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Find the probability that 5 different faces come up twice each if 6 side die is rolled ten times?

What methods should I apply here?

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3 Answers 3

up vote 3 down vote accepted

Hint:

First, how many ways are there to pick 5 of the 6 faces.

If I have chosen 5 different faces, how many different orders can they come in. What is the probability that they come in that order.

Now multiply everything together.

From this I get: $$\frac{10!}{2^{5} 6^9}$$

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where did the 10!/2^10 come from? –  meiryo May 24 '11 at 0:46
1  
@Meiryo: I was counting how many ways to arrange a,a,b,b,c,c,d,d,e,e in a row. Well there are 10! total, but we divide by 2! for each pair. So 10!/2^5. Aha! it seems my number needs to be modified! –  Eric Naslund May 24 '11 at 0:48
    
Ahh, I see now, thanks! –  meiryo May 24 '11 at 0:52
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Observe that, if five different faces come up twice each in ten rolls, then each of those faces comes up exactly twice and the sixth face doesn't appear at all. Let's see how many ways this can happen.

There are 6 ways to choose which face doesn't come up. For convenience, say "6" is the face that doesn't come up. We now want the rolls 1, 1, 2, 2, 3, 3, 4, 4, 5, and 5, but how many ways can we get these rolls? I'll leave that part to you. (Hint: Use multinomial coefficients.) So, 6 times whatever you get for that part tells you how many acceptable outcomes there are. Divide this by the total number of possible outcomes (6^10).

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I don't know how to use 'multinomial coefficients' can you elaborate further? –  meiryo May 24 '11 at 0:44
    
@meiryo: en.wikipedia.org/wiki/… –  El'endia Starman May 24 '11 at 0:55
1  
Sure. (FYI: They are sometimes called the "Mississippi numbers".) Suppose the 1, 1, 2, 2, etc. were ten distinct numbers. You would say immediately that there are 10! arrangements. How do we account for the repeats? Since the 1's are indistinguishable and they can be ordered in 2! ways, we should divide by 2!. Similarly for all the numbers, so we would get 10! / [(2!)^5] distinct arrangements. –  Austin Mohr May 24 '11 at 0:56
    
Understood, thank you so much Austin. –  meiryo May 24 '11 at 1:05
    
Less elegantly, the places for the $1$'s can be chosen in $\binom{10}{2}$ ways. For each such choice the places for the $2$'s can be chosen in $\binom{8}{2}$ ways, and so on. –  André Nicolas May 24 '11 at 1:33
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Just for kicks, here is a fun R simulation.

Running 10 repetitions of a 10,000 simulation run, we get an avg of 0.01178 ± 0.001

#  PARAMETERS:   #
# -------------- #
  set.seed(1)
  sides <- 6
  rolls <- 10

  # simulations to get a probability
  simulations <- 10000

  # repetitions to repeat the process, to get an average result
  reps <- 10


#  ROLL THE DIE  #
# -------------- #
  # This function is a single 10-roll sample
  OneTrial <- function(sides, rolls) { 
    outcome <- sample(sides, rolls, TRUE)

    # count the number of sides that came up exactly-two
    #    and check if that amounts to 5 faces 
    sum(table(outcome) == 2) == 5
  }

# --------------- #
#  MULTIPLE REPS  #
# --------------- #
results <- c()
for (i in 1:reps) {
  simulOutcomes <- replicate(simulations, OneTrial(sides, rolls))
  results[[i]]  <- sum(simulOutcomes) / simulations
}


mean(results)
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