OT: Math Question

Mustang · 06-10-2004, 11:32 AM

Figure this will be answered in 10 minutes here but.. just a quick statistic question.

Say you have 8 chips. On each chip is a number (for this example use 0,1,2,2,2,2,3,3)

You put the numbers in a hat and draw out 3. What I'm trying to figure out is the formula to get the total number of different combinations you can have (such as 0,1,2 or 2,2,2), the average total when you draw out 3 and to top it off, I need the percent of how often each total could occur. (Percent chance 6 will be the total...)

I need more of a generic formula instead of just the answer to the example because I need to compute different values based on a variable numbe of chips and numbers.

Any help is appreciated...

KevinNU7 · 06-10-2004, 11:35 AM

is 1,2,3 the same as 2,3,1?

Mustang · 06-10-2004, 11:39 AM

Quote:

Originally Posted by KevinNU7

is 1,2,3 the same as 2,3,1?

For the most part, yes. I don't care on when the chips were pulled out.. just that they were pulled out so.. 1,2,3 could be pulled hmm... 8 different ways for a total of 6.

sachmo71 · 06-10-2004, 11:44 AM

I hate math.

QuikSand · 06-10-2004, 12:23 PM

If you have 8 chips (before introducing the fact that some have the same value as others), there are 56 possible combinations of three chips that can be drawn without replacement. Some math texts refer to this using the letter "c" like this: 8c3

What you're basically calculating, generically, is this

8! / 3! * (8-3)!

...which inevitably boils down to:

the top 3 factors in 8! / 3!

...or 8x7x6 / 3x2x1 = 56.

In this case, you can essentially construct each possible outcome, which has a 1/56 chance of happening. Then, the next step I'd take would be to count the possible ways that you can get each possible total, from the 56 possible alternatives.

For example - to reach a sum of 3, you would have to draw the 0, the 1, and one of the four 2s = 1x1x4= 4 possible ways to do that. So, the chances of getting a sum of 3 is 4/56.

Same logic for each possible sum (which gets a little more complicated to count_ gives you a list of possible totals, with the probability for each. Then you take a weighted average to get your average result.

I think that's the easiest roadmap for this...

Bee · 06-10-2004, 12:25 PM

My suggested formula would be

x=1/3

06-10-2004, 11:35 AM	#2
KevinNU7 College Starter Join Date: May 2003 Location: Beantown	is 1,2,3 the same as 2,3,1? __________________ Boston Bashers - III.14 - (8347)

06-10-2004, 12:23 PM	#5
QuikSand lolzcat Join Date: Oct 2000 Location: Annapolis, Md	If you have 8 chips (before introducing the fact that some have the same value as others), there are 56 possible combinations of three chips that can be drawn without replacement. Some math texts refer to this using the letter "c" like this: 8c3 What you're basically calculating, generically, is this 8! / 3! * (8-3)! ...which inevitably boils down to: the top 3 factors in 8! / 3! ...or 8x7x6 / 3x2x1 = 56. In this case, you can essentially construct each possible outcome, which has a 1/56 chance of happening. Then, the next step I'd take would be to count the possible ways that you can get each possible total, from the 56 possible alternatives. For example - to reach a sum of 3, you would have to draw the 0, the 1, and one of the four 2s = 1x1x4= 4 possible ways to do that. So, the chances of getting a sum of 3 is 4/56. Same logic for each possible sum (which gets a little more complicated to count_ gives you a list of possible totals, with the probability for each. Then you take a weighted average to get your average result. I think that's the easiest roadmap for this... Last edited by QuikSand : 06-10-2004 at 12:23 PM.

06-10-2004, 11:44 AM	#4
sachmo71 The boy who cried Trout Join Date: Oct 2000 Location: TX	I hate math.

06-10-2004, 12:25 PM	#6
Bee Pro Starter Join Date: Oct 2000 Location: Fairfax, VA	My suggested formula would be x=1/3

Currently Active Users Viewing This Thread: 1 (0 members and 1 guests)