probability of rolling 5 octahedral dice, getting sum < 23

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sweetberry

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My first time so not sure how long before you get back. My problem: You roll 5 fair octahedral (8 sided) dice. Approximate the probability that the total showing will be "less" than 23? Use the Gaussian Distribution approximation or some other reasoning. Not sure if you know what Gaussian Distribution is? Not sure if you use the mean and standard deviation or the combinations rule on this? Can you help me get started thanks.
 
Re: probability roll 5 octahedral dice

Compute the mean and standard deviation for the total of the five dice. Then use the Gaussian (Normal) distribution to approximate the probability.
 
Thanks. So I come up with a mean of 4.5(can't remember how) and a S.D. of the square root of nxpxq or 272.93=16.52. Is that right so far? I don't remember how to do the normal distribution.
 
Then I took 22.5-4.5 divided by 16.52=1.09 and looked up the z score and it is .8621 so 86%. Is that It? Not sure if Im doing it right.
 
Your variance is incorrect. This is a discrete uniform distribution of 8 integers, not a binomial distribution. So the variance of the sum of 5 is \(\displaystyle 5 \times \frac {8^2-1}{12}\). The mean here is the mean of the sum, which is \(\displaystyle 5 \times \frac {8+1}{2}\).
 
Ok I need to finish this by tomorrow. Now do you use the binomial probabilities table to go up to 23 chances of rolling and the total times rolled would be?
 
It is not a binomial problem. Use the normal distribution as an approximation. Find the z-score for 23, then use the table.
 
Ok I took 23-mean of 22.5 divided by the S.D. of 5.124=.098 for a z score. From the table it is .5398 so is that correct or do I need to subtract from 1 which would be .4602? Thanks for all the help.
 
Another way is to use the generating function:

\(\displaystyle \left(\sum_{k=1}^{8}x^{k}\right)^{5}\)

When you expand this out (I would expand with a calculator), you can look at the coefficients

of the powers less than 23. Add them up. Indeed, your solution will be very close to 1/2.

The total number of outcomes is \(\displaystyle 8^{5}=32768\)

So, the probability is \(\displaystyle \frac{\text{answer from above}}{32768}\)

Just another way I thought I would mention.
 
We haven't used the generating function so I have NO idea how to do that. OK Plain English..Please how do you come up with 1/2?? Need an answer Soon thanks.
 
sweetberry said:
...how do you come up with 1/2?
Click to expand...
I'm sorry, but I'm not seeing where anybody said that the final answer was equal to one-half...? (For guidance, two of the tutors said that the answer should be "very close to 1/2", but that of course is a different thing from "equal to".)

Eliz.
 
I was kind of hinting by using caps: VERY CLOSE ; actual answer is exactly 1/2.

Approximation methods should thus = ~1/2
 
I understand, I just would like to know how you come up with that can you please SHOW ME. Thanks
 
sweetberry said:
I understand, I just would like to know how you come up with that can you please SHOW ME. Thanks
Click to expand...
You've been given explanations, formulas, set-ups, and everything-but-the-answer replies. How far have you gotten in using what you've been given so far? (If you really have no idea how to use anything you've been given, then I'm afraid you need much more intensive assistance than we can here provide, so I hope you're exaggerating.)

Please be complete, showing your work and reasoning so far. Thank you! :D

Eliz.
 
Denis said:
I was kind of hinting by using caps: VERY CLOSE ; actual answer is exactly 1/2.

Approximation methods should thus = ~1/2
Click to expand...
Since this is a discrete problem, \(\displaystyle sum < 23\) is the same as \(\displaystyle sum \le 22.5\), which is 1/2 by symmetry.
 
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