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My short computer code indicates that the total should be = 1440.
Now, here is my attempt to justify the above number mathematically!
We have 6 C 4 =15 ways of chosing the 1s and 2s.
We have 4 C 2 =6 ways of chosing the remaining 2 numbers.
The remaining 2 numbers (3, 4, 5, 6) =4 can be selected in 4 x 4 = 16 ways.
So, the overall total should be: 15 x 6 x 16 =1440 ways, which is in agreement with my computer code.

Therefore the probability should be: 1440 / 6^6 = 5 / 162.
Note: Some mathematician should check this!!.

Six 6-sided dice are rolled. What is the probability that exactly two of the dice show a 1 and exactly two of the dice show a 2? Express your answer as a common fraction.

My answer does not agree with yours guest.  Generally I have a lot of faith in the output of people's computer programs.  

I certainly do not have full confidence in my answer but the logic you have used to justify your programmng output does not make sense to me.

I think we need another adjudicator.  

Can I see you code please. It probably will not make sense to me, my programming skills are way our of date, but you never know maybe I wills see somthing there.  It is a good idea to include your code for these answers anyway.

Here is my logic:

1,1,2,2,x,y,

possibilities

x and y are both the same.     4*6!/2!2!2! = 4*90 = 360  ways

x and y are different.             4*3*6!/2!2! = 12*180 =   2160 ways

Total of 360+2160 = 2520  ways

Sample space = 6^6=46656

Prob = 2520/46656 = 35/648 = approx    0.054

.......

Hi Melody: Here is a short computer code which is very brief version of the full code. It calculates 16 combinations for for 1 of 6 numbers. To write a "full conditional statement", you have to repeat the "if condition" 8 x 6 x 2 =96 times !!!! So, I simply took a "shortcut" and arrived at 1440 as explained below the code:

a=0;b=1;c=1;d=1;e=1;f=1;g=1;p=0; cycle:n=b*100000+c*10000+d*1000+e*100+f*10+g; if(b==1 and c==1 and d==2 and e==2  and f!=1 and f!=2 and g!=1 and g!=2, goto loop, goto next);loop:printn," ",;p=p+1; next:g++;if(g<7, goto cycle, 0);g=1;f++;if(f<7, goto cycle, 0);g=1;f=1;e++;if(e<7, goto cycle,0);g=1;f=1;e=1;d++;if(d<7, goto cycle,0);g=1;f=1;e=1;d=1;c++;if(c<7, goto cycle,0);g=1;f=1;e=1;d=1;c=1;b++;if(b<7, goto cycle,0);g=1;f=1;e=1;d=1;c=1;b=1;a++;if(a==0, goto cycle,0);print"Total = ",p

OUTPUT: 112233  112234  112235  112236  112243  112244  112245  112246  112253  112254  112255  112256  112263  112264  112265  112266  Total =  16
This is only for 1 number. Then you cycle throug 6 numbers =16 x 6 = 96. and this is only one go for 15 combinations:6 C 4 = 15. And 15 x 96 =1440. This is how I arrived at 1440.

P.S. I found your mistake in this calculation:   x and y are different. 4*3*6!/2!2! = 12*180 =   2160 ways

The mistake is : when x, y are different, you only have 6 choices not 12 as follows: (3,4), (3,5), (3,6), (4,5), (4,6), (5,6) making your calculation =6 * 6! / 2!2! =1080. Then add it to your first calculation and we get:

1080 + 360 = 1440 which agrees with the computer output.

Sorry Melody: I thought I had found a mistake, which on second thought, is NOT a mistake !!!. I was taking "combinations of the 4 numbers, instead of their permutation", which you have it accurately counted. Shows you how much I know about combinations and permutations! I don't really know where the difference of 2520 - 1440 =1080 is !!.

