Two dice have 36 rolls and 11 sums, and a 20-move, 10-turn game tree has 10.24 trillion lines before the dice are counted
Our own calculation, 7 November 2026 (UTC), for the article "Expectimax vs Monte Carlo Tree Search for dice games". Two ways each: the dice outcomes by enumerating every ordered roll and again by the counting formula (6 minus the distance of the sum from 7); the tree sizes by exact integer power and again by logarithm. The two ways agree.
What we found
- Two six-sided dice give 36 equally likely ordered rolls and 11 distinct sums, from 2 to 12; a sum of 7 comes up 6 ways in 36 (16.7 percent), a sum of 2 or of 12 one way each (2.8 percent). The formula gives the same 36 rolls over the same 11 sums.
- A game with 20 possible moves a turn searched 10 turns deep has 20 to the power 10, 10,240,000,000,000 move sequences, 10.24 trillion, before any dice roll is counted; the logarithm way gives the same.
- If every turn also carries a chance node over the 11 sums, the tree has (20 times 11) to the power 10, about 2.7e+23 lines; over the 36 ordered rolls, about 3.7e+28. That is what "every chance node multiplies the tree by the number of distinct rolls" means in numbers.
- A Monte Carlo search with a budget of 100,000 simulations visits about 1e-08 of the 10.24 trillion dice-free lines: one in 102,400,000.
Check it yourself
| Sum of two dice | Ways in 36 | Probability | Percent |
|---|---|---|---|
| 2 | 1 | 1/36 | 2.8% |
| 3 | 2 | 1/18 | 5.6% |
| 4 | 3 | 1/12 | 8.3% |
| 5 | 4 | 1/9 | 11.1% |
| 6 | 5 | 5/36 | 13.9% |
| 7 | 6 | 1/6 | 16.7% |
| 8 | 5 | 5/36 | 13.9% |
| 9 | 4 | 1/9 | 11.1% |
| 10 | 3 | 1/12 | 8.3% |
| 11 | 2 | 1/18 | 5.6% |
| 12 | 1 | 1/36 | 2.8% |
Tree sizes: 20^10 = 10,240,000,000,000; (20 x 11)^10 = 265,599,227,914,240,000,000,000; (20 x 36)^10 = 37,439,062,426,244,874,240,000,000,000.
What this cannot tell you: which algorithm wins a particular dice game; that depends on the game's rules, the evaluation function and the time budget, none of which is measured here.