Dave Myers
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Here's the scenario:

I roll seven 6-sided dice. I want to know the probability of rolling a 1 on ANY of the dice.

Further, what is the probability of rolling two 1's with those seven dice? And separately what is the probability of rolling three 1's with those same seven dice?

Then do the same thing for six 6-sided dice and five 6-sided dice and four 6-sided dice, etc.

Any help would be greatly appreciated.

Thanks.
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Todd Snyder
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Let me find the formula, but here are the probabilities for each. To find a 1 on any you would take 1 - the probability of zero, so with 7 dice the probability to get at least 1 one is 0.720918352766347.


7 dice

numberof1s probability
0 0.279081647233653
1 0.390714306127115
2 0.234428583676269
3 0.078142861225423
4 0.0156285722450846
5 0.00187542866941015
6 0.000125028577960677
7 3.57224508459076E-06

6 dice

numberof1s probability
0 0.334897976680384
1 0.401877572016461
2 0.20093878600823
3 0.0535836762688615
4 0.00803755144032922
5 0.000643004115226337
6 2.14334705075446E-05

5 dice

numberof1s probability
0 0.401877572016461
1 0.401877572016461
2 0.160751028806584
3 0.0321502057613169
4 0.00321502057613169
5 0.000128600823045268

4 dice

numberof1s probability
0 0.482253086419753
1 0.385802469135802
2 0.115740740740741
3 0.0154320987654321
4 0.000771604938271605

3 dice

numberof1s probability
0 0.578703703703704
1 0.347222222222222
2 0.0694444444444444
3 0.00462962962962963

2 dice

numberof1s probability
0 0.694444444444444
1 0.277777777777778
2 0.0277777777777778

1 die

numberof1s probability
0 0.833333333333333
1 0.166666666666667

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myersd37 wrote:
I roll seven 6-sided dice. I want to know the probability of rolling a 1 on ANY of the dice.


This is the complement to rolling 1 on no die, or 1-(5/6)^7 = 0.72 = 72%.

myersd37 wrote:
Further, what is the probability of rolling two 1's with those seven dice? And separately what is the probability of rolling three 1's with those same seven dice?


This is the probability of rolling exactly 2 1s and 5 non-1s:

(1/6)^2 * (5/6)^5 = 0.011

multiplied by the number of ways to choose any 2 out of 7, which is 21, so the answer is 0.23 = 23%. I think!
 
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David "Brother" Eicher
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http://anydice.com/

The math is pretty complex, but this program ought to be able to help you out.

FWIW, 7D6 have a 72.09% chance of containing at least one 1.
 
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Jeremy Lennert
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And here's the exact code you can run in anydice:
http://anydice.com/program/37fc

(Technically that's rolling six-sided dice that say "1" on one side and "0" on the other 5 sides and adding up the results, but that comes out the same as counting the number of 1s rolled on conventional dice.)
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Todd Snyder
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ok, I think I got the formula right, and if not, someone will correct me. Pretty sure the parens are in the right place too...

n = number of dice
s = sides on the dice
x = number of a digit you want to have

** = raised to the power
! = factorial

((s-1)**(n-x) / (s**n)) * (n! / (( n-x)! * x!))

so if n = 3, s = 6, and x = 0, we have...

((6-1)**(3-0) / (6**3) * (3! / (3!*0!)) = (5**3 / 6 ** 3) * (6 / 6) = 125 / 216 = .578703

and if n = 3, s = 6, and x = 1, we have...

((6-1)**(3-1) / (6**3) * (3! / (2!*1!)) = (5**2 / 6 ** 3) * (6 / 2) = (25 / 216)*3 = .34722

you get the idea.
 
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Dave Myers
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Wow. Thanks for the quick responses. This was definitely out of my math range.

Anyone want to tackle the follow up question?

Same scenario (I want 1s), but this time I start with seven dice, keep all the 1s, roll the rest of the dice, keep the all the 1s, etc. (Yatzee style, maximum three rolls). What is the probability that I get a three of a kind of only 1s? What is the probability that I get two of a kind (pair of 1s only)? Now what happens when I start with six dice instead of seven. How does the probability change? What about if I start with 5 dice?

I know this is a much harder question, so I have no hope on my own. Help me, BGG forums, you're my only hope.
 
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Todd Snyder
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You are looking at 3 different probabilities so I don't think you can easily do it strictly with one formula, but you can use branching to find the answer to that. With 7 dice, you have the probabilities above.


7 dice

numberof1s probability
0 0.279081647233653
1 0.390714306127115
2 0.234428583676269
3 0.078142861225423
4 0.0156285722450846
5 0.00187542866941015
6 0.000125028577960677
7 3.57224508459076E-06


so after the first roll, you have the above probability for each outcome. Then on your second roll, you use the other tables as needed, and then multiply the probabilities together.

For example, what is the probability of getting zero 1s with 7 dice being able to reroll 2 times (3 rolls total). It would be .27908 * .27908 * .27908 = .02173

What about exactly 1 one? Well, in this case, there are 3 ways to go about it, you either get the 1 on the first roll, the 2nd, or the 3rd.

If you get it on the first roll, the probabilites are .3907 (1 out of 7) * .3349 (0 out of 6) * .3349 + .27908 (0 out of 7)* .3907 (1 out of 7) * .3349 (0 out of 6) + .27908 (0 out of 7) * .27908 (0 out of 7) * .3907 (1 out of 7) = .11076

Things get a little more complex with exactly 2, as you can get 2, 0, 0 or 0, 2, 0 or 0, 0, 2 or 1, 1, 0 or 1, 0, 1 or 0, 1, 1. And so on. Plop stuff in a spreadsheet to help with the math. Or write a simple program to do the estimations. Things work more or less the same with fewer (or more) dice.

 
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Franz Kafka
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myersd37 wrote:
Wow. Thanks for the quick responses. This was definitely out of my math range.

Anyone want to tackle the follow up question?

Same scenario (I want 1s), but this time I start with seven dice, keep all the 1s, roll the rest of the dice, keep the all the 1s, etc. (Yatzee style, maximum three rolls). What is the probability that I get a three of a kind of only 1s? What is the probability that I get two of a kind (pair of 1s only)? Now what happens when I start with six dice instead of seven. How does the probability change? What about if I start with 5 dice?

I know this is a much harder question, so I have no hope on my own. Help me, BGG forums, you're my only hope.


If you're specifically trying to get as many 1s as you can, we could modify the existing analysis.

With just one roll, the chance of a 1 on a specific die is 1/6. If you're doing three rolls, it would be 1/6 + (5/6)*(1/6) + (5/6)*(5/6)*(1/6) = 91/216.

Edit: it would take some rearranging of Todd's formula. Where he has
((s-1)**(n-x) / (s**n))
I might have written
((1-(1/s))**(n-x)) * (1/s)**x
instead.
Then we could substitute 91/216 for 1/s.
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Andrés Santiago Pérez-Bergquist
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The formula everyone is giving is known as the Binomial Distribution, and lets you figure out the probability of an event happening k out of n times in independent trials (such as separate identical dice) each of which has probability p of succeeding. (And, through simple addition, you can thus figure out the probability of at least k or at most k.)

http://en.wikipedia.org/wiki/Binomial_distribution
 
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