Monday, March 22, 2010

A Math Problem


Well, more precisely, a probabilities problem.

Some of you will already be familiar with my obsessive nature. Assuming you have forgotten same, I will refer you to my recent spate of Harpies posts.

My recent obsession is related to the Chainmail combat tables, and their relationship to original DnD. In ODnD, you are directed to either employ the alternative d20 combat table, found in the first LBB, or use the Chainmail combat tables. If you use the Chainmail mass-combat tables, you treat your heroic ODnD character as multiple combatants, based on the LBB charts provided -- by character class, and are permitted to roll multiple six-sided dice to determine hits/kills. The number of dice you roll increases as you increase in levels.

Thinking about that Chainmail mass-combat system led me to the above table. I was trying to determine the probability of a certain number of sixes being rolled, based on the number of six-sided dice you were permitted. I'm not sure why it matters, i'm just obsessive that way.

Sadly, I do not remember my permutations, combinations and probabilities module from senior high-school (it being several *cough* years since I graduated) so much of the above table was created manually. Horrors!

You read the table thusly: starting on the first row (labelled 1) if you roll 1d6, there is a 83% chance that no six will be rolled, and a 17% chance that one six will be rolled. Going to row 2, if you roll 2d6, there is a 69% chance that no sixes will be rolled, a 28% chance that one six will be rolled, and a 3% chance that two sixes will be rolled. And so on. The "n" in the table is where the probability is less than one one-hundredth of a percent, and I classified that probability as "negligible" (less that 1/2 of one percent is also pretty negligible, but I couldn't bear to drop those probabilities, for some obsessive reason).

My skills (and confidence in my calculations) started to fail me, as you can see by the incompleteness of the table. I'm hoping one of my more mathematically-inclined readers will remind me of the formula to calculate these probabilities, or, even more generously, fill in the blanks...

Incidentally, I was interested to see that the probability of rolling one six diminishes, once you roll more than six dice. That makes sense, since there will be an increasing probability that you will now roll two or more sixes.

4 comments:

Norman Harman said...

Sorry, no remember math neither. Instead I have puter brute force. 10min and 1,000,000 rolls per row later we get some sadly not very accurate data. Who cares about tenths of a percentage anyway?

Oh why is the pre tag not allowed, curse you formating heathans!

| | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 1 | 83.34 | 16.66 | 0.00 | 0.00 | 0.00 | 0.00 | 0.00 | 0.00 | 0.00 | 0.00 | 0.00 |
| 2 | 69.48 | 27.75 | 2.77 | 0.00 | 0.00 | 0.00 | 0.00 | 0.00 | 0.00 | 0.00 | 0.00 |
| 3 | 57.87 | 34.76 | 6.91 | 0.46 | 0.00 | 0.00 | 0.00 | 0.00 | 0.00 | 0.00 | 0.00 |
| 4 | 48.17 | 38.64 | 11.59 | 1.54 | 0.08 | 0.00 | 0.00 | 0.00 | 0.00 | 0.00 | 0.00 |
| 5 | 40.21 | 40.15 | 16.07 | 3.23 | 0.32 | 0.01 | 0.00 | 0.00 | 0.00 | 0.00 | 0.00 |
| 6 | 33.46 | 40.21 | 20.15 | 5.33 | 0.79 | 0.06 | 0.00 | 0.00 | 0.00 | 0.00 | 0.00 |
| 7 | 27.99 | 39.04 | 23.39 | 7.83 | 1.55 | 0.18 | 0.01 | 0.00 | 0.00 | 0.00 | 0.00 |
| 8 | 23.24 | 37.29 | 26.01 | 10.41 | 2.59 | 0.42 | 0.04 | 0.00 | 0.00 | 0.00 | 0.00 |
| 9 | 19.32 | 34.93 | 27.96 | 13.00 | 3.91 | 0.77 | 0.10 | 0.01 | 0.00 | 0.00 | 0.00 |
| 10 | 16.11 | 32.28 | 29.08 | 15.60 | 5.41 | 1.30 | 0.22 | 0.03 | 0.00 | 0.00 | 0.00 |

A Paladin In Citadel said...

You are a scholar and a gentleman!

My most sincere thanks, my wife will be pleased that I will stop mumbling "factorials, factorials!" in my sleep.

A Paladin In Citadel said...

Incidentally, looking at my notes, your calcs should be very accurate up to 7 dice. Would you believe their are 1,679,616 combinations with 8 dice, over 10 million with 9 dice, at over 60 million with 10 dice? Try figuring those probabilities manually!

Lior said...

The probability for k successes rolling n dice is

n! / (k!((n-k)!)) p^k q^{n-k}

Where p=1/6 is the probability of success and q=1-p=5/6 is the probability of failure.