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Combinatorics question involving 6 6-sided dice

wittysoup

New member
Feb 13, 2013
7
When rolling 6 6-sided dice, how many different ways can you have exactly 4 different numbers?

I tried solving this like so,

the first dice has a possible 6 numbers, the second has a possible 5, the third has a possible 4, and the fourth, 3. Then there are 2 remaining dice of which each has to be one of the previous 4 numbers so there are:

6*5*4*3*4*4 = 5760 ways

something tells me I am not thinking correctly or might be missing something because when I did this by iteration and got 9216. (though I might have missed something here too)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
When rolling 6 6-sided dice, how many different ways can you have exactly 4 different numbers?

I tried solving this like so,

the first dice has a possible 6 numbers, the second has a possible 5, the third has a possible 4, and the fourth, 3. Then there are 2 remaining dice of which each has to be one of the previous 4 numbers so there are:

6*5*4*3*4*4 = 5760 ways

something tells me I am not thinking correctly or might be missing something because when I did this by iteration and got 9216. (though I might have missed something here too)
Welcome to MHB, wittysoup! :)

It is somewhat more complex.

There are 2 patterns with different counts: aaabcd and aabbcd.

Pattern aaabcd occurs 6*1*1*5*4*3 times.
Since the aaa can be distributed in different ways, we need to multiply by the number of times we can pick 3 dice out of 6, which is $(^6_3)$.
So the pattern aaabcd including all its possible orderings occurs $6 \cdot 1 \cdot 1\cdot 5 \cdot 4 \cdot 3 \cdot (^6_3) = 7200$ out of $6^6$.

The pattern aabbcd is more complex still.
The base pattern occurs 6*1*5*1*4*3 times.
Multiply by $(^6_2)$ for the different locations of aa.
Multiply by $(^4_2)$ for the remaining different locations of bb.
Divide by 2! because aa can be swapped with bb..
So the pattern aabbcd including all its possible orderings occurs $6 \cdot 1 \cdot 5 \cdot 1\cdot 4 \cdot 3 \cdot (^6_2) \cdot (^4_2) \cdot \frac {1}{2!} = 16200$ out of $6^6$.

So a total of $7200 + 16200 = 23400$ combinations.
 
Last edited:

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
[JUSTIFY]You never mentioned it, but when considering dice rolls it's generally assumed that the rolls are unordered (they have no notion of order, which makes sense since the dice are presumably identical). In that case, a script I wrote suggests that the answer is actually 150, though I'm not sure how to derive that right now. If the dice rolls are ordered, then I Like Serena's answer is correct.[/JUSTIFY]
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
[JUSTIFY]You never mentioned it, but when considering dice rolls it's generally assumed that the rolls are unordered (they have no notion of order, which makes sense since the dice are presumably identical). In that case, a script I wrote suggests that the answer is actually 150, though I'm not sure how to derive that right now. If the dice rolls are ordered, then I Like Serena's answer is correct.[/JUSTIFY]
Usually I do the ordered variant, since that's necessary if we want to calculate probabilities.
But let's see what we get in an unordered variant...

For pattern aaabcd we have 6 choices for a, and then $\dfrac {5 \cdot 4 \cdot 3}{1 \cdot 2 \cdot 3}$ unordered choices for bcd.
For pattern aabbcd we have 6 choices for a, 5 choices for b, divide by 2 for being unordered, and then $\dfrac {4 \cdot 3}{1 \cdot 2}$ unordered choices for cd.

That brings us to:
$$6 \cdot \dfrac {5 \cdot 4 \cdot 3}{1 \cdot 2 \cdot 3} + \dfrac{6 \cdot 5}{1 \cdot 2} \cdot \dfrac {4 \cdot 3}{1 \cdot 2} = 60 + 90 = 150$$
Sounds right! ;)
 

wittysoup

New member
Feb 13, 2013
7
each dice is different here it seems from the question.. so 1 1 1 2 3 4 is different than 1 2 1 1 3 4.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
Because the dice are all the same, usually the problem is interpreted as order not mattering in my experience. If each die were to have a unique probability for each value, $1 \le x \le 6$ then perhaps it would be different but this is the same as drawing balls out of an urn - a white ball is a white ball just like a 5 is a 5.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
each dice is different here it seems from the question.. so 1 1 1 2 3 4 is different than 1 2 1 1 3 4.
So those are the ordered variants.
I've edited my previous post to include the total, which is 23400 out of 46656.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Because the dice are all the same, usually the problem is interpreted as order not mattering in my experience. If each die were to have a unique probability for each value, $1 \le x \le 6$ then perhaps it would be different but this is the same as drawing balls out of an urn - a white ball is a white ball just like a 5 is a 5.
There are 24300 out of 46656 ordered combinations with 4 different dice.
The corresponding probability is P(4 different dice)=0.50154.