By the way, I was able to list 1/3 of 1440 = 480 permutations, and here is the list. See if you can find any mistakes or duplicates in it:

(112233, 112234, 112235, 112236, 112243, 112244, 112245, 112246, 112253, 112254, 112255, 112256, 112263, 112264, 112265, 112266, 112332, 112342, 112352, 112362, 112432, 112442, 112452, 112462, 112532, 112542, 112552, 112562, 112632, 112642, 112652, 112662, 113223, 113224, 113225, 113226, 113232, 113242, 113252, 113262, 113322, 113422, 113522, 113622, 114223, 114224, 114225, 114226, 114232, 114242, 114252, 114262, 114322, 114422, 114522, 114622, 115223, 115224, 115225, 115226, 115232, 115242, 115252, 115262, 115322, 115422, 115522, 115622, 116223, 116224, 116225, 116226, 116232, 116242, 116252, 116262, 116322, 116422, 116522, 116622, 121233, 121234, 121235, 121236, 121243, 121244, 121245, 121246, 121253, 121254, 121255, 121256, 121263, 121264, 121265, 121266, 121332, 121342, 121352, 121362, 121432, 121442, 121452, 121462, 121532, 121542, 121552, 121562, 121632, 121642, 121652, 121662, 122133, 122134, 122135, 122136, 122143, 122144, 122145, 122146, 122153, 122154, 122155, 122156, 122163, 122164, 122165, 122166, 122331, 122341, 122351, 122361, 122431, 122441, 122451, 122461, 122531, 122541, 122551, 122561, 122631, 122641, 122651, 122661, 123213, 123214, 123215, 123216, 123321, 123421, 123521, 123621, 124213, 124214, 124215, 124216, 124321, 124421, 124521, 124621, 125213, 125214, 125215, 125216, 125321, 125421, 125521, 125621, 126213, 126214, 126215, 126216, 126321, 126421, 126521, 126621, 131232, 131242, 131252, 131262, 131322, 131422, 131522, 131622, 132132, 132142, 132152, 132162, 133212, 134212, 135212, 136212, 141232, 141242, 141252, 141262, 141322, 141422, 141522, 141622, 142132, 142142, 142152, 142162, 143212, 144212, 145212, 146212, 151232, 151242, 151252, 151262, 151322, 151422, 151522, 151622, 152132, 152142, 152152, 152162, 153212, 154212, 155212, 156212, 161232, 161242, 161252, 161262, 161322, 161422, 161522, 161622, 162132, 162142, 162152, 162162, 163212, 164212, 165212, 166212, 211233, 211234, 211235, 211236, 211243, 211244, 211245, 211246, 211253, 211254, 211255, 211256, 211263, 211264, 211265, 211266, 211332, 211342, 211352, 211362, 211432, 211442, 211452, 211462, 211532, 211542, 211552, 211562, 211632, 211642, 211652, 211662, 212133, 212134, 212135, 212136, 212143, 212144, 212145, 212146, 212153, 212154, 212155, 212156, 212163, 212164, 212165, 212166, 212331, 212341, 212351, 212361, 212431, 212441, 212451, 212461, 212531, 212541, 212551, 212561, 212631, 212641, 212651, 212661, 213123, 213124, 213125, 213126, 213312, 213412, 213512, 213612, 214123, 214124, 214125, 214126, 214312, 214412, 214512, 214612, 215123, 215124, 215125, 215126, 215312, 215412, 215512, 215612, 216123, 216124, 216125, 216126, 216312, 216412, 216512, 216612, 221133, 221134, 221135, 221136, 221143, 221144, 221145, 221146, 221153, 221154, 221155, 221156, 221163, 221164, 221165, 221166, 221331, 221341, 221351, 221361, 221431, 221441, 221451, 221461, 221531, 221541, 221551, 221561, 221631, 221641, 221651, 221661, 223113, 223114, 223115, 223116, 223131, 223141, 223151, 223161, 223311, 223411, 223511, 223611, 224113, 224114, 224115, 224116, 224131, 224141, 224151, 224161, 224311, 224411, 224511, 224611, 225113, 225114, 225115, 225116, 225131, 225141, 225151, 225161, 225311, 225411, 225511, 225611, 226113, 226114, 226115, 226116, 226131, 226141, 226151, 226161, 226311, 226411, 226511, 226611, 231231, 231241, 231251, 231261, 232131, 232141, 232151, 232161, 232311, 232411, 232511, 232611, 233121, 234121, 235121, 236121, 241231, 241241, 241251, 241261, 242131, 242141, 242151, 242161, 242311, 242411, 242511, 242611, 243121, 244121, 245121, 246121, 251231, 251241, 251251, 251261, 252131, 252141, 252151, 252161, 252311, 252411, 252511, 252611, 253121, 254121, 255121, 256121, 261231, 261241, 261251, 261261, 262131, 262141, 262151, 262161, 262311, 262411, 262511, 262611, 263121, 264121, 265121, 266121) Total = 480
 

