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Subject: Probabilities in dice rolling rss

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Leigh Caple
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My maths is failing me, can anyone help??

I know that when rolling 3 six-sided dice the chance of getting 3 sixes is 1 in 216, but what do the odds change to when rolling 4 dice and getting 3 sixes? What about 5 dice? Or 6 dice?

Again the odds for rolling 5s or 6s of 3 dice is 1 in 27 but how does that change if I were to roll 4 dice or 5 dice or 6 dice?
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Ethan McKinney
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First question: do you need exactly 3 sixes, or is it 3 or more sixes?
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David C
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Leighbob wrote:
My maths is failing me, can anyone help??

I know that when rolling 3 six-sided dice the chance of getting 3 sixes is 1 in 216, but what do the odds change to when rolling 4 dice and getting 3 sixes? What about 5 dice? Or 6 dice?

Again the odds for rolling 5s or 6s of 3 dice is 1 in 27 but how does that change if I were to roll 4 dice or 5 dice or 6 dice?


1/6 * 1/6 * 1/6 = 1/216

The odds of NOT getting a six on a dice is 5/6

So the odds of getting exactly 3 sixes on 5 dice is:

1/6 * 1/6 * 1/6 * 5/6 * 5/6 = 25/7776

as for getting AT LEAST 3 sixes...
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Ethan McKinney
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Leighbob wrote:
I know that when rolling 3 six-sided dice the chance of getting 3 sixes is 1 in 216, but what do the odds change to when rolling 4 dice and getting 3 sixes? What about 5 dice? Or 6 dice?


First, calculate the number of possible combinations, treating each die separately: 6*6*6*6=1296

Second, plot out all of the ways that you could get three or more sixes (I'm assuming that you want "or more," not "exactly.") Keeping each die in a "slot":

1-5 6 6 6 (this means that the first die could be anything from 1 to 5, so this represents five results)
6 1-5 6 6
6 6 1-5 6
6 6 6 1-5
6 6 6 6

We have four lines of result, each with five possibilities*. That's 20. We have one more result (all sixes), so there area total of 21 combinations that meet the criteria, for a probability of 21/1296, or about .016 (1.6%). The chance with one three dice is 1/216, or about 0.004 (0.4%).

By going from three dice to four, you've roughly quadrupled your chances.

Leighbob wrote:
Again the odds for rolling 5s or 6s of 3 dice is 1 in 27 but how does that change if I were to roll 4 dice or 5 dice or 6 dice?


Similar procedure as above:

1-4 5 5 5
1-4 6 5 5
1-4 5 6 5
1-4 5 5 6
1-4 6 6 5
1-4 6 5 6
1-4 5 6 6
1-4 6 6 6

Cycle through this with the "1-4" in each of the slots, then with all 5s, all 5s or 6s, and all 6s. Add them up, divide, you have the result.

You may notice patterns that make this much quicker.
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Richard Sampson
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bippi wrote:

1/6 * 1/6 * 1/6 = 1/216

The odds of NOT getting a six on a dice is 5/6

So the odds of getting exactly 3 sixes on 5 dice is:

1/6 * 1/6 * 1/6 * 5/6 * 5/6 = 25/7776

as for getting AT LEAST 3 sixes...


That is the probability of getting a single outcome (ie dice 1-3 are sixes and dice 4-5 are not). You need to multiply 5 choose 3 (10) to that to get the full probability, giving 250/7776.

For AT LEAST 3 sixes, just add up the probability of getting 3, 4, and 5 (250/7776 + 25/7776 + 1/7776 = 276/7776).
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Leigh Caple
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Thansk for the responses so far. To clarify I'm looking for 3 or more 6s on 4, 5 or 6 dice. Cheers!
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Richard Sampson
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4 dice: 20/1296 + 1/1296 = 21/1296 = ~1.6%

5 dice: 250/7776 + 25/7776 + 1/7776 = 276/7776 = ~3.5%

6 dice: 2500/46656 + 375/46656 + 30/46656 + 1/46656 = 2906/46656 = ~6.2%

Edit: removed typo, answers same
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Christian Sperling
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The free programm "SmallRoller" calculates any probabilities you want (without getting a headache):
http://www.fnordistan.com/smallroller.html


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Mark Chaplin
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ras2124 wrote:
4 dice: 20/1296 + 1/1296 = 21/1296 = ~1.6%

5 dice: 250/7776 + 25/7776 + 1/7776 = 276/7776 = ~3.5%

6 dice: 2500/46656 + 375/46656 + 30/46656 + 1/46656 = 2906/46656 = ~6.2%

Edit: removed typo, answers same


The odds are slimmer than I would have guessed.

