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1. Given 5 dice rolls that are independent from each other, what is the probability for the following results? (order of roll does not matter)

1. all 5 dice rolls are the same

2. 4 dice rolls are the same

3. the dice rolls are in sequence (1-5 or 2-6) -order does not matter

4. two pairs of dice are the same (ex: 1 1 4 4 3)

5. the result is one pair and the other three are the same (ex: 1 1 1 6 6)

So far, my understanding of the problem has been the ff:
1. 1/6^5
2. 150/6^5
3.(5!+5!)/6^5
4. 1800/6^5
5. 300/6^5

Is this correct?

2.

Given 5 dice rolls that are independent from each other, what is the probability for the following results? (order of roll does not matter)

1. all 5 dice rolls are the same

2. 4 dice rolls are the same

3. the dice rolls are in sequence (1-5 or 2-6) -order does not matter

4. two pairs of dice are the same (ex: 1 1 4 4 3)

5. the result is one pair and the other three are the same (ex: 1 1 1 6 6)

So far, my understanding of the problem has been the ff:
1. 1/6^5
No, this is the probability all 5 are the same specific number. That is, that all 5 are 1 or all 5 are 2, etc. Since there are 6 possible numbers the probability all 5 die are the same number is 5/6^5.

[quote]2. 150/6^5[//quote]
The probability the first 4 die are, say, "1" and the other die any other number is (1/6^5)(5/6)= 5/6^5. Since there are 5 "positions" in which the "1" might appear the probability of 4 "1"s and one other number is 25/6^5. But, again, there are 6 number the 4 die might be so the probability of 4 dice being the same number and one die being different is 5(25/6^5)= 125/6^5. How did you get 150?

Quote:
3.(5!+5!)/6^5
The probability the five die are "1, 2, 3, 4, 5" in that order is 1/6^5. There are 5! different orders so the probability of "1, 2, 3, 4, 5" in any order is 5/6^5. Of course, it is exactly the same for "2, 3, 4, 5. 6" so you are correct.

Quote:
4. 1800/6^5
The first die could be any thing. The probability that the next die is the same, so that the first two die are the same, is 1/6. The probability the third die is different from the first two is 5/6. The probability the fourth is the same as the third is 1/6. The probability the last is different from either of those is 4/6= 2/3. The probability of "AABBC" is (1/6)(5/6)(1/6)(4/6)= 4/6^4. But there are (2!2!)/5! different orders so this probability is (4(2!2!))/(6^4(5!)). That is 1/(480(6^4)) How did you get 1800?

Quote:
5. 300/6^5
The first die can be anything. The probability the next two dice are the same is 1/6^2. The probability the fourth die is anything other than that is 5/6. The probability the fifth die is the same as the fourth is 1/6. The probability of "AAABB" is (1/6^2)(5/6)(1/6)= 5/6^4. There are 5!/(3!2!)= 10 ways to order "AAABB" so the probability of three the same and two the same but different from the others is 50/6^4. Since 300/6= 50, that is the same as your answer.
Quote:
Is this correct?

For #2, I got 150 from the ff:

number of choices for the number showing on the dice = 6
number of ways of choosing which 4 dice the four of a kind will appear on = 5C4 = 5!/(1!4!) = 5
number of choices for the last number on the last dice = 6-1 = 5
number of ways to choose the last dice = 1C1 = 1

So the number of correct outcomes is 6*5*5*1 = 150

For #4, I got 1800 from the ff:

Number of choices for the number on first pair = 6
number of ways which two dice have the number =5C2 = 10
number of choices for number on second pair = 5
number of ways which two dice have the number = 3C2 = 3
number of choices for number on last dice = 4
number of ways of choosing last dice = 1C1 = 1
correct outcomes = 6*10*5*3*4*1 = 3600

However, this counts each pair twice (66554 is treated separate than 55664)
so i divide 3600 by 2 = 1800