dstebbins Posted June 6, 2006 Posted June 6, 2006 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.
ecoli Posted June 6, 2006 Posted June 6, 2006 this may help you http://www.leonelearningsystems.com/s09_calculating_probability.htm
dstebbins Posted June 6, 2006 Author Posted June 6, 2006 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?
the tree Posted June 6, 2006 Posted June 6, 2006 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.
Bignose Posted June 7, 2006 Posted June 7, 2006 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.
dstebbins Posted June 7, 2006 Author Posted June 7, 2006 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.
the tree Posted June 7, 2006 Posted June 7, 2006 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.
dstebbins Posted June 7, 2006 Author Posted June 7, 2006 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.
the tree Posted June 7, 2006 Posted June 7, 2006 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]
dstebbins Posted June 7, 2006 Author Posted June 7, 2006 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?
Bignose Posted June 8, 2006 Posted June 8, 2006 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.
dstebbins Posted June 8, 2006 Author Posted June 8, 2006 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.
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