View Full Version : a question of odds

Miredorf

09-12-2012, 14:33

Hello im not really sure if this is the proper place to post this, but since it is related to mathhammer i though itd be fitting?

The question is basically what is the chance of obtaining a double 6 with 6 dice. And then a triple 6 again with 6 dice.

I would also be very grateful if you could post here the formula for calculating them. Thanks in advance!

NixonAsADaemonPrince

09-12-2012, 15:04

Do you want the chance for getting just a double 6 (e.g 6,6,x,x,x,x) or the chance of getting a roll including a double 6 (e.g 6,6,6,6,x,x). In other words, does it matter if say 4 6s are rolled, does that still count as a double?

dementian

09-12-2012, 19:16

As in chance of 2 or more 6's? or 3 or more 6s?

NixonAsADaemonPrince

09-12-2012, 19:31

As in chance of 2 or more 6's? or 3 or more 6s?

Exactly, expressed with much improved clarity kind sir!

AntaresCD

10-12-2012, 19:26

Probability of 2 or more 6s (6 dice rolled):

It's actually easier to calculate 1 or less 6s, so let's do that, since prob 2+ 6s = 1 - prob 1- 6s.

Prob 1- 6s = prob exactly 0 6s + prob exactly 1 6

Prob exactly 0 6s = 6C0 * (5/6)^6 * (1/6)^0 = 1 * 0.3348... * 1 = 0.3348...

Prob exactly 1 6 = 6C1 * (5/6)^5 * (1/6)^1 = 6 * 0.4018... * 0.1666... = 0.4018...

Therefore: Prob 1- 6s = 0.3348... + 0.4018... = 0.7367...

Therefore Prob 2+ 6s = 1 - 0.7367... = 0.2632... or approximately 26.3%

Probability of 3 or more 6s (6 dice rolled):

Similarly, it's actually easier to calculate 2 or less 6s, so let's do that, since prob 3+ 6s = 1 - prob 2- 6s.

Prob 2- 6s = prob exactly 0 6s + prob exactly 1 6 + prob exactly 2 6s

Prob exactly 0 6s = 6C0 * (5/6)^6 * (1/6)^0 = 1 * 0.3348... * 1 = 0.3348...

Prob exactly 1 6 = 6C1 * (5/6)^5 * (1/6)^1 = 6 * 0.4018... * 0.1666... = 0.4018...

Prob exactly 2 6s = 6C2 * (5/6)^4 * (1/6)^2 = 15 * 0.4822... * 0.0277... = 0.2009...

Therefore: Prob 2- 6s = 0.3348... + 0.4018... + 0.2009... = 0.9377...

Therefore Prob 3+ 6s = 1 - 0.9377... = 0.0622... or approximately 6.2%

Hope that helps (and that my statistics weren't rusty or my transcription sloppy)!

russellmoo

11-12-2012, 01:37

Nice thinking Antares- I looked at this thread- realized how much math would go into calculating 2 or more 6's and thought it was not worth the effort- It didn't occur to me to calculate the other side of the probability and do some simple subtraction- your method is much easier and quicker-

Lord Inquisitor

11-12-2012, 01:46

http://vassarstats.net/binomialX.html

This is a site I use all the time for this sort of thing. What you're looking for is binomial theory (since there are two outcomes to each trial - either you get a 6 or you don't). While it's good to understand the maths behind it, the calculator is very handy and convenient.

So for example, if you want to know the odds of rolling two sixes out of six dice, use the following

n (total number of attempts) = 6

k (necessary number of passes) = 2

p (probability of success) = 1/6

Then calculate! This will give you the odds of "exactly two", or "two or more" as output.

And you can check that Antares's numbers are right. Which they are (despite a typo in his titles).

http://vassarstats.net/binomialX.html

This is a site I use all the time for this sort of thing. What you're looking for is binomial theory (since there are two outcomes to each trial - either you get a 6 or you don't). While it's good to understand the maths behind it, the calculator is very handy and convenient.

So for example, if you want to know the odds of rolling two sixes out of six dice, use the following

n (total number of attempts) = 6

k (necessary number of passes) = 2

p (probability of success) = 1/6

Then calculate! This will give you the odds of "exactly two", or "two or more" as output.

And you can check that Antares's numbers are right. Which they are (despite a typo in his titles).

Thank you for the link, I didn't know that site!

ElBeaver

11-12-2012, 15:51

AnyDice.com is also nice. But requires basic programming skills.

AntaresCD

11-12-2012, 20:59

And you can check that Antares's numbers are right. Which they are (despite a typo in his titles).

I was so busy making sure I transcribed the math properly that I forgot to check the titles, oops...

Miredorf

13-12-2012, 22:49

hey guys really thanks for all the answers =). All this comes because we wanted to make irresistible forces less common, but im seeing triple 6 is way too rare.. only 6% in 6 dice... :/

Chicago Slim

20-12-2012, 14:14

TOTAL POWER TABLE:

The statistical tool you're looking for is called a binomial distribution... Here's some probabilities:

Binomial distribution for two or more sixes on 2-6 dice:

2 0.027

3 0.074

4 0.131

5 0.196

6 0.263

And, you're correct, moving it to 3 sixes instead of two would hugely decrease the odds:

3 0.004

4 0.016

5 0.035

6 0.062

You're welcome!

As to the specific question of how common total power should be, I personally think that 20-26% on 5-6 dice is pretty reasonable: often enough that it comes up, but low enough percentage that you can never expect it...

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