I am interesting in producing a table of probabilities for dice rolls. These are standard 6 sided dice. What is the probability that for rolling X dice, Y dice will roll (hit) at least number Z or higher.
To hopefully simplify, I would like to fill out the following table. *edit – Table has been filled out thanks to amd's answer. Thanks!
EXACT Probabilities
#dice|# "hits"| # to roll or higher!
(X) | (Y) | (Z)
| | 2 | 3 | 4 | 5 | 6
--------------------------------------------------------------------
1 |1 | 83.33% | 66.67% | 50.00% | 33.33% | 16.67%
--------------------------------------------------------------------
2 |1 | 27.78% | 44.44% | 50.00% | 44.44% | 27.78%
2 |2 | 69.44% | 44.44% | 25.00% | 11.11% | 2.78%
--------------------------------------------------------------------
3 |1 | 6.94% | 22.22% | 37.50% | 44.44% | 34.72%
3 |2 | 34.72% | 44.44% | 37.50% | 22.22% | 6.94%
3 |3 | 57.87% | 29.63% | 12.50% | 3.70% | 0.46%
--------------------------------------------------------------------
4 |1 | 1.54% | 9.88% | 25.00% | 39.51% | 38.58%
4 |2 | 11.57% | 29.63% | 37.50% | 29.63% | 11.57%
4 |3 | 38.58% | 39.51% | 25.00% | 9.88% | 1.54%
4 |4 | 48.23% | 19.75% | 6.25% | 1.23% | 0.08%
--------------------------------------------------------------------
5 |1 | 0.32% | 4.12% | 15.63% | 32.92% | 40.19%
5 |2 | 3.22% | 16.46% | 31.25% | 32.92% | 16.08%
5 |3 | 16.08% | 32.92% | 31.25% | 16.46% | 3.22%
5 |4 | 40.19% | 32.92% | 15.63% | 4.12% | 0.32%
5 |5 | 40.19% | 13.17% | 3.13% | 0.41% | 0.01%
--------------------------------------------------------------------
6 |1 | 0.06% | 1.65% | 9.38% | 26.34% | 40.19%
6 |2 | 0.80% | 8.23% | 23.44% | 32.92% | 20.09%
6 |3 | 5.36% | 21.95% | 31.25% | 21.95% | 5.36%
6 |4 | 20.09% | 32.92% | 23.44% | 8.23% | 0.80%
6 |5 | 40.19% | 26.34% | 9.38% | 1.65% | 0.06%
6 |6 | 33.49% | 8.78% | 1.56% | 0.14% | 0.00%
Cumulative Probabilities
#dice|# "hits"| # to roll or higher!
(X) | (Y) | (Z)
| | 2 | 3 | 4 | 5 | 6
--------------------------------------------------------------------
1 |1 | 83.33% | 66.67% | 50.00% | 33.33% | 16.67%
--------------------------------------------------------------------
2 |1 | 97.22% | 88.89% | 75.00% | 55.56% | 30.56%
2 |2 | 69.44% | 44.44% | 25.00% | 11.11% | 2.78%
--------------------------------------------------------------------
3 |1 | 99.54% | 96.30% | 87.50% | 70.37% | 42.13%
3 |2 | 92.59% | 74.07% | 50.00% | 25.93% | 7.41%
3 |3 | 57.87% | 29.63% | 12.50% | 3.70% | 0.46%
--------------------------------------------------------------------
4 |1 | 99.92% | 98.77% | 93.75% | 80.25% | 51.77%
4 |2 | 98.38% | 88.89% | 68.75% | 40.74% | 13.19%
4 |3 | 86.81% | 59.26% | 31.25% | 11.11% | 1.62%
4 |4 | 86.81% | 59.26% | 31.25% | 11.11% | 1.62%
--------------------------------------------------------------------
5 |1 | 99.99% | 99.59% | 96.88% | 86.83% | 59.81%
5 |2 | 99.67% | 95.47% | 81.25% | 53.91% | 19.62%
5 |3 | 96.45% | 79.01% | 50.00% | 20.99% | 3.55%
5 |4 | 80.38% | 46.09% | 18.75% | 4.53% | 0.33%
5 |5 | 40.19% | 13.17% | 3.13% | 0.41% | 0.01%
--------------------------------------------------------------------
6 |1 | 100.00%| 99.86% | 98.44% | 91.22% | 66.51%
6 |2 | 99.93% | 98.22% | 89.06% | 64.88% | 26.32%
6 |3 | 99.13% | 89.99% | 65.63% | 31.96% | 6.23%
6 |4 | 93.77% | 68.04% | 34.38% | 10.01% | 0.87%
6 |5 | 73.68% | 35.12% | 10.94% | 1.78% | 0.07%
6 |6 | 33.49% | 8.78% | 1.56% | 0.14% | 0.00%
Edit: I have filled out my table above with @amd's proposed solution. Also, below is some more background information and my work so far. The game my friends and I are playing is Zombicide. In the game, you acquire different weapons, that have different statistics. A weapon rolls a certain number of dice, and hits on a certain roll of the dice or higher. So one weapon may roll 6 dice, but hit on only 5 and 6's, while another weapon may roll only 2 dice, but hits on 3, 4, 5, and 6's. My goal is to use the most effective weapon given the number of targets I want to hit.
When searching Google, I had originally found this chart depicting the odds. But I wanted to know the math involved as well.
My first work was to find the math to hit with one dice roll. I knew a hit of 6 would be a 16.7% probability. A hit of of 5 or 6 would be a 33% possiblity (16.7 + 16.7). Each additional dice face gives you an additional 16.7% chance to hit. I got stumped when trying to extrapolate this out for multiple dice.
What I ended up producing myself was an average number of hits table. I did this by taking the chance any particular die would hit and adding them together. So if you rolled 6 dice, hitting on 5 and 6's would on average get 0.66 hits. If you rolled 2 dice, and hit on 3, 4, 5, 6, you would average 1.33 hits.
# | hits on this +
dice|2 |3 |4 |5 |6
|----|----|----|----|-----
1 |0.83|0.67|0.50|0.33|0.17
2 |1.67|1.33|1.00|0.67|0.33
3 |2.50|2.00|1.50|1.00|0.50
4 |3.33|2.67|2.00|1.33|0.67
5 |4.17|3.33|2.50|1.67|0.83
6 |5.00|4.00|3.00|2.00|1.00
Best Answer
This is a fairly standard application of the binomial distribution, which is in fashion in games these days. It gives the probability of getting exactly $k$ successes in $n$ trials as $$ \binom n kp^k(1-p)^{n-k}, $$ where $p$ is the probability of a single trial succeeding.
In your case, the number of trials $n=X$, the number of dice that you’re rolling, and the probability of success $p$ is the probability of rolling $Z$ or higher, namely, $\frac{7-Z}6$.