I'd like to ask a similar question, if I may:

What are the odds of getting three or more 5 or 6 results on three, four, five, or six dice? Will tip!







 
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Christian Sperling
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According to SmallRoller:

Probability of rolling 5 or 6 at least 3 times on 3D6 = 3,70%

4D6 = 11,11%
5D6 = 20,99%
6D6 = 31,96%
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Leigh Caple
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Thanks everyone!
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Mitch T
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ras2124 wrote:
bippi wrote:

1/6 * 1/6 * 1/6 = 1/216

The odds of NOT getting a six on a dice is 5/6

So the odds of getting exactly 3 sixes on 5 dice is:

1/6 * 1/6 * 1/6 * 5/6 * 5/6 = 25/7776

as for getting AT LEAST 3 sixes...


That is the probability of getting a single outcome (ie dice 1-3 are sixes and dice 4-5 are not). You need to multiply 5 choose 3 (10) to that to get the full probability, giving 250/7776.

For AT LEAST 3 sixes, just add up the probability of getting 3, 4, and 5 (250/7776 + 25/7776 + 1/7776 = 276/7776).


Maybe I'm talking out of my ass here, but this seems incorrect. I've never taken a statistics class, but using this logic, rolling zero sixes is always the most likely outcome. Let's extend this out to say, 10,000 D6. Do you think rolling 0 sixes is the most likely outcome? I imagine as you start getting into big numbers, you'll see a classic bell curve if you plotted the various probabilities.

Actually, I see what the correction is that you made.
 
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Mitch T
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ras2124 wrote:
4 dice: 20/1296 + 1/1296 = 21/1296 = ~1.6%

5 dice: 250/7776 + 25/7776 + 1/7776 = 276/7776 = ~3.5%

6 dice: 2500/46656 + 375/46656 + 30/46656 + 1/46656 = 2906/46656 = ~6.2%

Edit: removed typo, answers same


So here's one then. With 4 dice for instance the apparent odds of rolling at least 3 is 1.6%. Logically, 98.4% should consist of two or less.

two or less: 625/1296+ 500/1296+ 125/1296= 1250/1296~ 96.4%. A two percent margin of error doesn't seem like much, but with theoretical statistics, it's pretty significant, especially because I imagine that margin of error goes up as the numbers increase. Where'd the extra two go?
 
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Lamb
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gulp Matrix math gulp
 
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Pelle Nilsson
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elbmc1969 wrote:

First, calculate the number of possible combinations, treating each die separately: 6*6*6*6=1296

Second, plot out all of the ways that you could get three or more sixes (I'm assuming that you want "or more," not "exactly.")


This method works great for 2 or 3 or 4 dice, but the number of combinations to consider grows quite enormously when you keep adding dice...

6 36 216 1296 7776 46656
279936 1679616 10077696 60466176 362797056 2176782336
13060694016 78364164096
 
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Pelle Nilsson
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Playing with R (with the dice library add-on):

3 or more 6s on 4d6:
> getEventProb(4, 1, 6, list(6:6, 6:6, 6:6))
[1] 0.01620370

3 or more 6s on 5d6:
> getEventProb(5, 1, 6, list(6:6, 6:6, 6:6))
[1] 0.03549383

3 5-6s on 3d6:
> getEventProb(3, 1, 6, list(5:6, 5:6, 5:6))
[1] 0.03703704

3 or more 5-6s on 4d6:
> getEventProb(4, 1, 6, list(5:6, 5:6, 5:6))
[1] 0.1111111

3 or more 5-6s on 5d6:
> getEventProb(5, 1, 6, list(5:6, 5:6, 5:6))
[1] 0.2098765

0 6s on 10000d6:
(5/6)^10000
[1] 0 (too small to be displayed at all...)