There are 150 out of 462 unordered combinations.
The corresponding proportion is 0.32468, which is different from the probability.

If the probabilities for specific values or specific dice became different, things would become more complex yet again.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
I agree with the two situations you proposed, but don't agree that the second can't be considered a probability.

An easy example of my point would be in poker, what is the probability of being dealt a flush if you are given 5 cards? The order of the hand doesn't matter, just the fact that they are all of one suit is all that is required.

If s = spades, then {As, 2s, 3s, 8s, Js} = {2s, As, 3s, 8s, Js}. The probability of a flush is well known.

Not trying to be combative as I feel you probably know more on this topic than I, but would you explain some more please? :)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
I agree with the two situations you proposed, but don't agree that the second can't be considered a probability.

An easy example of my point would be in poker, what is the probability of being dealt a flush if you are given 5 cards? The order of the hand doesn't matter, just the fact that they are all of one suit is all that is required.

If s = spades, then {As, 2s, 3s, 8s, Js} = {2s, As, 3s, 8s, Js}. The probability of a flush is well known.

Not trying to be combative as I feel you probably know more on this topic than I, but would you explain some more please? :)
In the case of a flush in poker it does not matter.
The number of ordered combinations is just 5! times the number of unordered combinations.
So the unordered proportion is the same as the ordered proportion.

Now consider for instance the chance on 6 sixes with 6 dice.
It is one possible combination out of 462 unordered combinations.
But the probability is much lower than 1 in 462.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
Hmm, I need some time to think about this but my immediate answer is that I wouldn't claim the unordered probability is

\(\displaystyle \frac{ \text{Unordered outcomes}}{ \text{Unordered possibilities}}\), rather \(\displaystyle \frac{ \text{Unordered outcomes}}{ \text{All possibilities}}
\)
Using your dice example you could write the solution at \(\displaystyle \left( \frac{1}{6} \right)^{6}\) or as \(\displaystyle \frac{1}{6^6}\).

Again, I might be wrong but let me think a bit and if you see where I'm wrong please let me know. :)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Hmm, I need some time to think about this but my immediate answer is that I wouldn't claim the unordered probability is

\(\displaystyle \frac{ \text{Unordered outcomes}}{ \text{Unordered possibilities}}\), rather \(\displaystyle \frac{ \text{Unordered outcomes}}{ \text{All possibilities}}
\)
Using your dice example you could write the solution at \(\displaystyle \left( \frac{1}{6} \right)^{6}\) or as \(\displaystyle \frac{1}{6^6}\).

Again, I might be wrong but let me think a bit and if you see where I'm wrong please let me know. :)
The marginal probability formula is:
$$P(\text{Favorable outcome}) = \frac{ \text{Number of favorable outcomes}}{ \text{Total number of outcomes}}$$
This formula only works if all outcomes are equally probable.
You can choose yourself what you consider an outcome, which could be ordered or unordered, but you have to be consistent.
Either choice works as long as the outcomes are equally likely.

In the case of a flush your unordered outcomes are equally likely.
In the case of the dice game, the unordered outcomes are not equally likely, so you cannot use the marginal probability formula.
 

wittysoup

New member
Feb 13, 2013
7
Okay, so assuming order is not important, we'll have 150 total possible combinations of dice containing exactly 4 of the 6 total numbers?
 

wittysoup

New member
Feb 13, 2013
7
Usually I do the ordered variant, since that's necessary if we want to calculate probabilities.
But let's see what we get in an unordered variant...

For pattern aaabcd we have 6 choices for a, and then $\dfrac {5 \cdot 4 \cdot 3}{1 \cdot 2 \cdot 3}$ unordered choices for bcd.
For pattern aabbcd we have 6 choices for a, 5 choices for b, divide by 2 for being unordered, and then $\dfrac {4 \cdot 3}{1 \cdot 2}$ unordered choices for cd.

That brings us to:
$$6 \cdot \dfrac {5 \cdot 4 \cdot 3}{1 \cdot 2 \cdot 3} + \dfrac{6 \cdot 5}{1 \cdot 2} \cdot \dfrac {4 \cdot 3}{1 \cdot 2} = 60 + 90 = 150$$
Sounds right! ;)
how did you get these fractions "[FONT=MathJax_Main]5[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]3[/FONT]" and "[FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]2[/FONT]" ?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
how did you get these fractions "[FONT=MathJax_Main]5[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]3[/FONT]" and "[FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]2[/FONT]" ?
They are $(^5_3)$ and $(^4_2)$.

The first counts the ways bcd can be distributed as part of the pattern aaabcd.
For "b" we have 5 remaining choices (after "a").
Then for "c" 4 remaining choices.
And for "d" 3 remaining choices.
Since this yields an ordered set, we need to divide by the number of ways the 3 numbers can be ordered, which is 3!.