Statistics Challenge

[austenandrews]

Okay, math wonks, I've got a problem I need solved.

 

In my game I frequently have lots of opponents attacking a couple of PCs. I want to speed up combat by rolling for several bad guys with one To Hit roll. Specifically I want to make one To Hit roll for a group of five (identical) opponents.

 

So let's say each bad guy has an 11-less chance to hit. If I were to roll 3d6 five times, what are the odds that all 5 would hit (rather small)? What are the odds that 4 out of 5 would hit (less small)? And so on, up to the odds that 1 will hit (greater than an 11-less chance)?

 

Ideally I'd like to make a table with "Base To-Hit Roll" across the top, "Chance For [1,2,3,4,5] Hits" down the side, and each cell containing a 3d6 result. That way I could calculate the chance for one bad guy to hit, roll 3d6 and compare the result to the table to see how many bad guys out of 5 were successful.

 

Unfortunately I have no idea what equations to use (and needless to say, it's way beyond the scope of hand-generating the percentages).

 

So who knows enough about statistics to give me a clue?

 

Thanks,

-AA

[Gary]

Re: Statistics Challenge

 

Okay, math wonks, I've got a problem I need solved.

 

In my game I frequently have lots of opponents attacking a couple of PCs. I want to speed up combat by rolling for several bad guys with one To Hit roll. Specifically I want to make one To Hit roll for a group of five (identical) opponents.

 

So let's say each bad guy has an 11-less chance to hit. If I were to roll 3d6 five times, what are the odds that all 5 would hit (rather small)? What are the odds that 4 out of 5 would hit (less small)? And so on, up to the odds that 1 will hit (greater than an 11-less chance)?

 

Ideally I'd like to make a table with "Base To-Hit Roll" across the top, "Chance For [1,2,3,4,5] Hits" down the side, and each cell containing a 3d6 result. That way I could calculate the chance for one bad guy to hit, roll 3d6 and compare the result to the table to see how many bad guys out of 5 were successful.

 

Unfortunately I have no idea what equations to use (and needless to say, it's way beyond the scope of hand-generating the percentages).

 

So who knows enough about statistics to give me a clue?

 

Thanks,

-AA

 

Use the binomial theorem. This is easy if all the opponents have the exact same to hit chances. In your example, the odds of X number of hits are as follows:

 

5 hits: (.625)^5 = .095

 

4 hits: 5!/4!1! * (.625)^4 * (.375) = .286

 

3 hits: 5!/3!2! * (.625)^3 * (.375)^2 = .343

 

2 hits: 5!/2!3! * (.625)^2 * (.375)^3 = .206

 

1 hit: 5!/1!4! * (.625) * (.375)^4 = .062

 

0 hits: (.375)^5 = .007

 

If you have any questions about the math, or about the binomial theorem, let me know. It should be trivial to setup a spreadsheet to handle any number of attackers, as long as they all have the same to hit rolls.

[austenandrews]

Re: Statistics Challenge

 

Cool! That might be exactly what I'm looking for, but it still seems like something is missing.

 

If the chance of 0 hits is .007, that means the chance of at least one hit is .993, correct? So if I roll a 17, that's still one hit?

 

If that assumption isn't off-base, how do I scale it to 2 hits, 3 hits, etc.? I guess I'm confused about "at least X hits" v.s. "exactly X hits."

 

-AA

[Gary]

Re: Statistics Challenge

 

Cool! That might be exactly what I'm looking for, but it still seems like something is missing.

 

If the chance of 0 hits is .007, that means the chance of at least one hit is .993, correct? So if I roll a 17, that's still one hit?

 

If that assumption isn't off-base, how do I scale it to 2 hits, 3 hits, etc.? I guess I'm confused about "at least X hits" v.s. "exactly X hits."

 

-AA

 

Just sum up the above numbers.

 

5 hits: .095

4+ hits: .095 + .286 = .381

3+ hits: .381 + .343 = .724

2+ hits: .724 + .206 = .930

1+ hits: .930 + .062 = .992

 

The numbers don't quite sum up to 1 because of rounding.

 

You can apply the same methodology to any number of attackers if you understand how the binomial theorem works.

[lemming]

Re: Statistics Challenge

 

The arbritrary rule I've used in the past is +2 OCV for every doubling of opponents and then use the usual autofire style of determining extra hits.

Not nearly as scientific as Gary.

[JDRook]

Re: Statistics Challenge

 

Just sum up the above numbers.

 

5 hits: .095

4+ hits: .095 + .286 = .381

3+ hits: .381 + .343 = .724

2+ hits: .724 + .206 = .930

1+ hits: .930 + .062 = .992

 

The numbers don't quite sum up to 1 because of rounding.

 

You can apply the same methodology to any number of attackers if you understand how the binomial theorem works.

 

That's pretty impressive. If I'm doing the numbers right, that means all 5 goons would hit on 6- and at least 1 would hit on 17-. This would certainly make fighting legions of goons quicker and possibly more effective.

 

If you make a spreadsheet for this, see if herogames.com would host it.

[austenandrews]

Re: Statistics Challenge

 

Awesome! That's exactly what I need, then.

