Probability question (throwing dice)

[EricVT]

Homework Statement

An unbiased six-sided dice is thrown 10 times. What is the probability that exactly 4 of anyone number alone will occur?

Homework Equations

Binomial equation
Combinatorial equations

The Attempt at a Solution

You would be interested in the probability of choosing any number 4 times out of the 10 throws:

6*(10 C 4) (1/6)^4 * (5/6)^6

You would also be interested in the chance of throwing that same number that has already been thrown 4 times a 5th, 6th, 7th, 8th, 9th and 10th time (since that side of the dice isn't removed, it could still come up in the remaining 6 throws).

...(6 C 1) (1/6)^1 * (5/6)^5
+ (6 C 2) (1/6)^2 * (5/6)^4
+ (6 C 3) (1/6)^3 * (5/6)^3
+ (6 C 4) (1/6)^4 * (5/6)^2
+ (6 C 5) (1/6)^5 * (5/6)^1
+ (6 C 6) (1/6)^6

And also the chance that one of the remaining 5 numbers could be thrown 4 times...

5 * (6 C 4)*(1/6)^4 * (5/6)^2


However, these last two sets of equations aren't entirely independent of one another...you could, for example, start by throwing four 3's...and then in your remaining 6 tosses you get another 3 and then four 5's and then a 6.

I can't quite figure out how to separate out these values into truly independent events in order to combine them and get to the answer.


Am I on the right path here or is there a simpler way to think about this that I have overlooked? I'm quite stuck here so any help would be greatly appreciated.

 

[e(ho0n3]

You would be interested in the probability of choosing any number 4 times out of the 10 throws:

6*(10 C 4) (1/6)^4 * (5/6)^6

The notation (10 C 4) is weird. I would write C(10, 4). What you wrote is not ''the probability of choosing any number 4 times out of the 10 throws'' as there is no probability associated with choosing 4 out of the 10 throws.

You would also be interested in the chance of throwing that same number that has already been thrown 4 times a 5th, 6th, 7th, 8th, 9th and 10th time (since that side of the dice isn't removed, it could still come up in the remaining 6 throws).

Why would I be interested in that? We're only interested if a number shows up exactly four times.

However, these last two sets of equations aren't entirely independent of one another...you could, for example, start by throwing four 3's...and then in your remaining 6 tosses you get another 3 and then four 5's and then a 6.

I can't quite figure out how to separate out these values into truly independent events in order to combine them and get to the answer.

I think I understand your problem, which I will state in more precise terms: Let S be the number ways of throwing a regular die 10 times. Let S(x) be the number of ways of throwing a regular die 10 times such that x appears exactly 4 times. One may think that the probability of getting 4 of exactly one number in 10 throws is [S(1) + S(2) + ... + S(10)] / S. The problem here is double counting, e.g. throwing four 1's followed by four 2's and then a 5 is counted in both S(1) and S(2). But this is easily fixed: let S(x, y) be the number of ways of throwing a regular die 10 times such that x and y appear exactly 4 times and subtract this from S(1) + S(2) + ... + S(10) accordingly.

 

[EricVT]

Thanks for the reply.

When I wrote (10 C 4) I was intending that to be read as "from 10 choose 4", or the number of ways of arranging 4 things from 10 (i.e. 4 successful dice throws from 10 total throws). What I wrote IS a probability, it is the number of sides on the dice that could be rolled 4 times multiplied by the number of ways of rolling a certain number 4 times from 10 throws, times the probability of rolling a certain number on a dice 4 times, times the probability of rolling a different number on a dice 6 times.

Sorry if the notation seems strange, it's the first time I ever wrote "from this number choose that number" on a message board, so I wasn't sure of the proper notation.

And you're right about the second point, my reasoning was incorrect.

Your third bit of advice did help me though. I ended up with the probability:

C(6,1)*C(10,4)*(1/6)^4*(5/6)^6 - C(6,2)*C(10,8)*(1/6)^8*(5/6)^2

0.3253 (32.53%)

That's the result I was looking for.

Thanks again.