How to generate a formula for dice rolling probabilities

[Drax]

Hi all, I'm new to the forum

I have a bit of a problem in working out how to calculate some probabilities involving multiple dice rolls for a wargame. All the websites I have consulted so far assume I want o roll all my dice and add them up, but that's not the case.

Essentially this is an OR operator rather than an AND operator. Assuming fair six-sided dice, numbered 1 to 6, I understand that if I roll one die, the chances of rolling a '6' is 1/5; the chances of rolling a 5 or higher (so 5 or 6) is 2/6 = 1/3 and so on. And that the chances of rolling 5 or greater and then another 5 or greater with the next roll is 1/3 x 1/3 or 1 \ 9

But what I'm trying to do here is to work out a formula so that if I roll a series of, say, four dice, what are the chances of rolling a 5 or 6 with any one of those rolls? Or, to put it another way, if I roll four dice at the same time, what are the chances of at least one of them having a 5 or a 6 showing?

Clearly it is not as simple as just adding 1/3 plus 1/3 plus 1/3 plus 1/3 because then the probability would be over 100%, which is of course incorrect. I could roll four '1's, for example. So it's and OR operator: Die 1 can roll a 5 or 6, OR die 2 can roll a 5 or 6, and so on up to die 4.

So what does this look like iin terms of formulae, please, and how is that formula derived?

Thanks in advance

 

[MarkFL]

Hello, and welcome to FMH!

I am going to use:

"if I roll four dice at the same time, what are the chances of at least one of them having a 5 or a 6 showing?"

I would use the complement rule, where we can say it is certain either all four die will be less than 5 OR at least 1 die will be 5 or 6 (event X):

[MATH]\left(\frac{2}{3}\right)^4+P(X)=1[/MATH]
Hence:

[MATH]P(X)=1-\left(\frac{2}{3}\right)^4=\frac{65}{81}[/MATH]

 

[JeffM]

Hi all, I'm new to the forum

I have a bit of a problem in working out how to calculate some probabilities involving multiple dice rolls for a wargame. All the websites I have consulted so far assume I want o roll all my dice and add them up, but that's not the case.

Essentially this is an OR operator rather than an AND operator. Assuming fair six-sided dice, numbered 1 to 6, I understand that if I roll one die, the chances of rolling a '6' is 1/5; the chances of rolling a 5 or higher (so 5 or 6) is 2/6 = 1/3 and so on. And that the chances of rolling 5 or greater and then another 5 or greater with the next roll is 1/3 x 1/3 or 1 \ 9

But what I'm trying to do here is to work out a formula so that if I roll a series of, say, four dice, what are the chances of rolling a 5 or 6 with any one of those rolls? Or, to put it another way, if I roll four dice at the same time, what are the chances of at least one of them having a 5 or a 6 showing?

Clearly it is not as simple as just adding 1/3 plus 1/3 plus 1/3 plus 1/3 because then the probability would be over 100%, which is of course incorrect. I could roll four '1's, for example. So it's and OR operator: Die 1 can roll a 5 or 6, OR die 2 can roll a 5 or 6, and so on up to die 4.

So what does this look like iin terms of formulae, please, and how is that formula derived?

Thanks in advance

First, the probability of rolling a 6 on a fair die is 1/6, not 1/5, but you probably knew that and merely made a typo.

Second, the probabilities of something based on rolling four fair dice simultaneously is exactly the same as rolling one fair die four times in succession.

There are formulas for what you want to do, but they may involve a bunch of arithmetic. It may be easier to learn the basic laws of probability.

[MATH]0 \le P(A) \le 1.[/MATH]
[MATH]A \text { is a certainty} \iff P(A) = 1.[/MATH]
[MATH]A \text { is an impossibility} \iff P(A) = 0.[/MATH]
[MATH]P(A \text { or } B) \equiv P(A) + P(B) - P(A \text { and } B). [/MATH]
[MATH]\therefore P(A \text { and } B) = 0 \implies P(A \text { or } B) = P(A) + P(B).[/MATH]
The one above is why you can say that the probability of a 5 or 6 is 1/3. The probability of a 6 is 1/6. The probability of a 5 is 1/6. The probability of a 5 and a 6 at the same time is 0 because that is impossible. So 1/6 + 1/6 - 0 = 1/3. When the probability of A and B occurring simultaneously is zero (or impossible), we say that they are mutually exclusive.

