Probability - Revision history

imported>Administrator: 21 revisions imported

2023-06-22T15:08:22Z

21 revisions imported

← Older revision	Revision as of 15:08, 22 June 2023
(No difference)

imported>Administrator: 1 revision imported

2023-06-19T03:12:48Z

1 revision imported

← Older revision	Revision as of 03:12, 19 June 2023
(No difference)

1d4chan>Derp commander: More work

2021-08-01T04:51:37Z

More work

← Older revision		Revision as of 04:51, 1 August 2021
Line 45:		Line 45:
	===Complex combinations/What are the odds of a specific number of things happening?===		===Complex combinations/What are the odds of a specific number of things happening?===

	Having saved the rogue, we now go to 40k. We have a squad of ten tactical marines firing bolters into 3 plague zombies that we really need cleared off a victory point. For those that stumbled in here without knowing 40k (welcome to /tg!), each marine needs to roll several d6s to take out a zombie. First, they need 3+ on a d6 to hit, then 3+ on a d6 to wound, then the zombie needs to fail a 5+ Feel No Pain roll. What are the chances those tac marines will wipe out the zombies? Should we point a valuable squad of assault marines at them for support?		Having saved the rogue, we now go to 40k. We have a squad of ten tactical marines firing bolters into 3 plague zombies that we really need cleared off a victory point. For those that stumbled in here without knowing 40k (welcome to /tg/!), each marine needs to roll several d6s to take out a zombie. First, they need 3+ on a d6 to hit, then 3+ on a d6 to wound, then the zombie needs to fail a 5+ Feel No Pain roll. What are the chances those tac marines will wipe out the zombies? Should we point a valuable squad of assault marines at them for support?

	This is going to have an easy part and a hard one. The easy part is the chance of ''one'' marine killing ''one'' zombie. A d6 has a 4/6 chance of getting a 3+, and also a 4/6 chance of missing a 5+. This marine, then, has a 4/6 * 4/6 * 4/6 = 64/216 = 8/27 = '''29% chance''' to clip a zombie from the group. We've got ten marines, so our 'expected' number of dead zombies is 10*.29, or 2.9, just a little under the three we want. That suggests we should probably call up the assault marines, as there's a good chance we won't clear that whole group out.		This is going to have an easy part and a hard one. The easy part is the chance of ''one'' marine killing ''one'' zombie. A d6 has a 4/6 chance of getting a 3+, and also a 4/6 chance of missing a 5+. This marine, then, has a 4/6 * 4/6 * 4/6 = 64/216 = 8/27 = '''29% chance''' to clip a zombie from the group. We've got ten marines, so our 'expected' number of dead zombies is 10*.29, or 2.9, just a little under the three we want. That suggests we should probably call up the assault marines, as there's a good chance we won't clear that whole group out.
Line 113:		Line 113:

	d10 vs d12? That's 9+8+etc / 10*12, so 45/120.		d10 vs d12? That's 9+8+etc / 10*12, so 45/120.

			It turns out, if ''Y'' is larger than ''X'', the formula for the probability is very simple:
			[[File:Probability Eq10.png\|x60px]]

	===What are the odds that adX beats bdY?===		===What are the odds that adX beats bdY?===
Line 153:		Line 156:

	add up all the results and you have your answer!<br/>		add up all the results and you have your answer!<br/>

			====In general====
			Now, I am a stats nerd with an undergrad education. I have spent several hours trying to figure out the exact probabilities for the sums of dice, which is probably more than any sane person would spend on this article. So I'm sure you don't care either. Plus, the math involves Greek letters, and I'm quite attached to my kneecaps.

			Fortunately, other stats nerds get tired of this shit too, so they discovered a good approximation: the normal distribution. This works well for 3dX and gets better the more dice you roll.

