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Roll 5 dice simultaneously

  1. Dec 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate
    a) P(exactly three dice have the same number)
    b) Calculate the conditional probability P(three of the dice shows six|two of the dice shows 5)

    2. Relevant equations


    3. The attempt at a solution
    a) Say we have 1 as the number we get for three dice. There is 1/6 chance of getting a 1 and 5/6 of getting other numbers. The total sample space is 6^5 for five rolls.
    Hence for P(getting three 1's)=[(1/6)^3*(5/6)^2]/[6^5]
    But that was just for getting three 1's and there are total of 6 numbers that can be used.
    So
    P(exactly three dice have the same number)=6*[(1/6)^3*(5/6)^2]/[6^5].
    This answer seems to be right, but I would like to know how you can solve this using the binomial coefficent.

    My attempt:
    3 in 5 rolls will result in a number, which can be expressed as 5C3. Each number has 1/6 chance of being that number. The others have the probability of 5/6. The sample space is still 6^5.
    Putting this all together, [6*(5C3)*(1/6)^3*(5/6)^2]/[6^5], which yields a wrong answer.

    b) The conditional probability P(three of the dice shows six|two of the dice shows 5) can be written as
    P(three of the dice show six|two of the dice show 5)=P(three show 6, two show 5)/P(two show 5).
    Why are the events not independent here? Each die is a different entity so can't we just say P(three shows 6, two show 5)=P(three shows 6)*P(two show 5)?
    Anyway, P(two shows 5)=[(1/6)^2*(5/6)^3]/[6^5]
    I am not sure how to find the probability of the intersection of three show 6 and two show 5. I tried using the propoerty that the numerator of conditional probability can be written as P(three show six)P(two show 5|three show 6) but that got me nowhere.

    Any help will be appreciated!
     
  2. jcsd
  3. Dec 7, 2015 #2

    mfb

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    This the probability for "the first 3 dice show the same number, the last two do not have this number" (or any other set of 3 specific dice). It does not matter which three dice show the same number, you have to take this into account.

    Also, you include the factors of 1/6 twice. Use 1/6 and 5/6, or use 1 and 5 and divide by 65, but not both together. This is also a problem in all other formulas.
    Apart from the wrong factors of 6 (see above), this approach is right.

    Same as above: the dice don't have a given order. A die that shows 5 cannot show 6 and vice versa.
    This is not right. Neither for "exactly two" nor for "at least two".
     
  4. Dec 7, 2015 #3
    Okay, I am having a bit of trouble understanding the choose function. It would be great if you can help me clear a problem:

    P(two dice show 5)
    So there are 5C2 possible ways to choose the two dice. And those dice have one choice (that they are showing 5), so for that part, it will be (5C2)*1. Then the three other dice can be any of the five other numbers. Here, the right method is 5*5*5, but why can't we use the choose function, 5C3, here? There are 5C3 ways of choosing the rest of the dice, and there are 5 choices for them, so (5C3)*5. Why is (5C3)*5=/=5*5*5?
    Using 5*5*5, P(two dice show 5)= [(5C2)*5^3]/[6^5]
     
  5. Dec 7, 2015 #4

    PeroK

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    To try to clear up this confusion, think of 3 dice, rather than 5. Suppose you want precisely two dice to show 5. There are ##3C2 = 3## ways to choose the two dice that are the same. These are:

    55X
    5X5
    X55

    Now, as we have chosen the two places with the 5's, there is no independent choice for the remaining place. It's already decided.

    It's the same when you have 5 dice. There are ##5C2## choices for where the two 5's are. But, once you have chosen these, the remaining three places are chosen also:

    55XXX
    5X5XX
    etc.
     
  6. Dec 7, 2015 #5

    mfb

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    "exactly two", yes.

    Note that you can also choose the 3 non-5 dice instead of the 2 dice that are 5, as (5 choose 3) = (5 choose 2). It is still just one choice that fixes both groups.
     
  7. Dec 7, 2015 #6
    That makes it much more clear! Thanks so much!

    Okay :) thank you!
     
  8. Dec 7, 2015 #7
    I'm still a bit puzzled about part b of the original question. I do not know how to find the probability of the intersection (the numerator) in the conditional probability. Any hints to how I should approach this?
     
  9. Dec 7, 2015 #8

    mfb

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    P(three show 6, two show 5)? This is easier than all the other probabilities you calculated so far.
     
  10. Dec 7, 2015 #9
    Yes, that one. I can do them individually but not sure how you can combine them.
     
  11. Dec 7, 2015 #10

    mfb

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    What do you mean by individually?

    How many options are there to have "6 6 6 5 5" as result?
     
  12. Dec 7, 2015 #11
    Doh! So it's just 5C2 or 5C3?? So the probability is (5C3)/(6^5)?
     
  13. Dec 8, 2015 #12

    mfb

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    Right.
     
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