[Math4u] Re: Help! - How to calculate odds...

Suppose you are like me and know nothing about probability.

One way to solve this would be to throw the die 1.0 gazillion sets of
6. This would take a long time. Let's call this certain number "c."
After the first throw, we will have

One c (3/8) gazillion times

Zero c (5/8) gazillion times.

After two throws, we could have

Two c's (3/8)^2 gazillion times

One c [(3/8)*(5/8)+(5/8)*(3/8)] gazillion times. = 2(3/8)(5/8)
gazillion times

Zero c's (5/8)^2 gazillion times.

Discovery, this looks like Pascal's triangle. Keep doing this until
you see the pattern.

1 toss: 1,1

2 toss: 1,2,1

3 toss: 1,3,3,1

4 toss: 1,4,6,4,1

5 toss: 1,5,10,10,5,1

6 toss: 1,6,15,20.15.6.1

In the past, I have done this on a spreadsheet. It looks to me like
after 6 tosses,

The odds of getting 6 c's is 1*[(3/8)^6] The first number is from
Pascal's triangle.

The odds of getting 5 c's is 6*[(3/8)^5*(5/8)]

The odds of getting 4 c's is 15*[(3/8)^4*(5/8)^2]

The odds of getting 3 c's is 20*[(3/8)^3*(5/8)^3] ANSWER

The odds of getting 2 c's is 15*[(3/8)^2*(5/8)^4]

The odds of getting 1 c's is 6*[(3/8)^1*(5/8)^5]

The odds of getting 0 c's is 1*[(3/8)^0*(5/8)^6]

Check

The odds of getting 6 c's is 0.002781

The odds of getting 5 c's is 0.027809

The odds of getting 4 c's is 0.115871

The odds of getting 3 c's is 0.257492 ANSWER

The odds of getting 2 c's is 0.321865

The odds of getting 1 c's is 0.214577

The odds of getting 0 c's is 0.059605

Total odds is ++++++++++1.0

I think it is amazing that such equations will add up to 1.0

Another way

The odds that the first 3 throws will get the desired number and the
next 3 won't get the desired number is

(3/8)^3*(5/8)^3

Now I interpreted the problem to say we want the correct number to
come up EXACTLY 3 times. If they mean at least 3 times, then we need
to add the odds of getting the correct number 3, 4, 5, or 6 times
above.

We need to multiply this number times the different ways of picking 3
correct numbers. The first correct number has 6 throws where it can
occur. The next one has 5 throws where it can occur. The last has 4
throws where it can occur. So we have 6*5*4. But some of these
combinations are indistinguishable from each other. These 3 correct
numbers can be interchanged 3*2 different ways. So we actually have
6*5*4/(3*2) different ways of getting 3 correct numbers with 6 throws
of the die. This equals 20 which is the same number we picked out of
Pascal's triangle with the first method!

So the answer is

6*5*4/(3*2)*[(3/8)^3*(5/8)^3]

=20*[(3/8)^3*(5/8)^3]

=0.257492 ANSWER

Now if I compare this to the "choose and pick" language that others
will write in, I will understand it even better.

Regards,

Brian

--- In Math4u@yahoogroups.com, "Sabrina" <gummiwulf@...> wrote:
>
> Hello everyone. :) I was wondering if I could get some assistance
on
> how to calculate the following:
>
> Say I have a 3/50 chance of rolling a certain number on a die (or
> generating a number in a random number generator, etc). I roll the
> die 6 times, and come up with that number 3 times. How do I figure
> out what the odds are of hitting that number 3 times in 6 rolls?
>
> Thank you so much!
> ~gummiwulf
>

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