			Looking at my notes, for an adX the mean (average) is [[File:Probability Eq11.png\|x50px]], and the "standard deviation" is [[File:Probability Eq12.png\|x50px]]. That's all you need [https://www.wolframalpha.com/input/?i=normal+distribution+calculator&assumption=%7B%22F%22%2C+%22NormalProbabilities%22%2C+%22z%22%7D+-%3E%22-1%22&assumption=%7B%22F%22%2C+%22NormalProbabilities%22%2C+%22mu%22%7D+-%3E%220%22&assumption=%7B%22FVarOpt%22%7D+-%3E+%7B%7B%22NormalProbabilities%22%2C+%22sigma%22%7D%2C+%7B%22NormalProbabilities%22%2C+%22z%22%7D%2C+%7B%22NormalProbabilities%22%2C+%22mu%22%7D%7D to get an answer from Wolfram Alpha], but with a few caveats: You can't ask it for the probability that you'll roll ''exactly'' a number (it will pedantically tell you 0), but you can ask it the probability of rolling ''less than or equal'' or ''more than or equal'' to some number, and it doesn't respect the range of the dice pool (it will say it's possible, if unlikely, to roll a negative number, for example).

			So if, for whatever reason, you're doing a check roll for 12 or better on a 4d6, know that the mean is (4+4*6)/2 = 14 and the standard deviation is the square root of 182/3 (about 7.8), so the chance of passing the check is about [https://www.wolframalpha.com/input/?i=probability+that+x+%3E%3D+12+given+x+is+normally+distributed+mean%3D14+standard+deviation%3D7.8 60%].

	==Statistics==		==Statistics==
	If you're lazy or dealing with very weird rolls (What's the probability I'll get at least 4 successes on 10 dice, given that 1s remove a success but I ignore the first 1?) you can forgo the math and just roll the dice a million times and count the outcomes. Can't roll dice a million times? Computer can. Check out [http://topps.diku.dk/torbenm/troll.msp Troll Dice Roller]. It can take a bit of doing to figure out the formula, and some examples will require you to download and run the program locally, but it's fast and easy to understand the results.		If you're lazy or dealing with very weird rolls (What's the probability I'll get at least 4 successes on 10 dice, given that 1s remove a success but I ignore the first 1?) you can forgo the math and just roll the dice a million times and count the outcomes. Can't roll dice a million times? Computer can. Check out [http://topps.diku.dk/torbenm/troll.msp Troll Dice Roller]. It can take a bit of doing to figure out the formula, and some examples will require you to download and run the program locally, but it's fast and easy to understand the results.

1d4chan>Derp commander: let me save what I have so far

2021-08-01T00:00:56Z

let me save what I have so far

1d4chan>Derp commander: /* Combination of Two/What are the odds of at least one thing happening in a set of two? */ latexify

2021-07-31T23:13:42Z

Combination of Two/What are the odds of at least one thing happening in a set of two?: latexify

← Older revision		Revision as of 23:13, 31 July 2021
Line 37:		Line 37:
	We can get to this probability in a number of other ways, too. The simplest might be one minus the chance that neither attack hits. We know that's the product of the two miss rates, so this works out to [[File:Probability Eq4.png\|x30px]], or 1-(.4*.4)=.84. Convenient.		We can get to this probability in a number of other ways, too. The simplest might be one minus the chance that neither attack hits. We know that's the product of the two miss rates, so this works out to [[File:Probability Eq4.png\|x30px]], or 1-(.4*.4)=.84. Convenient.

	To check, we could alternatively look at the three mutually exclusive ways we could succeed this attack. The rogue could A: hit on both dice, B: hit on the first but miss the second, or C: hit on the second but miss the first. For the numbers: ~~<nowiki> P(A)=.6.6=.36, P(B)=.6.4=.24, P(C)=.6*.4=.24, this P(A)+P(B)+P(C) = .36+.24+.24 = .84 </nowiki>~~. We're now pretty sure we know the rogue's success rate, and are ready with the fleeing boots for that 16% failure chance.		To check, we could alternatively look at the three mutually exclusive ways we could succeed this attack. The rogue could A: hit on both dice, B: hit on the first but miss the second, or C: hit on the second but miss the first. For the numbers:

			[[File:Probability Eq5.png\|x60px]]

			We're now pretty sure we know the rogue's success rate, and are ready with the fleeing boots for that 16% failure chance.

