How quickly can you decide on a strategy?

 

[PeroK]

PS if you want a bit of fun, try this: $$-6.5 + \sqrt{42.25 + N^2 + N}$$

 

[FactChecker]

I am really surprised that my incorrect solution gave the same answers that your correct approach did. I would have thought that the part I ignored (the loss of the first roll result) would have changed the answer more. I will have to think about that.

 

[Master1022]

I like your logical approach. What if the score not to be exceeded is $8$? How quickly can you decide on a strategy?

Hmm, so far I can only think of doing the problem the same way but this time ignoring the denominators as we only care about the sign. Gut instinct suggests that I would re-roll for $\leq 4$ and stop for $\geq 5$. Mathematically:

For 4: $$ \text{EV} \propto (+1) + (+2) + (+3) + (+4) + (2 \cdot -4) > 0 \rightarrow \text{re-roll} $$
For 5: $$ \text{EV} \propto (+1) + (+2) + (+3) + (3 \cdot -5) < 0 \rightarrow \text{don't re-roll} $$

However, I will keep thinking

 

[PeroK]

Hmm, so far I can only think of doing the problem the same way but this time ignoring the denominators as we only care about the sign. Gut instinct suggests that I would re-roll for $\leq 4$ and stop for $\geq 5$. Mathematically:

For 4: $$ \text{EV} \propto (+1) + (+2) + (+3) + (+4) + (2 \cdot -4) > 0 \rightarrow \text{re-roll} $$
For 5: $$ \text{EV} \propto (+1) + (+2) + (+3) + (3 \cdot -5) < 0 \rightarrow \text{don't re-roll} $$

However, I will keep thinking

That's it. There must be a cutoff point. Your best guess was right. If we put $N = 8$ into my formula for the cutoff, we get:$$-6.5 + \sqrt{42.25 + N^2 + N} = 4.19$$ which means $4$ is below the cutoff and $5$ is above.

And, for $N = 9$ we have:$$-6.5 + \sqrt{42.25 + N^2 + N} = 5$$which means $5$ is exactly on the cutoff.

 

[Master1022]

That's it. There must be a cutoff point. Your best guess was right. If we put $N = 8$ into my formula for the cutoff, we get:$$-6.5 + \sqrt{42.25 + N^2 + N} = 4.19$$ which means $4$ is below the cutoff and $5$ is above.

And, for $N = 9$ we have:$$-6.5 + \sqrt{42.25 + N^2 + N} = 5$$which means $5$ is exactly on the cutoff.

Oh I see - I didn't understand the formula in @FactChecker 's post, but now I see. I'll have a bit of a think about it (and how to derive it) and get back if I have any questions! Many thanks

 

[FactChecker]

Oh I see - I didn't understand the formula in @FactChecker 's post, but now I see. I'll have a bit of a think about it (and how to derive it) and get back if I have any questions! Many thanks

I went back and edited my post. I think my approach was wrong. I am surprised that the answer from it was the same as your answer, which I believe is correct.

 

[DaveC426913]

I went back and edited my post. I think my approach was wrong. I am surprised that the answer from it was the same as your answer, which I believe is correct.

Legit question: If it got the right answers, is it still wrong?

 

[FactChecker]

Legit question: If it got the right answers, is it still wrong?

Yes. My approach was just too simple. I think it did not correctly consider an important fact -- that the entire gain from the first roll was lost if the total exceeded 9.