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10 fair six-sided dice are rolled. What is the probability of obtaining exactly one pair each of the odd numbers, that is 11, 33, and 55? Thanks.

Guest Jul 13, 2018
 #1
avatar+152 
+1

Since 10 fair six-sided dice are rolled, the total outcomes is \(6^{10}=60466176\).

And, the pairs are: 11,33

11,55

So, 2*3=6 pairs....

Thus, the probability is \(\frac{6}{60466176}=\frac{3}{3023308}\).

azsun  Jul 14, 2018
 #2
avatar+94100 
+1

 

That is far too simple Azun. The 6 on the top is definitely not correct.   sad

It is good that you gave it a go though. 

Melody  Jul 14, 2018
 #3
avatar+94100 
+1

I will attempt to count them too ...

I know that six of the rumbers are 

1,1,3,3,5,5, and I am assuming that the others four must be 2,4 or 6 each.

 

2222113355 10!/(4!*2*2*2)=10!/(4!*8)=18900
2224113355 10!/(3!*8)=75600
2226113355 75600
2244113355 10!/(2*2*8)=113400
2246113355 10!/(2*8)=226800
2266113355 113400
2444113355 75600
2446113355 226800
2466113355 226800
2666113355 75600
4444113355 18900
4446113355 75600
4466113355 113400
4666113355 75600
6666113355 18900
           3*18900+6*75600+3*113400+3*226800 = 1530900

 

 

 

 

So I get the probability as       \(\frac{1530900}{6^{10} }= \frac{175}{6912} \approx 0.025318287037037\)       Note: this has been edited (1st edit)

 

 

That is if i have made no mistakes.

There would be a number of more sensible ways to do this.     laugh

Melody  Jul 14, 2018
edited by Melody  Jul 14, 2018
 #4
avatar
+1

Is the denominator 10! or 6^10?

Guest Jul 14, 2018
edited by Guest  Jul 14, 2018
 #5
avatar+94100 
0

You are right, the denominator must be 6^10

 

Thanks for pointing that out!      laugh laugh laugh

 

I have edited my answer accordingly.

Melody  Jul 14, 2018
 #6
avatar+1216 
+1

Here’s a “sensible way” smiley that hides the demonstration of the nuanced details.

 

\(\dfrac {\binom{10}{2}\cdot \binom{8}{2}\cdot \binom{6}{2}\cdot (3^4)}{6^{10}} = \dfrac{175}{6912} \approx 2.532\% \)

 

 

 

GA

GingerAle  Jul 14, 2018
 #7
avatar+94100 
+1

Thanks Ginger.

 

I am thinking about the logic behind Ginger's answer.

We have 10 different colored dice.

How many ways can I chose 2 (these will be 1)  10C2

Now there are 8 dice left, how many ways can i choose 2 (these will be 3)  8C2

Now there are 6 dice left, how many ways can I choose 2 (these will be 5)  6C2

Now there are 4 dice left. Each can be 2,4 or 6 so there are  3^4 posible combinations

 

So total number of possible combinations is   10C2 * 8C2 * 6C2 * 3^4 

 

nCr(10,2)*nCr(8,2)*nCr(6,2)*3^4 = 1530900    Cool !!  

 

This works and that is great.

I am always reluctant to use logic like this because sometimes it works and sometimes it doesn't and I can never understand why that is.  Maybe I understand now idk but historically I had a lot of problems with this. (historically for me it often led to double counting)

Melody  Jul 14, 2018
 #8
avatar
+1

Here's an alternative.

Consider the sequence 113355nnnn, where the n's represent any 2, 4 or 6.

Its probability is (1/6)^6*(1/2)^4.

The number of different orderings is 10!/(2!*2!*2!*4!).

The overall probability will be the product of these two.

That works out to be 175/6912, approximately 0.025318.

Guest Jul 14, 2018
 #9
avatar+94100 
0

Thanks guest.

I don't understand the 4!

They do not have to be all the same ???

Melody  Jul 15, 2018
 #12
avatar+1216 
+1

Hi Melody,

 

The (4!) is in there for the same reason it’s in your solution. Four of the dice are indistinguishable from each other, so their permutations are factored out of the (10!).

 

To clarify, though the four dice appear to be distinguishable from each other because 2, 4, and 6 are distinguishable, they are treated as an indistinguishable group, that is, they are only distinguishable from the 1, 3, and 5. So this question has four groups: three groups of 2 indistinguishable element, each consisting of 1s, 3s, and 5s, and one group four that is an indistinguishable collective of NOT 1s, 3s, or 5s.

 

 

GA

GingerAle  Jul 15, 2018
edited by GingerAle  Jul 15, 2018
 #15
avatar+94100 
+1

Thanks Ginger   wink

Melody  Jul 15, 2018
 #10
avatar
+1

Just picture all the 2's and 4's with extra dots turning them all into 6's, say.

To all intents and purposes they're the same, same as the 1's, 3's and 5's.

Guest Jul 15, 2018
 #11
avatar+94100 
0

Ok, thanks, I think I get it.

I hope I can use it appropriately in the future. :)

Melody  Jul 15, 2018
 #14
avatar+1216 
+1

Just picture all the 2's and 4's with extra dots turning them all into 6's, say.

To all intents and purposes they're the same, same as the 1's, 3's and 5's.

 

Twos, fours, and sixes, Oh MY! This sounds like analysis using magical incantations and charms. “Snake eyes and boil, slick lies and toil, these ducks are now pigs.”  The ducks aren’t really pigs, but some members of the viewing audience may think they are.  This is entertaining, but not as much as using telekinetic powers to roll snake eyes or not snake eyes.

 

---------

DC Thomas, my cat, demonstrated his telekinetic powers for rolling snake eyes. This was very impressive to watch. He rolled snake eyes on about 60% of the attempts. The dice seemed to roll with normal expectation when he wasn’t using his telekinetic powers. This went on for a while, until Mr. Peabody, my dog, caught him switching out the loaded dice. The fight was on...

 

DC knows he’ll get caught when he uses loaded dice playing craps, or embezzles Monopoly money when he’s the banker. He does this just to pissoff Mr. Peabody –he likes to fight. Mr. Peabody always obliges. Long ago, I started securing the fragile artwork, and put the veterinarian on speed-dial, whenever we play games.

 

   

GA

GingerAle  Jul 15, 2018
edited by GingerAle  Jul 15, 2018
edited by GingerAle  Jul 15, 2018

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