	===Complex combinations/What are the odds of a specific number of things happening?===		===Complex combinations/What are the odds of a specific number of things happening?===

1d4chan>Derp commander: /* Combination of Two/What are the odds of at least one thing happening in a set of two? */ note

2019-04-14T19:28:39Z

Combination of Two/What are the odds of at least one thing happening in a set of two?: note

← Older revision		Revision as of 19:28, 14 April 2019
Line 33:		Line 33:
	Calculating this rogue's chances is a matter of combining the rules we've looked at above. To get the expected number of hits we can just add the chances from each d20: <nowiki> P({rolling 9+ on a d20})=.6</nowiki>, so we would expect <nowiki> 2*.6=1.2 </nowiki> successes, and if we wanted expected damage we could stop there.		Calculating this rogue's chances is a matter of combining the rules we've looked at above. To get the expected number of hits we can just add the chances from each d20: <nowiki> P({rolling 9+ on a d20})=.6</nowiki>, so we would expect <nowiki> 2*.6=1.2 </nowiki> successes, and if we wanted expected damage we could stop there.

	However, this is clearly bullshit, because the rogue can't hit twice on one attack. To compensate for this, we subtract the chance of getting 9+ on ''both'' rolls. This gives us the real formula we want: [[File:Probability Eq3.png\|x30px]]. This rogue, then, is looking at a .6+.6-.36='''.84 chance to hit''', or only a 16% chance to miss that crucial backstab.		However, this is clearly bullshit, because the rogue can't hit twice on one attack. To compensate for this, we subtract the chance of getting 9+ on ''both'' rolls. This, known as the inclusion-exclusion principle, gives us the real formula we want: [[File:Probability Eq3.png\|x30px]]. This rogue, then, is looking at a .6+.6-.36='''.84 chance to hit''', or only a 16% chance to miss that crucial backstab.

	We can get to this probability in a number of other ways, too. The simplest might be one minus the chance that neither attack hits. We know that's the product of the two miss rates, so this works out to [[File:Probability Eq4.png\|x30px]], or 1-(.4*.4)=.84. Convenient.		We can get to this probability in a number of other ways, too. The simplest might be one minus the chance that neither attack hits. We know that's the product of the two miss rates, so this works out to [[File:Probability Eq4.png\|x30px]], or 1-(.4*.4)=.84. Convenient.

1d4chan>Derp commander: /* Combination of Two/What are the odds of at least one thing happening in a set of two? */ add two more eqns

2019-04-14T19:21:46Z

Combination of Two/What are the odds of at least one thing happening in a set of two?: add two more eqns

← Older revision		Revision as of 19:21, 14 April 2019
Line 33:		Line 33:
	Calculating this rogue's chances is a matter of combining the rules we've looked at above. To get the expected number of hits we can just add the chances from each d20: <nowiki> P({rolling 9+ on a d20})=.6</nowiki>, so we would expect <nowiki> 2*.6=1.2 </nowiki> successes, and if we wanted expected damage we could stop there.		Calculating this rogue's chances is a matter of combining the rules we've looked at above. To get the expected number of hits we can just add the chances from each d20: <nowiki> P({rolling 9+ on a d20})=.6</nowiki>, so we would expect <nowiki> 2*.6=1.2 </nowiki> successes, and if we wanted expected damage we could stop there.

	However, this is clearly bullshit, because the rogue can't hit twice on one attack. To compensate for this, we subtract the chance of getting 9+ on ''both'' rolls. This gives us the real formula we want: ~~<nowiki> P({success}) + P({success}) - P({success})*P({success})</nowiki>~~. This rogue, then, is looking at a .6+.6-.36='''.84 chance to hit''', or only a 16% chance to miss that crucial backstab.		However, this is clearly bullshit, because the rogue can't hit twice on one attack. To compensate for this, we subtract the chance of getting 9+ on ''both'' rolls. This gives us the real formula we want: [[File:Probability Eq3.png\|x30px]]. This rogue, then, is looking at a .6+.6-.36='''.84 chance to hit''', or only a 16% chance to miss that crucial backstab.

