math-probability-help!
math-probability-help!
my science project question is "are you more likely to roll a sum of at least 18 with 3 ten-sided dice or 5 six-sided dice?" I know that the 3 ten-sided dice have 1000 possible outcomes, and I made a list of all of the favorable outcomes (212) I can't figure out how to find all of the favorable outcomes for the 5 six-sided dice. I tried to list them, but I can't seem to figure out a pattern to it, besides I'm sure there will be over 1000. Am I doing this all wrong? Is there another way?
In terms of the 5 6-siders, there are 6^5 possible combinations. That's 7776.
There are a number of ways to determine the number of favorable combinations. The minumum sum is 5 and the maximum is 30, of course. Since dice like this are normally distributed, that puts the average at 17.5 (vs. 16.5 for 3d10).
One way to solve this kind of problem is to determine the standard deviation (sigma) of the normal curve for both cases, and figure out how many sigmas away from the center 18 is. There are numerous places that can help you with this kind of a method, just google for "standard deviation dice". One particularly interesting page is: http://www.rpg.net/news+reviews/columns ... sep05.html
If you'd rather calculate the answer exactly, the you're correct that you need to figure out the number of favorable results. This can be quite complicated in general, though I'm sure there's a solution out there to be found. I'm not sure what the best way to do it is for 5d6.
Let's see... how about inductively...
When the first die is a 1, the remaining 4 dice need to sum to 17. Similarly when the first 2 dice are 1, the remaining 3 need to sum to 16. 3d6 is a tractable number of combinations to figure out by hand...
If you made a table of the possible values of the first 2 dice, it would only have 36 rows, with corresponding required values for the remaining 3d6 (and the number of favorable outcomes for those). If you summed these favorable outcomes for the "leftover" 3d6, you'd have the number of favorable outcomes for 5d6.
This wouldn't of course, be practical for larger numbers of dice, though if you think about it some more, you may be able to see how to reduce this to an equation... Let us know if that isn't a big enough hint...
There are a number of ways to determine the number of favorable combinations. The minumum sum is 5 and the maximum is 30, of course. Since dice like this are normally distributed, that puts the average at 17.5 (vs. 16.5 for 3d10).
One way to solve this kind of problem is to determine the standard deviation (sigma) of the normal curve for both cases, and figure out how many sigmas away from the center 18 is. There are numerous places that can help you with this kind of a method, just google for "standard deviation dice". One particularly interesting page is: http://www.rpg.net/news+reviews/columns ... sep05.html
If you'd rather calculate the answer exactly, the you're correct that you need to figure out the number of favorable results. This can be quite complicated in general, though I'm sure there's a solution out there to be found. I'm not sure what the best way to do it is for 5d6.
Let's see... how about inductively...
When the first die is a 1, the remaining 4 dice need to sum to 17. Similarly when the first 2 dice are 1, the remaining 3 need to sum to 16. 3d6 is a tractable number of combinations to figure out by hand...
If you made a table of the possible values of the first 2 dice, it would only have 36 rows, with corresponding required values for the remaining 3d6 (and the number of favorable outcomes for those). If you summed these favorable outcomes for the "leftover" 3d6, you'd have the number of favorable outcomes for 5d6.
This wouldn't of course, be practical for larger numbers of dice, though if you think about it some more, you may be able to see how to reduce this to an equation... Let us know if that isn't a big enough hint...
../ray\..
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Re: math-probability-help!
What you are attempting is beyond the scope of pencil and paper math.egrove wrote:my science project question is "are you more likely to roll a sum of at least 18 with 3 ten-sided dice or 5 six-sided dice?" I know that the 3 ten-sided dice have 1000 possible outcomes, and I made a list of all of the favorable outcomes (212) I can't figure out how to find all of the favorable outcomes for the 5 six-sided dice. I tried to list them, but I can't seem to figure out a pattern to it, besides I'm sure there will be over 1000. Am I doing this all wrong? Is there another way?
A basic statistics text will tell you how to solve this problem, but the math is likely beyond middle school level. You'll need at a miniumum the ability to manage fractions, algebra and understand what a factorial series is.
The other approach you could take is empirical (just roll the dice a lot and measure the results). This is easier with an electronic die-roller, like a random number generator in an Excel spreadsheet.
I'm not sure this is a good science fair project, as posed, because it is something you can work out if you have the right formula. Now you might want to test the formula against reality, by comparing results on actual rolled dice versus predicted percentages. I'd check with your science instructor to be sure your experimentis heading the right way.
In the meantime, you might want to check this site out and see if you understand it:
http://mathforum.org/probstat/probstat.html