4 Sided dice - flawed logic? If we had a 4 sided dice (numbers 1,2,3,4 on the faces) and rolled it 6 times and recorded the results.  What would be the probability that we rolled the same number 3 or more times? (Such as rolling 1,1,1,4,3,2 or 1,1,1,1,4,4 etc).  I need to avoid the problem of counting some results twice.  For example, if I rolled 1,2,1,2,1,2  I want to count this only once, not twice.  
My attempted solution went as follows.
Probability of getting exactly 3 of the same number 
$${6 \choose 3}×(4/4×1/4×1/4×1/4×3/4×3/4×3/4) = .5273$$
Probability of getting exactly 4 of the same number 
$${6 \choose 4}×(4/4×1/4×1/4×1/4×1/4×3/4×3/4) = .1319$$
Probability of getting exactly 5 of the same number 
$${6 \choose 5}×(4/4×1/4×1/4×1/4×1/4×1/4×3/4) = .0176$$
Probability of getting exactly 6 of the same number 
$${6 \choose 6}×(4/4×1/4×1/4×1/4×1/4×1/4×1/4) = .0010$$
Since we have found the individual probabilities of exactly each number, we can sum this up to find the probability of "at least". Which we get .6778.  
The answer as I have come to find out through computer programs however is closer to .648   I am unsure of where I went wrong in my logic.
 A: Assuming the dice used are fair, a restatement of the question is "What is the opposite probability a quaternary sequence of length 6 has no digit repeated 3 or more times?"
To do this, count how many quaternary sequences of length 6 have no digit repeated 3 or more times.
By the pigeon-hole principle, there will be exactly two digits which appear twice, and exactly two digits which appear once, or there will be three digits each of which appearing twice.  (e.g. 112234 for first scenario or 112233 for second scenario).
How many sequences of the first type?


*

*Pick the two duplicated numbers: $\binom{4}{2}$ (this forces the selection of the nonduplicated numbers)

*Pick the locations of the larger of the duplicated numbers: $\binom{6}{2}$

*Pick the locations of the smaller of the duplicated numbers: $\binom{4}{2}$

*Pick the location of the larger of the nonduplicated numbers: $\binom{2}{1}$ (the final number goes in the only remaining slot)


For a total of $\binom{4}{2}\binom{6}{2}\binom{4}{2}\binom{2}{1} = 1080$ sequences of the first type.
How many sequences of the second type?


*

*Pick the three duplicated numbers: $\binom{4}{3}$

*Pick the locations of the largest of the numbers: $\binom{6}{2}$

*Pick the locations of the second largest: $\binom{4}{2}$ (this forces the location of the remaining numbers)


For a total of $\binom{4}{3}\binom{6}{2}\binom{4}{2} = 360$ sequences of the second type.
How many quaternary sequences in total?
$4^6 = 4096$
The probability of not getting a digit repeated at least three times then is: $\frac{1080+360}{4096} = \frac{45}{128}\approx \%35.16$
And so the probability of getting a digit repeated at least three times is: $\frac{83}{128}\approx \%64.84$
A: Are you double counting the issue you brought up? (Counting 1,2,1,2,1,2 twice)
How many ways can that happen?  ${4\choose2}=6$ ways to pick the two numbers. $\frac{6!}{3!3!}=20$ ways the two numbers can be arranged.  So that is $\frac{120}{4^6}$ that was double counted.  If you subtract that from your answer it seems close to the simulated result.