[MATH]P(A) * P(B \text { given } A) \equiv P(A \text { and } B) \equiv P(B) * P(A \text { given } B).[/MATH]
Now what you are thinking about is called the binomial distribution. We define two states, success and failure, and consider a number of trials that are independent. In your case success would be a 5 or a 6 and failure would be anything less than 5. Got the idea?

[MATH]P(\text {exactly } k \text { successes in } n \text { trials}) = \dfrac{n!}{k! * (n - k)!} * p^k * (1 - p)^{(n-k)}, \text { where}[/MATH]
[MATH]p = \text { the probability of success in a single trial, and}[/MATH]
[MATH]i = 0 \implies i! =1 \text { whereas if } i \text { is an integer } > 0 \implies i! = i * (i - 1)!.[/MATH]
What the last definition means is that

0! = 1, 1! = 1, 2! = 2 * 1 = 2, 3! = 3 * 2! = 3 * 2 = 6, 4! = 4 * 3! = 24, etc.

So in four rolls to get exactly one 5 or 6 has a probability of

[MATH]\dfrac{4!}{1! * (4 - 1)!} * \left (\dfrac{1}{3} \right)^1 * \left (1 - \dfrac{1}{3} \right)^{(4 - 1)} =[/MATH]
[MATH]\dfrac{4 * \cancel{3!}}{1! * \cancel{3!}} * \dfrac{1}{3} * \left ( \dfrac{2}{3} \right)^3 = \dfrac{4}{3} * \dfrac{8}{27} = \dfrac{32}{81}.[/MATH]
When you want to know what the probability of at least one success in four trials is, you can calculate the the probabilities of exactly one, exactly two, exactly 3, and exactly 4, and then simply add them all up (which you can do because they are all mutually exclusive).

This can get you into a lot of arithmetic. There are some shortcuts (and Mark has given you one that works for the at least one case but not for the at least two case), but I suggest you use a spreadsheet to calculate all the probabilities involved to avoid arithmetical and conceptual errors. The probabilities for none, one, two, three, and four successes will add up to 1 if you made no mistakes.

There: that is about two weeks of a high-school course in probability theory in one fell swoop.

Good luck.

 

[Drax]

Gentlemen, thank you so much. That's really helpful, and I will be able to input my numbers into the formulae and get it to work. Jeff, yes, the 1/5 was indeed a typo as you rightly surmised!

I'm actually going to use ten-sided dice, but in my example I used six-sided to keep things familiar, but I appreciate the maths is pretty much the same, just a question of plugging in the required numbers for a d10 (ten-sided die) rather than the d6 (6-sided die).

As background, the scenario is that I have four attack helicopters attacking the target, a single helicopter will hit the target with a 9 or a 10, and since there are four helicopters, I get four chances to hit the target. One hit is enough to destroy it, so I only need one hit; more hits would have no additional effect. So (at least) one 9 or 10 is needed, with four 'chances' to get that 9 or 10. And the helicopters get only one roll each.

And yes, I am going to use a spreadsheet. I like spreadsheets so much that I get teased for it

Thanks again guys and if I run into more problems is it ok if I ask more questions?

Cheers
Tony (Drax)

 

[MarkFL]

Hello, Tony!

Yes, please feel free to ask any questions you may have.

 

[Drax]

Thanks. I also appreciate the explanations you both gave; I need the formulae yes, but I do like to be able to see why things work the way they do. It's very satisfying to see the derivations because they explain why the formulae work. Thank you both.

 

[pka]

Thanks. I also appreciate the explanations you both gave; I need the formulae yes, but I do like to be able to see why things work the way they do. It's very satisfying to see the derivations because they explain why the formulae work. Thank you both.

If you look at this expansion. The expansion tell us that tossing four dice the number of ways to get different sums.
The term \(\displaystyle 125x^{12}\) says there are one hundred and twenty-five ways to roll a sum of twelve or \(\displaystyle \mathcal{P}(S=12)=\frac{125}{6^4}\)