	We can get to this probability in a number of other ways, too. The simplest might be one minus the chance that neither attack hits. We know that's the product of the two miss rates, so this works out to ~~<nowiki> 1-( P({miss}) * P({miss}) )</nowiki>~~, or 1-(.4*.4)=.84. Convenient.		We can get to this probability in a number of other ways, too. The simplest might be one minus the chance that neither attack hits. We know that's the product of the two miss rates, so this works out to [[File:Probability Eq4.png\|x30px]], or 1-(.4*.4)=.84. Convenient.

	To check, we could alternatively look at the three mutually exclusive ways we could succeed this attack. The rogue could A: hit on both dice, B: hit on the first but miss the second, or C: hit on the second but miss the first. For the numbers: <nowiki> P(A)=.6.6=.36, P(B)=.6.4=.24, P(C)=.6*.4=.24, this P(A)+P(B)+P(C) = .36+.24+.24 = .84 </nowiki>. We're now pretty sure we know the rogue's success rate, and are ready with the fleeing boots for that 16% failure chance.		To check, we could alternatively look at the three mutually exclusive ways we could succeed this attack. The rogue could A: hit on both dice, B: hit on the first but miss the second, or C: hit on the second but miss the first. For the numbers: <nowiki> P(A)=.6.6=.36, P(B)=.6.4=.24, P(C)=.6*.4=.24, this P(A)+P(B)+P(C) = .36+.24+.24 = .84 </nowiki>. We're now pretty sure we know the rogue's success rate, and are ready with the fleeing boots for that 16% failure chance.

1d4chan>Derp commander: /* Notation/How do I write this stuff? */ eq0

2019-04-14T18:57:31Z

Notation/How do I write this stuff?: eq0

← Older revision		Revision as of 18:57, 14 April 2019
Line 13:		Line 13:
	While 3/6 is numerically equivalent to 1/2, 50%, 0.5, or just .5, you do lose information regarding the total possible outcomes in simplifying fractions. However, it can be more informative to say 20% when saying "rolling a 17+ on a d20," though. Unless you're in a probability class, the way you present your answer won't matter.		While 3/6 is numerically equivalent to 1/2, 50%, 0.5, or just .5, you do lose information regarding the total possible outcomes in simplifying fractions. However, it can be more informative to say 20% when saying "rolling a 17+ on a d20," though. Unless you're in a probability class, the way you present your answer won't matter.

	The probability of an event E happening is written as ~~<nowiki>P({E})</nowiki>~~. This is always a number between 0 and 1 (including 1 and 0). If you get something other than a number between 0 and 1, you're doing something wrong.		The probability of an event E happening is written as [[File:Probability Eq0.png\|x30px]]. This is always a number between 0 and 1 (including 1 and 0). If you get something other than a number between 0 and 1, you're doing something wrong.

	===Inverse/What are the odds of something not happening?===		===Inverse/What are the odds of something not happening?===

1d4chan>Derp commander: /* Compound/What are the odds of several things happening? */ Add eqns

2019-04-07T09:11:33Z

Compound/What are the odds of several things happening?: Add eqns

← Older revision		Revision as of 09:11, 7 April 2019
Line 19:		Line 19:

	===Compound/What are the odds of several things happening?===		===Compound/What are the odds of several things happening?===
	The probability of any mutually exclusive events happening is the sum of the probabilities of each event. That is, since you cannot simultaneously roll a 1, 2, or 3 on a [[d6]], they are mutually exclusive events. The probability of rolling a 1, 2 or 3 on a [[d6]] is ~~<nowiki>P({1})+P({2})+P({3})~~=1~~/6+1/6+1/6=3/6</nowiki>~~.		The probability of any mutually exclusive events happening is the sum of the probabilities of each event. That is, since you cannot simultaneously roll a 1, 2, or 3 on a [[d6]], they are mutually exclusive events. The probability of rolling a 1, 2 or 3 on a [[d6]] is [[File:Probability Eq1.png\|frameless\|upright=1.25]].

	For multiple events happening in a specific order over multiple trials, (for instance, the odds of rolling a d20 and getting four natural twenties in a row) take all the odds and multiply them together. So, the odds of rolling one natural twenty is 1/20. Two in a row? That's 11 / 2020, or 1/400. Three in a row? 1/8000. Four? 1/160000. Getting a 9 on a d10 and then getting a 75 or higher on a d100 is 1/10*25/100 or 25/1000 or 1/40.		For multiple events happening in a specific order over multiple trials, (for instance, the odds of rolling a d20 and getting four natural twenties in a row) take all the odds and multiply them together. So, the odds of rolling one natural twenty is 1/20. Two in a row? That's 11 / 2020, or 1/400. Three in a row? 1/8000. Four? 1/160000. Getting a 9 on a d10 and then getting a 75 or higher on a d100 is 1/10*25/100 or 25/1000 or 1/40.

	What does that mean for your chances of rolling a nat 20 if you just rolled one? Absolutely fucking nothing. Your dice don't remember what you just rolled, and if they did they wouldn't care. This is known as the Lack of Memory property, and is obvious from each event (roll) being independent. In notation, it would be ~~<nowiki>P({20}~~\|~~{20})</nowiki>~~, or "the probability of rolling a 20, given that a 20 has already been rolled."		What does that mean for your chances of rolling a nat 20 if you just rolled one? Absolutely fucking nothing. Your dice don't remember what you just rolled, and if they did they wouldn't care. This is known as the Lack of Memory property, and is obvious from each event (roll) being independent. In notation, it would be [[File:Probability Eq2.png\|frameless\|upright=0.35]], or "the probability of rolling a 20, given that a 20 has already been rolled."

	Keep in mind that this is NOT the same as saying "rolling a d20 two times and getting at least one natural 20." The probability of this happening is actually 9.75%. Why? See below.		Keep in mind that this is NOT the same as saying "rolling a d20 two times and getting at least one natural 20." The probability of this happening is actually 9.75%. Why? See below.

2602:306:C5A2:8E50:D1C0:C95D:6CE7:F4C1: Info on using Troll Dice Roller /statistical averaging for figuring out probability.

2018-05-24T06:49:33Z

Info on using Troll Dice Roller /statistical averaging for figuring out probability.

← Older revision		Revision as of 06:49, 24 May 2018
Line 149:		Line 149:

	add up all the results and you have your answer!<br/>		add up all the results and you have your answer!<br/>

			==Statistics==
			If you're lazy or dealing with very weird rolls (What's the probability I'll get at least 4 successes on 10 dice, given that 1s remove a success but I ignore the first 1?) you can forgo the math and just roll the dice a million times and count the outcomes. Can't roll dice a million times? Computer can. Check out [http://topps.diku.dk/torbenm/troll.msp Troll Dice Roller]. It can take a bit of doing to figure out the formula, and some examples will require you to download and run the program locally, but it's fast and easy to understand the results.

← Older revision		Revision as of 00:00, 1 August 2021
Line 53:		Line 53:
	We have ten marines and we're looking to put at least three unsaved wounds downrange, so we're looking for the number of ways to score precisely 3 successes in 10 tests. To make matters a little worse, we actually also need the number of ways to score 4 successes, or 5, or 6, or more, up to ten. Inflicting more wounds is not a problem for us. To work this out, we'll need the probability for a given scenario of hits, and the number of ways that scenario can occur.		We have ten marines and we're looking to put at least three unsaved wounds downrange, so we're looking for the number of ways to score precisely 3 successes in 10 tests. To make matters a little worse, we actually also need the number of ways to score 4 successes, or 5, or 6, or more, up to ten. Inflicting more wounds is not a problem for us. To work this out, we'll need the probability for a given scenario of hits, and the number of ways that scenario can occur.

	Let's look at inflicting exactly one wound. This requires one marine to succeed and all the others to fail. If we said that marine were the first guy in the squad we could tell how likely that is by multiplying everything together. We know the first marine succeeds and '''all''' the others fail, so this is ~~P(success)P(failure)P(failure)P(failure)etc..~~. one success and nine fails, in that order. P(failure) is just 1-P(success), so we've got .29 and .71.		Let's look at inflicting exactly one wound. This requires one marine to succeed and all the others to fail. If we said that marine were the first guy in the squad we could tell how likely that is by multiplying everything together. We know the first marine succeeds and '''all''' the others fail, so this is [[File:Probability Eq6.png\|x30px]] one success and nine fails, in that order. P(failure) is just 1-P(success), so we've got .29 and .71.