Some more fun you can have with that library:
> getTotalProbs(2, 6)
$probabilities
Total Probability Ways to Roll
[1,] 2 0.02777778 1
[2,] 3 0.05555556 2
[3,] 4 0.08333333 3
[4,] 5 0.11111111 4
[5,] 6 0.13888889 5
[6,] 7 0.16666667 6
[7,] 8 0.13888889 5
[8,] 9 0.11111111 4
[9,] 10 0.08333333 3
[10,] 11 0.05555556 2
[11,] 12 0.02777778 1

$average
[1] 7


Turned out the dice library was a bit too slow when the number of dice got above 9 or so, so switched to calculating probabilities using built-in functions instead to plot the probabilities to roll at least 3 6s for up to 100 dice (hope I got this right):

> y=c();for (x in 3:100) {y[x]=pbinom(x-3,x,5/6);};plot(y)

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Bernard
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ras2124 wrote:
bippi wrote:

1/6 * 1/6 * 1/6 = 1/216

The odds of NOT getting a six on a dice is 5/6

So the odds of getting exactly 3 sixes on 5 dice is:

1/6 * 1/6 * 1/6 * 5/6 * 5/6 = 25/7776

as for getting AT LEAST 3 sixes...


That is the probability of getting a single outcome (ie dice 1-3 are sixes and dice 4-5 are not). You need to multiply 5 choose 3 (10) to that to get the full probability, giving 250/7776.

For AT LEAST 3 sixes, just add up the probability of getting 3, 4, and 5 (250/7776 + 25/7776 + 1/7776 = 276/7776).


For my game design I am also trying to figure out what the changes are to get at least three sixes when you throw with 5 dice.

I understand you have to add up the probabilities for getting exactly 3, 4 and 5 sixes.

25/7776 as probability for getting 4 sixes out of 5 dice I understand

1/7776 as probability for getting 5 sixes out of 5 dice I understand

But can someone explain to me how you get to 250 possibilities when you need 3 sixes out of 5 dice? I can't seem to get the plot that belongs to that number.
 
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Franz Kafka
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Bernaar wrote:
But can someone explain to me how you get to 250 possibilities when you need 3 sixes out of 5 dice? I can't seem to get the plot that belongs to that number.


If you specify the order of the results (such as "six, non-six, non-six, six, six"), you can just multiply the probabilities out, as others have noted:

1/6 * 5/6 * 5/6 * 1/6 * 1/6 = 25/7776

However, you want to know all the possible results with three sixes, so you have to consider all the possible orders. There are ten distinct ways to order those results. We can describe them by just labeling the spots with non-sixes:
1&2, 1&3, 1&4, 1&5, 2&3, 2&4, 2&5, 3&4, 3&5, 4&5
The example I wrote above would have been the "2&3".

Hope this helps.
 
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Bernard
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JosefK wrote:
Bernaar wrote:
But can someone explain to me how you get to 250 possibilities when you need 3 sixes out of 5 dice? I can't seem to get the plot that belongs to that number.


If you specify the order of the results (such as "six, non-six, non-six, six, six"), you can just multiply the probabilities out, as others have noted:

1/6 * 5/6 * 5/6 * 1/6 * 1/6 = 25/7776

However, you want to know all the possible results with three sixes, so you have to consider all the possible orders. There are ten distinct ways to order those results. We can describe them by just labeling the spots with non-sixes:
1&2, 1&3, 1&4, 1&5, 2&3, 2&4, 2&5, 3&4, 3&5, 4&5
The example I wrote above would have been the "2&3".

Hope this helps.


Thanks for your reply. I still would like to see the plot of the 3 sixes out of 5 dice.

This 4 sixes out of 5 dice plot is clear to me:

1-5 6 6 6 6 = 5
6 1-5 6 6 6 = 5
6 6 1-5 6 6 = 5
6 6 6 1-5 6 = 5
6 6 6 6 1-5 = 5


But with the 3 sixes out of 5 dice you get something like this:

1-5 1-5 6 6 6 = ?
6 1-5 1-5 6 6 = ?
6 6 1-5 1-5 6 = ?
6 6 6 1-5 1-5 = ?
1-5 6 6 6 1-5 = ?

I assume the question marks have the value 50 (I know because the outcome is 250), but can somebody explain the math behind it and show me why the outcome is 50?

 
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Antonio Giannotta
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Bernaar wrote:

Thanks for your reply. I still would like to see the plot of the 3 sixes out of 5 dice.

This 4 sixes out of 5 dice plot is clear to me:

1-5 6 6 6 6 = 5
6 1-5 6 6 6 = 5
6 6 1-5 6 6 = 5
6 6 6 1-5 6 = 5
6 6 6 6 1-5 = 5


But with the 3 sixes out of 5 dice you get something like this:

1-5 1-5 6 6 6 = ?
6 1-5 1-5 6 6 = ?
6 6 1-5 1-5 6 = ?
6 6 6 1-5 1-5 = ?
1-5 6 6 6 1-5 = ?

