Suppose I have a die with x number of sides, and I roll it y times. Is there a probability equation I can use to figure out the odds of it rolling a certain side a certain fraction of the times?

 

For example, if I have an 8-sided die, and I roll it 500 times. What equation can I use to figure out if it will roll a seven exactly 54 out of those 500 times. Thanks ahead of time.

 

EDIT: Please understand that I am NOT asking you for the answer. Even if you give me an equation along with it, I want to figure it out myself. I don't want to appear as someone who is too lazy to do my own math homework, especially with the math skills I have (I work at a fast-food place as a summer job, and I can sometimes tell how much to charge a person and how much change to give them before I even finish entering it into the calculator!) Once you give me the equation, I'll try and figure it out and post what I get here, and you can tell me if I'm right or wrong.

this may help you

 

http://www.leonelearningsystems.com/s09_calculating_probability.htm

Look up the binomial distribution

this may help you

 

http://www.leonelearningsystems.com/s09_calculating_probability.htm

Please, sir, I am not asking for basic probability, or at least it doesn't seem that way to me. I can figure the probability of one roll. I'm trying to figure the probability of a number of successful rolls in a number of rolls total when I have a certain probability.

 

For example, that link you give me talks about the possibilities of rolling two sixes being one in thirty-six. That's just one roll. What would be the odds of it rolling two sixes exactly twenty times in one hundred rolls?

I think you'll find that your awnser is right there.

What's the probability of rolling a c on an x sided die? (1/x).

What's the probability of doing that twice? (1/x)².

What's the probability of doing that thrice? (1/x)³.

I repeat myself: look up the binomial distribution: http://mathworld.wolfram.com/BinomialDistribution.html

 

The example you gave is equivalent to 54 successes in 500 trials, where a success has a 1 in 8 chance. The binomial distribution lets you compute exactly that.

I repeat myself: look up the binomial distribution: http://mathworld.wolfram.com/BinomialDistribution.html

 

The example you gave is equivalent to 54 successes in 500 trials' date=' where a success has a 1 in 8 chance. The binomial distribution lets you compute exactly that.[/quote']

I'm sorry, but the way they word that is a little confusing. I have never been good at reading textbook-style math walkthroughs. I just finished 11th grade, and I have always had to have my teacher explain to me what needs to be done rather than consult the textbook.

 

It's easy to do it in my head if I know exactly what to do. It's just figuring out exactly what to do that is the problem.

Well o.k. the probability [math]P[/math] of a success with the probability [math]p[/math], happening [math]n[/math] times out of [math]N[/math] attempts is expressed as:

[math]P_{p}(n|N)[/math]

and can be worked out as:

[math]P_{p}(n|N)=^{N}C_n\times p^n \times q^{N-n}[/math]

Where [math]q[/math] is the probaility of a failure or [math]1-p[/math]

You should be able to do that on your calculator.

Of course, ask if you need any more.

so what do the subscripts and | signs mean? I thought | was just a verticle line to type in case you were making a picture out of text. I guess not.

There is not a single item in any character set which exists purely for the sake of ASCII art, in this context the pipe, |, is dividing the two variables and telling you what is to be done. In much the same way most characters have no meaning without context.

 

Maybe it'd be clearer if I were to substitute in your example.

You want to work out:

[math]P_{\frac{1}{8}}(54|500)[/math]

Which is:

[math]{}^{500}C_{54}\times (\frac{1}{8})^{54} \times (\frac{7}{8})^{446}[/math]

There is not a single item in any character set which exists purely for the sake of ASCII art' date=' in this context the pipe, |, is dividing the two variables and telling you what is to be done. In much the same way most characters have no meaning without context.

 

Maybe it'd be clearer if I were to substitute in your example.

You want to work out:

[math']P_{\frac{1}{8}}(54|500)[/math]

Which is:

[math]{}^{500}C_{54}\times (\frac{1}{8})^{54} \times (\frac{7}{8})^{446}[/math]

Okay, can you show me step-by-step how to work that?

This page, http://mathworld.wolfram.com/Combination.html, shows you how to compute the combination. (Note that the tree used a superscript, then a subscript and Mathword uses both subscripts, but they are the same thing... read "n choose k" in this case "500 choose 54"). I am hoping that you can compute the powers yourself using a calculator or computer.

Well, thank you for your help. I'll save the email I got as a reply notification so I can come back to this thread later when I have time, but thank you for everything.