	The probability of any scenario like this, then, is just ~~<nowiki> P(success)^{number of successes} * P(failure)^{number of failures}~~, or in this case, .29.71.71.71.71.71.71.71.71.71 = .29.71^9 = </nowiki> .29*.0458 = '''.0133'''. Wow, it sure got tedious in here. So, this 1.3% chance is the chance of scoring exactly one kill in exactly this way.		The probability of any scenario like this, then, is just [[File:Probability Eq7.png\|x30px]], or in this case, <nowiki>.29.71.71.71.71.71.71.71.71.71 = .29.71^9 = </nowiki> .29*.0458 = '''.0133'''. Wow, it sure got tedious in here. So, this 1.3% chance is the chance of scoring exactly one kill in exactly this way.

	We know there ten scenarios exactly like this one, differing only in which marine got the kill. We don't care about the difference, so those ten scenarios with 1.33% each give us 10*.0133 = .133, or a '''13.3% chance to kill exactly one zombie'''.		We know there ten scenarios exactly like this one, differing only in which marine got the kill. We don't care about the difference, so those ten scenarios with 1.33% each give us 10*.0133 = .133, or a '''13.3% chance to kill exactly one zombie'''.
Line 69:		Line 69:
	Know what a factorial is? In math, the exclamation point pulls double duty both as a point of excitement and also as notation for 'multiply this number by everything below it'. 5!, for example, is 54321. Everyone learns about them in middle school, then nobody ever uses them. ...until now.		Know what a factorial is? In math, the exclamation point pulls double duty both as a point of excitement and also as notation for 'multiply this number by everything below it'. 5!, for example, is 54321. Everyone learns about them in middle school, then nobody ever uses them. ...until now.

	~~TcS = T! / (S! * (T-S)! )~~		[[File:Probability Eq7a.png\|x90px]]

	So, 10C3 = 10! / (3! * 7!) = (10987654321)/((321)(7654321)). Cool thing is, the sequence 7654321 will always show up in the numerator and the denominator. We can cross it right off.		So, 10C3 = 10! / (3! * 7!) = (10987654321)/((321)(7654321)). Cool thing is, the sequence 7654321 will always show up in the numerator and the denominator. We can cross it right off.
Line 80:		Line 80:


	Alright, we've got our tool. Back on track. Ten marines, three zombies. The probability of any scenario that delivers '''S''' wounds from '''T''' marines with a per-marine kill rate of '''P''' is ~~'''P^S*(1-P)^(T-S)'''~~. There are '''TcS''' ways for that scenario to occur. We're down to the good stuff:		Alright, we've got our tool. Back on track. Ten marines, three zombies. The probability of any scenario that delivers '''S''' wounds from '''T''' marines with a per-marine kill rate of '''P''' is [[File:Probability Eq8.png\|x30px]]. There are '''TcS''' ways for that scenario to occur. We're down to the good stuff:

	Probability of inflicting exactly S wounds = ~~'''TcS * P^S*(1-P)^(T-S)'''~~		Probability of inflicting exactly S wounds = [[File:Probability Eq9.png\|x30px]].