I assume the question marks have the value 50 (I know because the outcome is 250), but can somebody explain the math behind it and show me why the outcome is 50?



The arrangements are:

1) x x 6 6 6

2) x 6 x 6 6

3) x 6 6 x 6

4) x 6 6 6 x

5) 6 x x 6 6

6) 6 x 6 x 6

7) 6 x 6 6 x

8) 6 6 x x 6

9) 6 6 x 6 x

10)6 6 6 x x


Working this out, the 1-5 part is irrelevant, we are in the realms of binomial expansions here. Forgive me if I use more mathematical notation.

3 sixes out of 5 dice is (p+q)^5, where p is the probability of a six (1/6) and q is the probability of not a six (5/6, represented by an x above).

(p + q)(p + q)(p + q)(p + q)(p + q) = 1

If we multiply out these five brackets we find every possible combination of results, hence equal to 1. (You can consider each pair of brackets to represent a die and the p represents the outcome "thrown a six") Thus p^6 (ppppp) is the result of 5 sixes.

If you're still with me, you can see there is only one way to get 5 sixes, so there is only one p^5. We'd therefore calculate the probability of 5 sixes out of 5 dice as:

1 * (1/6)^5 or (1/6)(1/6)(1/6)(1/6)(1/6) or 1/7776

On the other hand, there are 5 different ways of making (p^4)q (that is, 4 sixes and one not six):

q p p p p
p q p p p
p p q p p
p p p q p
p p p p q

So we can see that the probability of 4 sixes out of 5 dice is:

5 * p^4 * q
= 5 * (1/6)^4 * (5/6)
= 5 * (1/1296) *(5/6) = 25/7776


There are 10 ways of arranging 3 out of 5 sixes (see above) so the probability of 3 out of 5 dice is:

10 * p * p * p * q * q
= 10 * p^3 * q^2
= 10 * (1/6)^3 * (5/6)^2 = 250/ 7776

Therefore (well done for sticking with this) the probability of 3 or more sixes out of five dice is (250 + 25 + 1)/7776
= (276/7776) = 0.035494 = 3.5%

Sorry if that's too much information. I hope that helps.

(The SmallRoller program mentioned by another poster seems to give an incorrect fraction but the percentage is correct to 6 d.p.)
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A L D A R O N
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dave
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Leigh, if you haven't used Excel or some equivalent spreedsheet to do this, you should. It's quite easy to cut and paste your way to for a large matrix and then use various commands (like countif) to figure out the probabilities you want. Not particularly elegant, but it works great and doesn't take much time and really gives you a feel for the numbers.

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Mark Chaplin
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Aldaron wrote:



Great site!



 
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Pelle Nilsson
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dave65tdh wrote:
Leigh, if you haven't used Excel or some equivalent spreedsheet to do this, you should. It's quite easy to cut and paste your way to for a large matrix and then use various commands (like countif) to figure out the probabilities you want. Not particularly elegant, but it works great and doesn't take much time and really gives you a feel for the numbers.


I would (as in my last post) recommend R. I used to use excel, and various other tools (and pen and paper of course) but eventually I started looking for something good and R is what I found (so far). It's a bit of overkill, but since it is free and the advanced stuff doesn't distract you unless you go look for it, I think it is an excellent choice to do simple calculations like those above).

Just installing R (http://www.r-project.org/) and launch it (double-click the icon) and you can immediately solve the probability of rolling 3 6's on 5 dice by typing in:
dbinom(3, 5, 1/6)

To calculate the probability of rolling more than 2 6's run:
pbinom(2, 5, 1/6, FALSE)

To get a description of what those and related functions do, run:
?pbinom

Now, that's all you need to solve a lot of dice problems, even without having any idea what R is or what else it can do (which is a billion of things), or what "binom" means.

Installing the dice and prob packages for instance will give you some more nice functions that might be a bit easier to use and make it easier to solve much more complex problems without having to resort to manually counting permutations (although it is always good to do that as well now and then to sanity-check results).

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Pelle Nilsson
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To be fair I bet there is something similar to pbinom and dbinom in Excel (and other spreadsheet apps) as well, so you can probably type in something similar to the things I mentioned in a cell to get the answer, if you can figure out how to do it.
 
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