Guest's first answer is correct

Here is a Monte Carlo program (in MathCad) that supports Guest #1's result:

Of course, since the Monte Carlo program depends on random numbers the result it returns may be slightly different each time it runs.  However, it does seem to provide strong support for Guest #1.

Hello Alan: That conclusion is extremely surprising in light of having just completed the individual permutations of each of the following 16 groups. Because I permuted each of the 16 groups separately, I don't think the possibility of duplication is likely. Finally, I assembled the entire list and the computer counted them as the total of these 16 individual permutations and both agreed as to the total of =2,520 permutations!
So, Alan, where could the overcounting possibly be? Any ideas where to look for possible duplications?
1 - 112233 = 90 permutations
2 - 112234 = 180p 
3 - 112235 = 180p 
4 - 112236 = 180p 
5 - 112243 = 180p 
6 - 112244 = 90p 
7 - 112245 = 180p 
8 - 112246 = 180p 
9 - 112253 = 180p 
10 -112254 = 180p 
11 -112255 = 90p 
12 -112256 = 180p
13 -112263 = 180p 
14 -112264 = 180p 
15 -112265 = 180p 
16 -112266 = 90p
Here is a complete list of all 90 permutations of the first group:112233. See if you can find anything wrong with it. Thanks.
{1, 1, 2, 2, 3, 3}, {1, 1, 2, 3, 2, 3}, {1, 1, 2, 3, 3, 2}, {1, 1, 3, 2, 2, 3}, {1, 1, 3, 2, 3, 2}, {1, 1, 3, 3, 2, 2}, {1, 2, 1, 2, 3, 3}, {1, 2, 1, 3, 2, 3}, {1, 2, 1, 3, 3, 2}, {1, 2, 2, 1, 3, 3}, {1, 2, 2, 3, 1, 3}, {1, 2, 2, 3, 3, 1}, {1, 2, 3, 1, 2, 3}, {1, 2, 3, 1, 3, 2}, {1, 2, 3, 2, 1, 3}, {1, 2, 3, 2, 3, 1}, {1, 2, 3, 3, 1, 2}, {1, 2, 3, 3, 2, 1}, {1, 3, 1, 2, 2, 3}, {1, 3, 1, 2, 3, 2}, {1, 3, 1, 3, 2, 2}, {1, 3, 2, 1, 2, 3}, {1, 3, 2, 1, 3, 2}, {1, 3, 2, 2, 1, 3}, {1, 3, 2, 2, 3, 1}, {1, 3, 2, 3, 1, 2}, {1, 3, 2, 3, 2, 1}, {1, 3, 3, 1, 2, 2}, {1, 3, 3, 2, 1, 2}, {1, 3, 3, 2, 2, 1}, {2, 1, 1, 2, 3, 3}, {2, 1, 1, 3, 2, 3}, {2, 1, 1, 3, 3, 2}, {2, 1, 2, 1, 3, 3}, {2, 1, 2, 3, 1, 3}, {2, 1, 2, 3, 3, 1}, {2, 1, 3, 1, 2, 3}, {2, 1, 3, 1, 3, 2}, {2, 1, 3, 2, 1, 3}, {2, 1, 3, 2, 3, 1}, {2, 1, 3, 3, 1, 2}, {2, 1, 3, 3, 2, 1}, {2, 2, 1, 1, 3, 3}, {2, 2, 1, 3, 1, 3}, {2, 2, 1, 3, 3, 1}, {2, 2, 3, 1, 1, 3}, {2, 2, 3, 1, 3, 1}, {2, 2, 3, 3, 1, 1}, {2, 3, 1, 1, 2, 3}, {2, 3, 1, 1, 3, 2}, {2, 3, 1, 2, 1, 3}, {2, 3, 1, 2, 3, 1}, {2, 3, 1, 3, 1, 2}, {2, 3, 1, 3, 2, 1}, {2, 3, 2, 1, 1, 3}, {2, 3, 2, 1, 3, 1}, {2, 3, 2, 3, 1, 1}, {2, 3, 3, 1, 1, 2}, {2, 3, 3, 1, 2, 1}, {2, 3, 3, 2, 1, 1}, {3, 1, 1, 2, 2, 3}, {3, 1, 1, 2, 3, 2}, {3, 1, 1, 3, 2, 2}, {3, 1, 2, 1, 2, 3}, {3, 1, 2, 1, 3, 2}, {3, 1, 2, 2, 1, 3}, {3, 1, 2, 2, 3, 1}, {3, 1, 2, 3, 1, 2}, {3, 1, 2, 3, 2, 1}, {3, 1, 3, 1, 2, 2}, {3, 1, 3, 2, 1, 2}, {3, 1, 3, 2, 2, 1}, {3, 2, 1, 1, 2, 3}, {3, 2, 1, 1, 3, 2}, {3, 2, 1, 2, 1, 3}, {3, 2, 1, 2, 3, 1}, {3, 2, 1, 3, 1, 2}, {3, 2, 1, 3, 2, 1}, {3, 2, 2, 1, 1, 3}, {3, 2, 2, 1, 3, 1}, {3, 2, 2, 3, 1, 1}, {3, 2, 3, 1, 1, 2}, {3, 2, 3, 1, 2, 1}, {3, 2, 3, 2, 1, 1}, {3, 3, 1, 1, 2, 2}, {3, 3, 1, 2, 1, 2}, {3, 3, 1, 2, 2, 1}, {3, 3, 2, 1, 1, 2}, {3, 3, 2, 1, 2, 1}, {3, 3, 2, 2, 1, 1} = 90 permutations