	Exactly three wounds, ten marines, three zombies? 10c3 * (.29^3)(.71^(10-3)). [https://www.wolframalpha.com/input/?i=combination(10,3)(.29%5E3)*(.71%5E(10-3)) Wolfram Alpha, away!] '''26.6%'''		Exactly three wounds, ten marines, three zombies? 10c3 * (.29^3)(.71^(10-3)). [https://www.wolframalpha.com/input/?i=combination(10,3)(.29%5E3)*(.71%5E(10-3)) Wolfram Alpha, away!] '''26.6%'''
Line 88:		Line 88:
	Ever so close. That's the chance of exactly three unsaved wounds. We would be fine with four, or five, or any other number above three. So we need to either [https://www.wolframalpha.com/input/?i=sum+combination(10,j)(.29%5Ej)(.71%5E(10-j)),+j%3D3+to+10 add up everything above S=3], or [https://www.wolframalpha.com/input/?i=1-sum+combination(10,j)(.29%5Ej)(.71%5E(10-j)),+j%3D0+to+2 add the results for zero, one, and two wounds and subtract that from one]. '''59% chance.''' This game is going to be close, but the marines are more likely than not to win.		Ever so close. That's the chance of exactly three unsaved wounds. We would be fine with four, or five, or any other number above three. So we need to either [https://www.wolframalpha.com/input/?i=sum+combination(10,j)(.29%5Ej)(.71%5E(10-j)),+j%3D3+to+10 add up everything above S=3], or [https://www.wolframalpha.com/input/?i=1-sum+combination(10,j)(.29%5Ej)(.71%5E(10-j)),+j%3D0+to+2 add the results for zero, one, and two wounds and subtract that from one]. '''59% chance.''' This game is going to be close, but the marines are more likely than not to win.

	'''TL;DR''' The general form of all this is that if you have a number of tries T each with a success rate of P, the chance of getting exactly S successes is ~~'''TcS * (P^S)((1-P)^(T-S))'''~~. If you're looking for at least S successes, add the results of this formula for everything between S and T. Or, if you're a real human with things to do, [https://www.wolframalpha.com/input/?i=sum+(combination(T,j)(P%5Ej)*((1-P)%5E(T-j))),+j%3DS+to+T+where+P%3D.29,+S%3D3,+T%3D10 make a machine do it].		'''TL;DR''' The general form of all this is that if you have a number of tries T each with a success rate of P, the chance of getting exactly S successes is [[File:Probability Eq9.png\|x30px]]. If you're looking for at least S successes, add the results of this formula for everything between S and T. Or, if you're a real human with things to do, [https://www.wolframalpha.com/input/?i=sum+(combination(T,j)(P%5Ej)((1-P)%5E(T-j))),+j%3DS+to+T+where+P%3D.29,+S%3D3,+T%3D10 make a machine do it].

	This is pretty easy to modify. Just change the inputs a little bit to try things out. [https://www.wolframalpha.com/input/?i=sum+(combination(T,j)(P%5Ej)((1-P)%5E(T-j))),+j%3DS+to+T+where+P%3D(2%2F32%2F32%2F3),+S%3D3,+T%3D20 Marines are rapid firing]? [https://www.wolframalpha.com/input/?i=sum+(combination(T,j)(P%5Ej)((1-P)%5E(T-j))),+j%3DS+to+T+where+P%3D(2%2F32%2F3),+S%3D3,+T%3D10 Plague zombies are normal cultists]? [https://www.wolframalpha.com/input/?i=sum+(combination(T,j)(P%5Ej)((1-P)%5E(T-j))),+j%3DS+to+T+where+P%3D(2%2F35%2F6*2%2F3),+S%3D4,+T%3D8 Eight devastators firing S8 AP3 krak missiles at a T6 3+ Carnifex with 5+ cover]? Enjoy.		This is pretty easy to modify. Just change the inputs a little bit to try things out. [https://www.wolframalpha.com/input/?i=sum+(combination(T,j)(P%5Ej)((1-P)%5E(T-j))),+j%3DS+to+T+where+P%3D(2%2F32%2F32%2F3),+S%3D3,+T%3D20 Marines are rapid firing]? [https://www.wolframalpha.com/input/?i=sum+(combination(T,j)(P%5Ej)((1-P)%5E(T-j))),+j%3DS+to+T+where+P%3D(2%2F32%2F3),+S%3D3,+T%3D10 Plague zombies are normal cultists]? [https://www.wolframalpha.com/input/?i=sum+(combination(T,j)(P%5Ej)((1-P)%5E(T-j))),+j%3DS+to+T+where+P%3D(2%2F35%2F6*2%2F3),+S%3D4,+T%3D8 Eight devastators firing S8 AP3 krak missiles at a T6 3+ Carnifex with 5+ cover]? Enjoy.