dormammu Joined: Nov 11, 2015
• Posts: 2
November 11th, 2015 at 5:40:02 AM permalink
In a game where 2 players roll a variable number of dice, and the object is to roll the highest number on any individual die, what is the general formula for determining the probability of win/lose/tie on a given roll? The dice are not added up in this game, only the highest number showing in each player's hand is taken.

Example: On a particular turn, Player 1 rolls 4 dice, and Player 2 rolls 6 dice. What is the probably that Player 1 will roll a higher number than Player 2, or that it will be a tie.

An additional twist, which makes the problem more interesting: If they do tie, how can the formula be altered such that a greater number of dice showing the higher number acts as a tie breaker.

Example: On a particular turn, Player 1 rolls 5 dice getting 1, 2, 3, 5, 5. Player 2 rolls 6 dice, getting 1, 2, 2, 2, 4, 5. Player 1 wins. Joined: Jul 3, 2015
• Posts: 30
November 11th, 2015 at 7:40:51 AM permalink
Quote: dormammu

Example: On a particular turn, Player 1 rolls 4 dice, and Player 2 rolls 6 dice. What is the probably that Player 1 will roll a higher number than Player 2, or that it will be a tie.

Without the twist, P(plr 1 win) = P(roll 6)*P(villain lower) + P(roll 5)*P(villain lower) + ... + P(2)*P(villain lower)

= [1 - (5/6)^4] * (5/6)^6 +
(5/6)^4 * [1 - (4/5)^4] * (4/6)^6 +
(4/6)^4 * [1 - (3/4)^4] * (3/6)^6 +
(3/6)^4 * [1 - (2/3)^4] * (2/6)^6 +
(2/6)^4 * (1 - .5^4) * (1/6)^6

=~ 20.06%

P(tie) = [1 - (5/6)^4] * [1 - (5/6)^6] +
(5/6)^10 * [1 - (4/5)^4] * [1 - (4/5)^6] +
(2/3)^10 * [1 - (3/4)^4] * [1 - (3/4)^6] +
.5^10 * [1 - (2/3)^4][1 - (2/3)^6] +
(1/3)^10 * (1 - .5^4)(1 - .5^6)

=~ 42.52%

You can generalize that by replacing the exponents 4 and 6 with variables.
dormammu Joined: Nov 11, 2015
• Posts: 2
November 11th, 2015 at 5:45:10 PM permalink
Thanks for your quick reply. At the same time I posted here, I also asked my son (who is very good at math/statistics) to solve this - and got this:

Sum[X=1-6]{[(X/6)^n - ((X - 1)/6)^n] * [1 - (X/6)^m]}

where the number of dice for Player 1 is m, and the number of dice for Player 2 is n.

Given this, the chance of winning when both players have two dice is:
P(4) = (16/36 - 9/36) * 20/36 = 140/1296
P(5) = (25/36 - 16/36) * 11/36 = 99/1296
P(6) = (36/36 - 25/36) * (1 - 36/36) = 11/36 * 0 = 0
35/1296 + 96/1296 + 135/1296 + 140/1296 + 99/1296 = 505/1296

It follows that since both have an equal chance of winning in this case, the chance of tying must be 1 - (P(win) * 2) = 286/1296 (without the additional constraint)

The formula for tying in the general case would be P(Tie) = P(Opponent gets X as highest) * P(You get X as highest) = Sum[X=1-6]{[(X/6)^n - ((X - 1)/6)^n] * [(X/6)^m - ((X - 1)/6)^m]}

To test our theory, let's calculate it out given the 2 die scenario:
1/36 * 1/36 = 1/1296
3/36 * 3/36 = 9/1296
5/36 * 5/36 = 25/1296
7/36 * 7/36 = 49/1296
9/36 * 9/36 = 81/1296
11/36 * 11/36 = 121/1296

Add 'em all up, and you get 286.

So, if my son's answer is: Sum[X=1-6] { [(X/6)^n - ((X - 1)/6)^n] * [1 - (X/6)^m] }

Then yours would be expressed: 1 - (Sum[X=1-6] { (X/6)^n * [1 - ((X-1)/X)^n] * ((X-1)/6)^m })

Both work:) Thanks again.