In a certain game, two unfair six-sided dice are rolled. One dice is red, the other is blue. Both dice have 1,2,3,4,5,6 as values. These dice are rolled at the same time. Both dice have 10 life points each.

Upon rolling, if red shows 2, 2 points are deducted from the blue's life points. Likewise, if blue shows 3, 3 points are deducted from red's life points. Whatever number appears from red, is deducted from blue life points, and vice versa.

We keep rolling both dice, until one dice' life points gets to 0. Last dice standing, wins.

What is the expected number of rolls, for this game? If given the chance to bet which dice will win, what's the chance that red wins? Also, what's the chance that blue wins?

At first I took this as a P(A U B) = P(A) + P(B) - P(A&B) but I am not getting the correct results. I do get P(A) and P(B), but have been stuck getting P(A&B). I may not be using the right formula anyway.

P(A) is the probability space for red, P(B) is for blue.

Sample probability distribution: P(A) -- {1,2,3,4,5,6} = {0.1, 0.15, 0.2, 0.25, 0.05, 0.25} P(B) -- {1,2,3,4,5,6} = {0.15, 0.2, 0.1, 0.25,0.25,0.05}

Any help appreciated, thanks.

  • $\begingroup$ So, if red rolls a 1,3,4,5, or 6 it doesn't matter and nothign happens? Similarly if blue rolls anything other than a 3 nothing happens? You talk about $A$ and $B$, what are those meant to represent? You say these dice are unfair, what are the probabilities associated with the dice? $\endgroup$
    – JMoravitz
    Jul 29, 2022 at 13:33
  • $\begingroup$ Ultimately, this can be phrased as a markov chain and standard techniques for markov chains apply. Since red only "deals damage" in increments of two, we might as well have phrased it as blue having "five hits worth of health" and similarly red having "four hits worth of health", leading to a relatively smaller state-space of $23$ game states ($5\times 4$ states where the game is still continuing, plus the end states for red wins, blue wins, and tie). From each mid-game state you can transition to one of four (or three) next states depending on who (if anyone) dealt damage this turn. $\endgroup$
    – JMoravitz
    Jul 29, 2022 at 13:39
  • 2
    $\begingroup$ If you haven't concocted the question yourself, can you give the exact wording ? $\endgroup$ Jul 29, 2022 at 13:41
  • $\begingroup$ If red rolls a 1,3,4,5, or 6, blue life points also deduct 1,3,4,5,6, and vice versa. $\endgroup$
    – mavjrulled
    Jul 29, 2022 at 17:02
  • 2
    $\begingroup$ Besides the fact that you specified unfair dice, without giving details, the expected number of rolls can be alternatively computed, with no knowledge of Markov theory, purely by recursion. Personally, despite the fact that I am ignorant of Markov theory, I still strongly suspect that Markov theory is much the preferable approach. If interested in the recursion approach, leave a comment. Also, please respond to the comment of Just a User (above). $\endgroup$ Jul 29, 2022 at 19:04

1 Answer 1


You can deduct the health of each dice with its own score and then say that whoever reaches zero first wins.

Basically you have to calculate the expected number of trials it takes for a dice to get to zero or less starting from 10. Let e(10) be the expected number.

e(10) = sum( e(10 - x) + 1 )* pr(x) ; for x in 1 to 6. This can be easily calculated given pr(x) .

Then expected number of trials would be minimum of both values calculated.


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