Thanks Alan: I ran them on WolframAlpha and, in my experience, it NEVER lists duplicates! Please be good enough to look at this Wolfram page where I ran the 180 permutations of:112234:

https://www.wolframalpha.com/input/?i=permutations+of+%7B1%2C+1%2C+2%2C+2%2C+3%2C+4%7D

P.S. if you find  it difficult to read, you may press the "plain text" button, which might be easier to read. Thanks.

Thanks Alan. You are absolutely right. I just found out this about Wolfram: when your enter a set of numbers such as:112243, the first thing the computing engine does is to put them in "numerical order" and then calculate the permutations on them in numerical order. in other words, it treats:112234 exactly the same as:112243, hence the duplication in reversed groups of: 34 and 43, 35 and 53, 36 and 63, 45 and 54, 46 and 64, 56 and 65.
And that is the difference of 1,080 =6 groups x 180

Thanks Alan and guest.  

Some time maybe I will work out why mine is wrong.

It is always good to have the correct answer anyway.  

What would we do without your Monty Carlo simulations.  They really ae very useful !

I arrived at the now accepted result, but via a different method.

Think of a qualifying sequence as any containing exactly two ones,two twos, and two "others".

Every qualifying sequence has the same probability of occurring, (1/6).(1/6).(1/6).(1/6).(4/6).(4/6)=16/(6^6).

The order doesn't matter.

The number of qualifying sequences will be 6!/(2^3), the three twos to remove the duplication of the 1's, the 2's, and the "others".

The total probability will be the product of these. 

That works out to be 5/162, approximately 0.030864.