# Throwing dice 7 times

1. Jul 9, 2011

### binjip

Throwing a die 7 times

1. The problem statement, all variables and given/known data

Throw a die 7 times.

i) What is the probability that you get number 6 twice and all other outcomes once. (e.g. one possible set of outcomes would be 6, 6, 5, 4, 3, 2, 1)

ii) What is the probability that you get all the numbers of a die? (e.g. 6, 5, 4, 3, 2, 1, x where x is {1, ..., 6})

2. Relevant equations

P() = # relevant outcomes / # of all possible outcomes

3. The attempt at a solution

i)

There are 67 possible ways of arranging the set of 7 outcomes (# of permutations).

There 6*5*4*3*2*1*1 = 6! ways of arranging the numbers under the given conditions.

P(two 6, and all other outcomes) = 6! / 67 = 5! / 66

ii) # of permutations is 67. No change here.

The # of ways to arrange the numbers under give conditions changes:

6*6*5*4*3*2*1 = 6*6!

P(all 6 numbers of a die) = 6*6! / 67 = 5! / 65

Is this correct? Would appreciate any comments. Thanks

Last edited: Jul 9, 2011
2. Jul 9, 2011

### Klockan3

Re: Throwing a die 7 times

This is wrong, try calculating the probability of getting all 6 sides with just 6 dice and you would see why. You are missing some dice outcomes.

3. Jul 9, 2011

### binjip

Hi Klockan3,

ok, suppose I had only 6 turns throwing a die (or as you suggested having only 6 dice).

There would be 6*5*4*3*2*1 possible ways of arranging the numbers so that I have each number exactly once. This is 6!. So the result would be 6! / 6^6.

Now, I argued that I have an additional turn in which I can throw whatever comes up. So the denominator becomes 6^7. The nominator changed to reflect the possibility of getting any number - I multiplied by 6.

Ok, you are saying that I'm missing some dice outcomes. What I can think of is that the possibility of throwing any number extends to each of the 7 turns (or, alternatively, to each of the 7 dice). So I would multiply by 7.

This would change the result to 7*6*6*5*4*3*2*1 / 6^7.

Do you agree?

4. Jul 9, 2011

### Ray Vickson

Re: Throwing a die 7 times

Think of tossing 7 balls into 6 bins. You want the number of ways of having two balls in bin 6 and one ball in bins 1-5 each. You can do it sequentially: first think of it as bin 6 and not bin 6. How many ways are there of having 2 balls in bin 6 and 5 balls in bins not 6? For each such way we are tossing 5 balls in bins 1-5. How many ways are there of doing that? Altogether, you ought to end up with the *multinomial coefficient* C(7;1,1,1,1,1,2).

RGV

5. Jul 9, 2011

### Klockan3

I am sorry I didn't see this earlier, thought that the two sixes in the first statement had to be next to each other for some reason... So it is also wrong.

6. Jul 10, 2011

### binjip

Re: Throwing a die 7 times

Thank you Ray, I didn't think of it this way, but it seems much clearer now.

Ok, so I have 7! / (1!*1!*1!*1!*1!*2!) = 7!/2 = 2520 ways of arranging the 7 balls into 6 bins.

There are 6^7 of arranging the balls when there are no conditions.

P({1,...5} comes up once and 6 twice) = (7! / 2) / 6^7 = 0.009.

Now the second problem was the probability that I have 5 numbers once and one number twice. This could be stated as putting 7 balls into 6 bins and adding the multinomial coefficient for each case.

P(2-6 once, 1 twice) + P(1, 3-6 once, 2 twice) + ... + P(1-5 once, 6 twice) = 6 * C(7; 1,1,1,1,1,2) = 6*7!/2 = 15,120

So the final probability is (6*7!/2) / 6^7 = .054

Do you agree? Thanks.

Last edited: Jul 10, 2011
7. Jul 10, 2011

### binjip

Re: Throwing a die 7 times

Ok, I run a simulation and the results are correct. Thanks.