 

Let me see if I've got the numbers down. For six opponents:

 

6 hits: (.625)^6 = .006

5 hits: 6!/5!1! * (.625)^5 * (.375) = .21

4 hits: 6!/4!2! * (.625)^4 * (.375)^2 = .32

3 hits: 6!/3!3! * (.625)^3 * (.375)^3 = .26

2 hits: 6!/2!4! * (.625)^2 * (.375)^4 = .12

1 hit: 6!/1!5! * (.625) * (.375)^5 = .03

0 hits: (.375)^6 = .003

 

6 hits: .006 = 3

5 hits: .22 = 7-less

4 hits: .54 = 10-less

3 hits: .80 = 12-less

2 hits: .92 = 14-less

1 hit: .95 = 15-less

 

Hmm, my numbers seem to be a little off. I just quickly plugged them into a hand calculator, probably made some mistakes. But are my formulas above correct?

 

And for different to-hit rolls, I just insert different success & failure chances for .625 and .375?

 

This will be a wonderful tool when it's finished.

 

-AA

[ghost-angel]

Re: Statistics Challenge

 

this is all well and cool, and probably extremely useful. Especially from within a time saving POV.

 

But I'd like to point out that the laws of probability reset with each use. And that some players may be unhappy with the above concept. I'd ask them before putting something like the above probabilities into action.

 

This looks most useful when running a convention game where speed really is of the essence, I don't know if I'd do this for home game use.

 

just my $.02

[Gary]

Re: Statistics Challenge

 

Awesome! That's exactly what I need, then.

 

Let me see if I've got the numbers down. For six opponents:

 

6 hits: (.625)^6 = .006

 

Your formulas are correct. Here is your error. .625^6 = .06, not .006.

 

And for different to-hit rolls, I just insert different success & failure chances for .625 and .375?

 

Correct.

[austenandrews]

Re: Statistics Challenge

 

That's pretty impressive. If I'm doing the numbers right' date=' that means all 5 goons would hit on 6- and at least 1 would hit on 17-. This would certainly make fighting legions of goons quicker and possibly more effective.[/quote']

 

Especially if you factor in a Multiple Attacker Bonus -- with 5 goons you'd get a -4 DCV modifier, which means 4 OCV goons attacking an 8 DCV hero.

 

Though it shouldn't actually be more effective, since the numbers should come out the same. Except that it's quicker, so GMs might be more inclined to throw lots of goons at the PCs.

 

The only hitch in this scheme is Blocking & Missile Deflection. How can you determine which attacks the hero attempts to block? (I suppose you could calculate the odds on that, too, but that would truly make my brain hurt.)

 

-AA

[austenandrews]

Re: Statistics Challenge

 

Your formulas are correct. Here is your error. .625^6 = .06' date=' not .006.[/quote']

 

Sweet! Thanks a bunch. I'll work on that spreadsheet and let everyone know when it's done.

 

-AA

[Sean]

Re: Statistics Challenge

 

A possibly heretical thought is to use percentile dice instead of 3d6 for generating your random number when you've got a table of probability worked out for hits. Ducks angry looks from the d6 purists. Then you can round off to 1% increments which should be sufficiently accurate for all but the most rabid rules lawyer player.

[Dust Raven]

Re: Statistics Challenge

 

The arbritrary rule I've used in the past is +2 OCV for every doubling of opponents and then use the usual autofire style of determining extra hits.

Not nearly as scientific as Gary.

Lemming, I like your idea a LOT better than these others. Primarily because I can't follow them (and to think I'm a mathmatician). I don't see any need to figure out the exact probability, especially since it's a game the simulates fiction.

 

I think I'd go with a +2 per 5 agents though, instead of with doublings.

[lemming]

Re: Statistics Challenge

 

Lemming, I like your idea a LOT better than these others. Primarily because I can't follow them (and to think I'm a mathmatician). I don't see any need to figure out the exact probability, especially since it's a game the simulates fiction.

 

I think I'd go with a +2 per 5 agents though, instead of with doublings.

Thanks. The +2/5 agents should work fine. I was doing doublings due to hordes of agents (think using the dumped contents of a box of cheerios) when I first used the rule. Since then, maybe 16 agents at a time...

[Talon]

Re: Statistics Challenge

 

See this thread for another method (make sure you read to the end ):

 

http://www.herogames.com/forums/showthread.php?t=3005

[Fireg0lem]

Re: Statistics Challenge

 

For blocking, roll the block chances first and then reduce the number of goons based on that. I believe you would figure out how many succesful blocks the hero would get in the exact same way; how many 11- rolls out of 5 will the hero make?

[austenandrews]

Re: Statistics Challenge

 

For blocking' date=' roll the block chances first and then reduce the number of goons based on that. I believe you would figure out how many succesful blocks the hero would get in the exact same way; how many 11- rolls out of 5 will the hero make?[/quote']

 

Good point. Yes, multiple Block/MD rolls can use the same charts. By reducing the number of goons, you don't have to worry about whether or not those goons would have hit, which was my initial concern. Nice catch.

 

Also the charts are useful for groups attacking groups.

 

I've worked up charts for groups of 2, 3, 4 and 5 opponents. All of the above scenarios (many v.s. one, Block/MD, many v.s. many) will likely occur in my game tonight. Tomorrow I'll let you guys know how it goes. If the system works, I'll post the charts I've got.

 

-AA

 

EDIT: Oops, I just realized that I can't use the charts for Blocking & MD, of course, because each roll builds up a -2. Have to roll each one separately. But your method of reducing the number of goons is perfect.

[austenandrews]

Re: Statistics Challenge

 

Quick update: I used the charts in my game last night and they worked like a charm! I even had an encounter between 2 PCs, 40-odd "good" NPCs and 50 mortlings (zombies). Using the charts and average instead of rolled damage, the battle came in neatly under an hour.

 

Later I'll figure out how to post the charts here.

 

-AA