Say you were to “host” a single elimination tournament of 8 dice, each with a different number of sides: 4, 6, 8, 10, 12, 20, 24, and 30. In each round, the dice that rolls the highest wins (re-roll both on ties). All dice are numbered normally from 1 to the number of sides except the 10-sided, which is 0-9. The dice are seeded and placed into a bracket for the tournament (30-sided is 1-seed, 24-sided is 2-seed, etc.) (10-sided is the 5th seed, and 8-sided is the 6th seed). Here’s what it would look like:


The “l”s are used as spacing.

Assume all dice are perfectly fair and have an equal chance of rolling each number. What is the chance each dice has of winning the tournament?

(If you’re wondering why I’m asking this, my friend gave it to me as a problem to see if I could solve it. I couldn’t and went back to him asking for the answer only to find out he didn’t know either)

In a more general sense, if you have an 8x8 matrix that contains the probability that one dice/team will beat another (based on seed, so row 1 col 2 is the chance that the 1 seed beat the 2 seed), what is the chance of each seed winning the tournament?

Feel free to just answer the specific case (with the dice) if the general case is too complicated.

  • $\begingroup$ Welcome to MSE. It is in your best interest that you use MathJax. $\endgroup$ – José Carlos Santos Jun 23 '18 at 0:04
  • 1
    $\begingroup$ The singular is "die" not "dice". $\endgroup$ – saulspatz Jun 23 '18 at 0:06
  • $\begingroup$ @saulspatz Well, see here :) $\endgroup$ – Saad Jun 23 '18 at 3:06

When two normal dice with $m$ and $n$ sides ($m \ge n$) face each other, the probability for the outcomes are as follows:

$$P(X_m = X_n) = \frac{n}{m} \frac{1}{n} = \frac{1}{m}$$ $$P(X_m \lt X_n) = \frac{1 + \ldots + n-1}{mn} = \frac{n(n-1)}{2mn} = \frac{n-1}{2m}$$ $$P(X_m \gt X_n) = 1 - P(X_m = X_n) - $P(X_m \lt X_n) = \frac{2m - n - 1}{2m}$$

Since a draw is not possible (the dice will be rolled again until the dice have a different outcome), this scenario should not be considered. If $D_{n, m}$ is the event in which the $n$-sided die wins from the $m$-sided die, we get:

$$P(D_{n, m}) = \frac{P(X_m \lt X_n)}{1 - P(X_m = X_n)} = \frac{\frac{n - 1}{2m}}{1 - \frac{1}{m}} = \frac{n - 1}{2m - 2}$$

$$P(D_{m, n}) = 1 - P(D_{n, m}) = \frac{2m - n - 1}{2m - 2}$$

When a special $s$-sided die with values ranging from $0$ to $s - 1$ is introduced, the formula changes slightly. If this die competes with a regular $m$-sided die, we can distinguish two cases:

  1. $m \gt s$

    $$P(X_m = X_s) = \frac{s - 1}{ms}$$ $$P(X_m \lt X_s) = \frac{(s - 1)(s - 2)}{2ms}$$ $$P(D_{s, m}) = \frac{\frac{(s - 1)(s - 2)}{2ms}}{1 - \frac{s - 1}{ms}} = \frac{(s - 1)(s - 2)}{2ms - 2s - 2}$$

  2. $s \gt m$

    $$P(X_s = X_m) = \frac{1}{m}$$ $$P(X_s \lt X_m) = \frac{1 + \ldots + s}{ms} = \frac{(s + 1)s}{2ms} = \frac{s + 1}{2m}$$ $$P(D_{m, s}) = \frac{\frac{s + 1}{2m}}{1 - \frac{1}{m}} = \frac{s + 1}{2m - 2}$$

Using this formulas, we can now calculate the probability that a given dice reaches the next round. For instance, when the $30$-sided die competes with the $4$-sided die, we find:

$$P(D_{4, 30}) = \frac{4 - 1}{2 \cdot 30 - 2} = \frac{3}{58} \approx 0.0517$$

Then, using all probabilities for the dice to reach the second round, we can do the same for the third round. For instance, the probability of the $8$-sided to reach the third round equals:

$$P(D_{8, 20}) \left[P(D_{6, 20}) P(D_{8, 6}) + P(D_{24, 6}) P(D_{8, 24})\right] = \frac{8 - 1}{2 \cdot 20 - 2} \left[\frac{6 - 1}{2 \cdot 24 - 2}\left(1 - \frac{6 - 1}{2 \cdot 8 - 2}\right) + \left(1 - \frac{6 - 1}{2 \cdot 24 - 2}\right)\frac{8 - 1}{2 \cdot 24 - 2}\right]$$

$$= \frac{7}{38}\left[\frac{5}{46}\frac{9}{14} + \frac{41}{46}\frac{7}{46}\right] \approx 0.0379$$

We can do the same thing for the other dice, and, based on these results, calculate the probability for each die to win the third round and thus the tournament. Doing the required calculations, we finally find:

$$P(30) \approx 0.5093$$ $$P(24) \approx 0.2713$$ $$P(20) \approx 0.1568$$ $$P(12) \approx 0.0432$$ $$P(10) \approx 0.0095$$ $$P(8) \approx 0.0067$$ $$P(6) \approx 0.0024$$ $$P(4) \approx 0.0008$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.