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.

Guest Dec 31, 2019

#1**+2 **

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!!.

Guest Dec 31, 2019

#2**+1 **

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

Melody Dec 31, 2019

#4**+1 **

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.**

Guest Dec 31, 2019

edited by
Guest
Dec 31, 2019

#6**+2 **

Thanks for all that information, I do appreciate your efforts.

But I do not think that is correct.

The way i have done it,

Is to think of every dice as being distinguishable. Like they are all different colours so

if x and y are different (3,4) is different from (4,3)

So what I think is,

You have found how many combinations there are that satisfy the conditions **given that the dice are indistinguishable.**

then

You have divided by by the total number of ways **WHEN the dice are different from each other**, (distinguishable)

So I think you have mixed identical dice in the numberator with different looking dice in the denominator.

So I still think my answer is more likely to be correct than your is.

Melody
Dec 31, 2019

#8**+2 **

Hi Melody: Bravo!!! You were right after all. I simply computed the permutations of the (3,4,5,6) =4 x 4 =16 and presto got the same number that you did. I believe my mistake was in using 6 C 4, which I believe now to be wrong, because 4 are 2 separate entities(1,1 and 2,2).

112233 = ........................................................90 permutations

112234 112243 =..........................................360p

112235 112244 112253 =...............................450p

112236 112245 112254 112263=..................720p

112246 112255 112264= ...............................450p

112256 112265=..............................................360p

112266=..............................................................90p

**GRAND TOTAL =2,520 PERMUTATIONS.**

Guest Jan 1, 2020

#9**+2 **

Thanks Guest,

I am glad that we are now in agreement. That is always comforting.

Melody
Jan 1, 2020

#17**+1 **

You should know, Mr. BB, that using “brovos” for a woman is often considered mockery. The correct form is “brave” and “brava” is the singular form.

In any case, Melody's *brave* should be directed here, https://web2.0calc.com/questions/six-6-sided-dice-are-rolled-what-is-the-probability, where Melody and I *monkey around* with this exact question.

GA

GingerAle
Jan 2, 2020

#5**+1 **

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 Dec 31, 2019

edited by
Guest
Dec 31, 2019

#11**+3 **

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.

Alan Jan 1, 2020

#12**+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

Guest Jan 1, 2020

#14**0 **

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.**

Guest Jan 1, 2020

#16**0 **

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

Guest Jan 1, 2020

#19**0 **

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 !

Melody Jan 5, 2020

#20**0 **

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.

Tiggsy Jan 5, 2020