Probabilities and rolling dice

I was wondering if anyone could tell me how I could determine the probability of getting a certain sum or better on multiple dice, each with a different number of sides.?

Comicsanctum said:

I was wondering if anyone could tell me how I could determine the probability of getting a certain sum or better on multiple dice, each with a different number of sides.?

Click to expand...

Do you have a specific problem you need to do?

srmichael said:

Do you have a specific problem you need to do?

Click to expand...

No, I was just curious about the method used. I have several dice of different shapes and though I know how to determine the probability if the dice were all the same type, I have no clue where to start from when they are all different shaped.

Comicsanctum said:

No, I was just curious about the method used. I have several dice of different shapes and though I know how to determine the probability if the dice were all the same type, I have no clue where to start from when they are all different shaped.

Click to expand...

Suppose one die has N sides numbered n=1-N and another has M sides numbered m=1-M. Since the two dice are independent events, probabilities multiply:
1st die, P(n) = 1/N
2nd die, P(m) = 1/M
P(1st=n and 2nd=m) = (1/N)(1/M) = 1/(NM)

Now you have to "count the ways" of getting a specific result. The total possible results of throwing the two dice is NM. Probability is the ratio of the number of ways to get a "success" divided by the the total possible (which is NM).

If you want to work it out for a specific N and M, repost showing us your work.

DrPhil said:

Suppose one die has N sides numbered n=1-N and another has M sides numbered m=1-M. Since the two dice are independent events, probabilities multiply:
1st die, P(n) = 1/N
2nd die, P(m) = 1/M
P(1st=n and 2nd=m) = (1/N)(1/M) = 1/(NM)

Now you have to "count the ways" of getting a specific result. The total possible results of throwing the two dice is NM. Probability is the ratio of the number of ways to get a "success" divided by the the total possible (which is NM).

If you want to work it out for a specific N and M, repost showing us your work.

Click to expand...

So, if I have a six sided and an eight sided die, it would be (1/6)(1/8)=1/48. Meaning there are 48 different summed results and the odds of getting any one of them is 1 in 48, or about 2%. Is there a way to "count the ways" without actually boxing out the results and literally counting them? Both for an individual number and a set of numbers such as >5?

Comicsanctum said:

So, if I have a six sided and an eight sided die, it would be (1/6)(1/8)=1/48. Meaning there are 48 different summed results and the odds of getting any one of them is 1 in 48, or about 2%. Is there a way to "count the ways" without actually boxing out the results and literally counting them? Both for an individual number and a set of numbers such as >5?

Click to expand...

I would resort to counting - but look out for developing patterns.

N=6, M=8, m+n > 5:
if n=1, m=5,6,7,8
if n=2, m=4,5,6,7,8
if n=3, m=3,4,5,6,7,8
if n=4, m=2,3,4,5,6,7,8
if n=5,6. m=1,2,3,4,5,6,7,8
P = 38/48 = 0.792

Check, N=6, M=8, m+n < 5
n=1, m=1,2,3,4
n=2, m=1,2,3
n=3, m=1,2
n=4, m=1
P = 10/48 = 0.208 ok

Comicsanctum said:

So, if I have a six sided and an eight sided die, it would be (1/6)(1/8)=1/48. Meaning there are 48 different summed results and the odds of getting any one of them is 1 in 48, or about 2%. Is there a way to "count the ways" without actually boxing out the results and literally counting them? Both for an individual number and a set of numbers such as >5?

Click to expand...

Actually, boxing out the results and literally counting them, is a simple way if you have a spreadsheet. Also, it provides a way to explain these concepts, if you are a teacher.

Example for > 5

Total = T = 6*8 = 48
Win = W = 48-10 = 38
Win% = 100*38/48 = 79%

Comicsanctum said:

So, if I have a six sided and an eight sided die, it would be (1/6)(1/8)=1/48. Meaning there are 48 different summed results and the odds of getting any one of them is 1 in 48, or about 2%. Is there a way to "count the ways" without actually boxing out the results and literally counting them? Both for an individual number and a set of numbers such as >5?

Click to expand...

No. This is wrong.

Yes, there are forty-eight possibilities of equal probability, but that does not mean their sums are also equally probable. Let's see why.

How many ways can you get 14. One, 8 on the 8-die and 6 on the 6-die. 1 possibility gone. P(14) = 1/48.

How many ways can you get 13. Two, 8 on the 8-die and 5 on the 6-die or 7/6. 3 gone. P(13) = 2/48 = 1/24.

How many ways can you get 12. Three. 8/4, 7/5 on the 6-die, 6/6. 6 gone. P(12) = 1/16.

How many ways can you get 11. Four. 8/3, 7/4, 6/5, or 5/6. 10 gone. P(11) = 1/12.

How many ways can you get 10. Five. 8/2, 7/3, 6/4, 5/5, and 4/6. 15 gone. P(10) = 5/48.

How many ways can you get 9. Six. 8/1, 7/2, 6/3, 5/4, 4/5, 3/6. 21 gone. P(9) = 1/8.

How about 8. Six. 7/1, 6/2, 5/3, 4/4, 3/5, 6/2. 27 gone. P(8) = 1/8.

How about 7. Six. 6/1, 5/2, 4/3, 3/4, 2/5, 1/6. 33 gone. P(7) = 1/8.

How about 6. Five. 5/1, 4/2, 3/3, 2/4, 1/5. 38 gone. P(6) = 5/48.

How about 5. Four. 4/1, 3/2, 2/3, 1/4. 42 gone. P(4) = 1/12.

How about 4. Three. 3/1, 2/2, 1/3. 45 gone. P(4) = 1/16.

How about 3. Two. 1/2, 2/1. 47 gone. P(3) = 1/24.

How about 2. One. 1/1. All forty-eight possibilities are accounted for. P(1) = 1/48.

A way to check is to see if the probabilities add up to 1 exactly.

JeffM said:

No. This is wrong.

Yes, there are forty-eight possibilities of equal probability, but that does not mean their sums are also equally probable. Let's see why.
.

Click to expand...

Jeff is pointing out what you see on the diagonals of the diagram above.
For example, there are 5 ways to get SUM=6.
SUM=6 is 5 times as likely as SUM = 2

Bob Brown MSEE said:

Jeff is pointing out what you see on the diagonals of the diagram above.
For example, there are 5 ways to get SUM=6.
SUM=6 is 5 times as likely as SUM = 2

Click to expand...

So there is no short cut. Ok, I understand the process now. Thank you all for your help!

Comicsanctum said:

So there is no short cut. Ok, I understand the process now. Thank you all for your help!

Click to expand...

The answer there depends on what "short cut" means.
Suppose we have a regular die numbered one to six.
Then we have a eight sided die numbered with even numbers 2 to 16.
Also we have a ten sided die numbered with odd numbers 1 to 19.

I ask a computer to expand a polynomial: Look Here.

You can see that one of the terms is \(\displaystyle 18x^{29}\).
That tells us that there 18 ways to throw the sum of 29.
So the probability of getting a sum of 29 is \(\displaystyle \dfrac{18}{480}\)

If you wanted to know the probability of throwing a sum greater that 29, then add up all the coefficients of terms having powers greater than 29. Divide that by the 480 possible triples.