I have this question that I'm kinda stuck on.

It's a game scenario in which we set up an expectimax tree. In the game, you have 3 dice with sides 1-4 that you roll at the beginning. Then, depending on the roll, the player can choose one of the dice to reroll or not reroll anything. Points are assigned like so:

  • 10 points if there's 2 of a kind
  • 15 if there's 3 of a kind
  • 7 if there's a series like 1-2-3 or 2-3-4
  • Otherwise, or if the sum is higher than the rewards from above, the score = sum of the rolls

For additional context, this is an example expectimax tree I came up with, for the case that the player rolled a 1,2,4 and is considering rerolling or not: enter image description here

Now lets introduce a new agent -- a robot that's supposed to help the human player. we assume

  • the human player choses any action with uniform probability regardless of the initial roll
  • there's a robot that, given a configuration of dice and the human's desired action, actually implements the action with probability 1-p and overrides it with a "no reroll" order with probability p>0. It has no effect if the human's decision is already to not reroll.

For that scenario, I came up with this expectimax tree: enter image description here

Now for the part I'm actually stuck on -- lets define A, B, C, and D as the expected reward of performing actions "reroll die 1", "reroll die 2", "reroll die 3", and "no reroll." How do we find $R_H$, the expected reward for the human acting without the robot's help, and $R_{AH}$ the expected reward for if the robot helps? We can't use p in the expression, we only have access to A,B,C,D and we're supposed to write it in the form $X + Y_p$

*EDIT: I asked again and the question was worded weirdly. They said we should definitely use p. What was meant by not using p is X and Y themselves can't contain p. But Y will be multiplied by p in the final simplified form.

For $R_{H}$ I think the answer should be $\frac{(A + B + C + D}{4}$ because of uniform distribution over A-D.

I'm supposing that $R_{AH}$ would be $\frac{(A + B + C)(1-p) + D + Dp}{4}$? Because the robot doesn't override with probability $1-p$, but he can only override A-C and does with probability $p$.

I think something feels slightly wrong about my answer but I'm not sure what.

  • 1
    $\begingroup$ Are you sure you are not allowed to use p? The expected reward R_AH is a function of p. $\endgroup$
    – Cohensius
    Feb 24, 2021 at 7:29
  • $\begingroup$ Hi. Could you please put your main specific question in the title? "Expectimax World Problem — Dice Rolling Game" is not a question, but the title of an article. $\endgroup$
    – nbro
    Feb 24, 2021 at 13:34
  • 1
    $\begingroup$ @Cohensius I asked the instructors and they clarified that you can use p. What was meant by not using p is X and Y themselves can't contain p. But Y will be multiplied by p in the final simplified form. It was a slight wording confusion. I added my best guess at a solution. However, I'm not 100% sure, and would appreciate confirmation :) $\endgroup$
    – Manny
    Feb 24, 2021 at 17:06
  • $\begingroup$ @nbro done! sorry $\endgroup$
    – Manny
    Feb 24, 2021 at 17:06

1 Answer 1


If I understand,

  • With probability $p$ the robot select no re-roll (action $D$).
  • With probability $1-p$ the human uniformly select an action between $A,B,C,D$.

Thus the expected reward for if the robot helps is

$$R_{AH}= p\cdot D+(1-p)\frac{A+B+C+D}{4} = \frac{(A+B+C)(1-p)+D+3pD}{4}$$


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