Facebook Page
Twitter
RSS
Thanks Thanks:  0
+ Reply to Thread
Results 1 to 3 of 3
  1. MHB Apprentice

    Status
    Offline
    Join Date
    Aug 2019
    Posts
    4
    Thanks
    1 time
    Thanked
    0 times
    #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?
    Last edited by marcadams267; September 14th, 2019 at 05:52.

  2. # ADS
    Circuit advertisement
    Join Date
    Always
    Posts
    Many
     

  3. MHB Journeyman
    MHB Math Helper

    Status
    Offline
    Join Date
    Jan 2012
    Posts
    966
    Thanks
    293 times
    Thanked
    1,010 time
    Thank/Post
    1.046
    #2
    Quote Originally Posted by marcadams267 View Post
    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 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 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 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 Quote:
    Is this correct?

  4. MHB Apprentice

    Status
    Offline
    Join Date
    Aug 2019
    Posts
    4
    Thanks
    1 time
    Thanked
    0 times
    #3 Thread Author
    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

Similar Threads

  1. Probability of winning dice game
    By TheFallen018 in forum Basic Probability and Statistics
    Replies: 3
    Last Post: May 30th, 2018, 10:10
  2. Dice rolling - running total probability
    By lfdahl in forum Challenge Questions and Puzzles
    Replies: 8
    Last Post: January 23rd, 2018, 10:24
  3. Dice Probability (5 Sided Dice and 6 Sided Dice)
    By coolguy56 in forum Basic Probability and Statistics
    Replies: 1
    Last Post: March 4th, 2015, 13:26
  4. Dice Problem
    By veronica1999 in forum Basic Probability and Statistics
    Replies: 3
    Last Post: June 28th, 2013, 16:24
  5. dice probability
    By jacks in forum Basic Probability and Statistics
    Replies: 2
    Last Post: August 3rd, 2012, 10:42

Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards