Homework Help: Dice Probability

1. Nov 3, 2005

j00borg

I have absolutely no idea on how to calculate probabilities. So some help, would be greatly appreciated.

What are the odds, in fraction form, of rolling with six die and getting a 6 6 6 6 4 1 on the first roll?

2. Nov 4, 2005

James R

The probability is always the number of "favorable" outcomes divided by the number of possible outcomes.

You're rolling six dice. Each one has a 1/6 probability of showing a specific number after the roll.

So, the chances of getting the above combination when you roll all 6 dice is $1/6^6[/tex]. This is for the situation where you want a "6" on the first die, "6" on the second, etc. etc, and "1" on the last die. But maybe you want the numbers 6,6,6,6,4,1 to show up on ANY of the dice, as long as 4 dice show the number "6" and the other 2 show "4" and "1". In that case, we need to know how many ways that can happen. The answer is that there at 30 ways to arrange the given 6 numbers. So, the probability in that case would be [itex]30/6^6$.

3. Nov 11, 2005

method

Thanks James this is the best way I've seen this explained.
All of this if very clear other than the:
I was wondering what the equation for this number would be?
I am working on a proability problem with D10 dice and knowing the way you got 30 in an equation would be huge, huge help.
Thanks

4. Nov 11, 2005

HallsofIvy

Didn't we just have this problem- or one like it?
Imagine that the dice are "labeled" so that you can distinguish between them:
How many different ways can you "order" 61, 62, 63, 64, 4, 1? Well, you could choose any of the 6 numbers to go first: 6 ways. For each of those you could choose any of the 5 remaining numbers, then 4, then 3, then 2, the put in the last: there are 6(5)(4)(3)(2)= 6! ways. HOWEVER, many of those are really the same, they just have the (labled) 6's rearranged. How many would there be if we don't have the lables and can't distinguish between them? There are 4! ways to rearrange just 4 things (same argument as above: 4(3)(2)(1)). Since there are 4! ways of rearranging a single (unlabled) order to get one (labled) arrangement, we must divide by 4! to correct for that: 6!/4!= (6(5)(4)(3)(2))/(4(3)(2)(1))= 6*5= 30.

In general, if you have N objects, n1 of them the same, n2 of them the same, ..., nm of them the same, with n1+ n2+ ...+ n3= N, there are
$$\frac{N!}{n_1!n_2!...n_m!}$$

In the above problem, N= 6, n1= 4 (there are 4 6's), n2= 1 (there is 1 4), n3= 1 (there is 1 1).
In the above, N= 6

Last edited by a moderator: Nov 12, 2005
5. Nov 11, 2005

nickrock23

ok if you have 2 10-sided dice and you want to know the probability of you getting 1 successful roll (8,9,10) with one unsuccessful roll (1,2,3,4,5,6,7), and likewise if you had more than 2 10-sided dice, and wanted to know the probability of 1 successful roll with the rest unsuccessful, this is the equation:

s * (u to the (d-1) power) = x/(s+u) to the d power

s=3 u=7 d=dice x=unknown
s represents successful integers, u unsuccessful

for 2 dice:

3* (7 to the 1st power) = 21
(3+7) to the 2 power = 100
the odds of you rolling 1 successful roll along with another unsuccessful roll is 21 in 100. you can make a simple chart of all the combinations, it will come out to 21 out of 100, trust me.

for 3 dice:

3 * (7 to the d-1 power) = 3-1 (or 7 squared) = 3*49=147
(s+u)cubed = 1000
therefore the odds of hitting one successful roll with 2 unsuccesful rolls is 147/1000

and so on. your odds basically go from 1 in 5, to 1.5 in 10, to 1 in 10, to .024 in 10, to .016 in 10 etc.

2 dice = 21/100
3 dice = 149/1000
3 dice = 1029/10,000
4 dice = 2401/100,000
5 dice = 16,807/1,000,000

Last edited: Nov 11, 2005
6. Nov 11, 2005

Tide

nick,

Unfortunately, you cannot have a 10-sided die with equal probabilities for each face (there are only 5 regular solids). Better go with a dodecahedron! :)

Last edited: Nov 11, 2005
7. Nov 11, 2005

nickrock23

ive never seen a 10 sided die to be honest

8. Nov 11, 2005

method

Here's a D10.

http://en.wikipedia.org/wiki/Ten-sided_die
http://en.wikipedia.org/wiki/Trapezohedron
The dice are labled 1,2,3,4,5,6,7,8,0 (0 meaning 10)

It would be great to have a chart that said if your dice pool is 8 (or n amount) what is the probability that you will throw 0,1,2,3,4,5,6,7 or 8 successes. Successes would be an 8,9 or 0.

In these games you can have small or large dice pools. 1-20 possibly and you may only need 3-5 successes.

If probability is FE over PS (FE Favorable Events 8,9,0 in our case and PS Possible States)
PS could be x to the n power (with x = Number of sides on the die and n = number of dice thrown.)
So for 2 dice we would see PS= 10 to the second power = 100.
The PS for 3 dice would be 10 to the third power = 1,000, etc...

What I don't quite understand the equation to find FE (Or the amount of time only 1 success will show up out of that 100 and also the number of time 2 successes will show up.)

Last edited by a moderator: May 2, 2017
9. Nov 16, 2005

James R

Tide:

There's no problem having a 10-sided die with equal probabilities on each face. But you're right that it can't be a regular polyhedron. 10-sided dice are readily available and commonly used in role playing games. They look a bit like cut diamonds, with irregular "edges", made up of two groups of 5 faces kind of wedged together. (I'm sure there'll be picture somewhere.)

10. Nov 16, 2005

shmoe

You could also double label an icosahedron to simulate a D10. I'm the right variety of nerd to own many 10 sided dice though.

You can make the table you want easily enough. In your case the probability any single dice is a Success is 3/10, 3 events 8, 9 or 0, out of 10. The probability of a failure is the 7/10.

If you have n dice and you want the probability of exactly m successes (so we have n-m failures), this is given by $$(3/10)^{m}(7/10)^{n-m}$$, times the number of ways of selecting the m dice to be successes. The number of ways of doing this is "n choose m", this is assuming the dice are not labeled or the order rolled doesn't matter (like rolling m identical dice in one shot). This is given by:

$$\frac{n!}{m!(n-m)!}$$

Last edited: Nov 16, 2005
11. Nov 16, 2005

Tide

Good points! I had regular solids on my mind and should have realized there are other solids with suitable symmetry.

12. Oct 30, 2010

elmitico

Sorry for reviving such and old thread, but it really goes in the direction of my problem.

I have exactly the the same situation, but instead of considering 8,9 and 0 success, i have to consider 8=1success, 9=2successes and 0=3 successes.

I need to compute the probability of getting X successes in n throws.

the problem is that there are several different ways of getting X successes. For example, say you want 3 successes in 3 rolls. This can be accomplished with:

three 8's
one 8 and one 9
one 10.

so i have to add the probability of each of these events.

generalizing the solution given in this thread, i can compute the probability of of getting m8 eights, m9 nines and m10 tens in n throws. No problem there.

But when you go to a big amount of dices, this way of thinking becomes impractical, as thinking in all the ways you can get the desired number of successes and adding up the individual probabilities is to cumbersome.

Can anyone help me out with this?

Thanks!

Last edited: Oct 30, 2010