I play a board game and need to be able to figure out in my head the probability of making a hit given certain scenarios. Each unit rolls a certain number of dice and makes a hit for every die below a certain number. So a unit with (3,2) rolls 3 dice and hits on a 2 or less (and gets a hit for every die with a 2 or less). I need to figure out on the fly the probability of at least one hit, at least 2 hits, etc… I thought maybe there was a way to figure out the wrong way (that is, add 33% chance 3 times) and then modify that number somehow. Can anyone help me figure out a way to do this fast, easily, in my head?
For (N,m), what is the range of N over which you are interested? (That is, how many dice might you be rolling?) This would help guide the heuristic. Also, to confirm: you are interested in the probability of at least m hits, not the probability of exactly m hits, right?
Generally, the units will roll between 1 and 4 die/dice. And yes, I’m interested in “at least,” not exactly. Thanks.
Well, those numbers of dice are quite small, so perhaps we should just cover all the cases exactly first, and then worry about simplifying for easier memorization or calculation later:
Letting p be the probability of making a hit with just one die, and q = 1 - p be the probability of failing to make a hit with just one die:
The probability that every die hits is p^n, where n is the number of dice. Similarly, the probability that at least one die hits is 1 - q^n.
The probability that at most one die fails to hit is p^n + n * q * p^(n - 1). This is the only sort of messy one, and perhaps cries most for a simple approximation…
Finally, the probability that, out of four dice, two hit and two fail is: 6(pq)^2.
This covers every question you might care to ask about up to 4 dice. Now to simplify it…
During game play, do you actually want the probability, or are you more interested in whether or not the chance is better than, say, two-thirds of achieving a certain number of hits?
In other words…
You have asked for a quick way to answer: What is the probability of achieving H or more hits with N dice for a hit threshold of M pips. However, the situation is simpler if really all you want to know is: How large does M have to be if I want to be at least (roughly) 70% certain of achieving my goal of H hits with N dice.
A simple lookup table for this would be:
Num. Hits Thus, require Or written so as to
Dice Needed M at least... show a pattern...
1 1 4 4
2 1,2 3,5 3 5
3 1,2,3 2,4,6 2 4 6
4 1,2,3,4 2,3,5,6 2 3 5 6
The actual probabilities for these pip thresholds are:
D H M Prob
1 1 4 0.67
2 1 3 0.75
2 2 5 0.69
3 1 2 0.70
3 2 4 0.74
3 3 6 1.00
4 1 2 0.80
4 2 3 0.69
4 3 5 0.87
4 4 6 1.00
You could relax the ones that go to M=6 down to M=5 if you’re okay dropping to 50-60% for those two.
So… is this the sort of test you’d actually do with any probabilities you would end up calculating (in which case the threshold table might suffice)? Or, do you need actual probabilities to feed into, say, an expectation value calculation?
There seems to be a problem with this:
If q= 1 - p
then pq will be identical for p=1/6 and p=5/6 and for p=2/6 and p=4/6
because 1-5/6=1/6 and 1-5/6=1/6 (same for 2/6 and 4/6)
therefore, you would have the same odds for rolling under 2 as under 6, 38%.
I think maybe you forgot a term in the equation, either that or I’m interpreting wrong.
I don’t know the right formula for this, but I’m working on addressing the OP and have made some progress with the following charts:
For at least 1 hit (yes, there is rounding, intentionally):
number of dice across the top, number to roll below down the left
1 2 3 4
2 1/6 3/9 5/12 8/16
3 2/6 5/9 8/12 13/16
4 3/6 7/9 11/12 15/16
5 4/6 8/9 1 1
6 5/6 1 1 1
at least 2 hits:
1 2 3
2 0 0 1/12
3 0 1/9 3/12
4 0 2/9 6/12
5 0 4/9 9/12
6 0 6/9 11/12
to be continued…
The idea will be to express the results in fractions that use the same denominator for a given number of dice and then watch how the numberator changes as you change the ‘to-hit’ roll and the number of hits required. It should be easier to remember than percentages or some other options.
Plus I’m just curious now.
Whoops, I gave the probability of exactly hitting two and missing two (which does have the observed symmetry), when I was supposed to give the odds of hitting at least two.
Alright, the probability of hitting at least two = 1 - [the probability of missing at most one, but with p and q switched].
So, to summarize:
Probability of hitting all: p^n
Probability of hitting at least one: 1 - q^n
Probability of hitting all but one: p^n (1 + nq/p)
Probability of hitting at least two: 1 - q^n (1 + np/q)
answered while posting 
Or summarized even more:
Probability of hitting all: p^n
Probability of missing at most one: p^n (1 + nq/p)
Probability of hitting at least k (i.e., probability of not missing >= all but k - 1): 1 - Probability of hitting >= all but (k - 1) with p and q swapped
That should do it, I think. Exact, fairly easy to memorize and apply, and covers all the cases.
OK, so here are the fraction charts:
Number of dice rolled are listed across the top.
Number needed to roll below is listed down the left.
If you need at least 1 hit:
1 2 3 4
2 1/6 3/9 5/12 8/16
3 2/6 5/9 8/12 13/16
4 3/6 7/9 11/12 15/16
5 4/6 8/9 1 1
6 5/6 1 1 1
If you need at least 2 hits:
1 2 3 4
2 0 0 1/12 2/16
3 0 1/9 3/12 7/16
4 0 2/9 6/12 11/16
5 0 4/9 9/12 14/16
6 0 6/9 11/12 1
If you need at least 3 hits:
1 2 3 4
2 0 0 0 0
3 0 0 0 2/16
4 0 0 2/12 5/16
5 0 0 4/12 9/16
6 0 0 7/12 14/16
If you need at least 4 hits:
1 2 3 4
2 0 0 0 0
3 0 0 0 0
4 0 0 0 1/16
5 0 0 0 3/16
6 0 0 0 8/16
It’s a bit late, I don’t think my brain can handle trying to pull patterns out of this right now - but maybe there’s something there.
Minor note: The way the OP worded it, they might want the left labelled “Number to roll <= than” rather than “Number to roll < than”; i.e., from 1 through 5.
Major note: I don’t think your charts are quite correct… For example, you list it as impossible to get all 4 hits of 4 dice to be below 3 (i.e., <= 2); why would that be impossible? It has probability 1/81.
I think I’ve got what the OP is after here - charts can easily be changed if not.
As to point #2 - it’s certainly not impossible, but it is rounded to zero on the basis of the denominator being used. Essentially, anything that is listed as zero has a probability of less than 1/2 of 1 over the denominator for that amount of dice.
I’m certain that the player could determine that if he has four dice, he could roll 4 of anything without needing a chart
The chart just says it ain’t likely.
Oh, so the chart was only approximate? Sorry, I missed that.
My bit about the OP’s labelling standards came from:
Though I’m sure he can cope with any way of presenting it.
Yeah, the OP’s two statements do kind of conflict in terms of defining the roll needed. But it’s no big deal - if you wanted 5 or less, instead of less than 5, then you would just look at the row labeled 6 instead of the row labeled 5. Or better yet, just relabel them, subtracting 1 from each row label.
To put the above comments in rather simpler language, instead of thinking about hitting, think about missing.
If you’re rolling 3 6-sided dice and hitting on a 1 or a 2, you have a chance per die of missing of 4/6 (rolls of 3, 4, 5, 6), which reduces to 2/3. So, for all to miss, the chance is 2/3 x 2/3 x 2/3 = 8/27. Thus the chance of at least one hitting is 1 - (8/27) or roughly 2/3.
Thanks for all the responses. I’ve been away so haven’t had a chance to respond in kind. Your questions about my intent have made me reexamine what exactly I’m trying to estimate. Ultimately, I want to quickly estimate the odds of winning an engagement but I’m thinking that’s gonna be pretty hard to do. For those interested here are the details of a battle:
Each fighting unit has a Speed (A,B,C), a Strength (1-4), and an Attack (1-4) Value.
Speed determines who rolls first. A units roll before B, etc… Defender rolls A’s before Attacker rolls A’s,etc…
Strength determines how many die are rolled.
Attack determines the maximum value that will score a hit.
Each hit reduces the strength (die rolls) of the enemy unit with the highest current strength by one. The defender has a significant advantage since he will roll first on Speed ties and therefore have a chance to reduce the number of enemy die on the return attack.
Typically there will be 3 units per side in any battle.
I was thinking something along the lines of giving a numerical value to Speed and creating a function that gives an overall strength of a piece. It would certainly be rough, but at least it would be objective.