How Many Ways Can You Roll Six Dice with Repeating Numbers?

[ArcanaNoir]

Ah, I see, vela was using it differently than I was.
Sorry.

So now what?

 

[I like Serena]

I just realized why vela suggested to start from the other side, but we're not there yet to see why this is the case.
We can continue the way I was going, and show you where things get complex.
But in retrospect I think it's better to first get the solution with as little confusion as possible.

So let's follow vela's pattern of exactly 4 the same and the other 2 as don't care.
That is: XXXX YZ or XXXX YY.

How many possibilities did you have for that?

 

[ArcanaNoir]

(6/6)(1/6)(1/6)(1/6)(5/6)(5/6)

 

[I like Serena]

Next is the PEPPER analogy.

How many ways can you shuffle the letters XXXXYZ?
Note that these are different outcomes, but all have exactly 4 the same dice.

 

[ArcanaNoir]

6!/4! ?

 

[I like Serena]

6!/4! ?

Yes.

However, we still need to distinguish XXXXYZ and XXXXYY.
Do you know how many ways you can shuffle XXXXYY?

The chance on exactly 4 the same would be:
(number of ways to shuffle XXXXYZ) x P(XXXXYZ) + (number of ways to shuffle XXXXYY) x P(XXXXYY).

 

[ArcanaNoir]

(6/6)(1/6)(1/6)(1/6)(5/6)(4/6)= 5/1944
(6/6)(1/6)(1/6)(1/6)(5/6)(1/6)=5/7776

$$\frac{(6!)(5)}{(4!)(1944)}+\frac{(6!)(5)}{(4!)(2!)(7776)}= \frac{25}{288}$$

 

[I like Serena]

You've got it!

Now you can find the chance of "at least 4 the same" by summing this with "exactly 5 the same" and "all the same".

 

[ArcanaNoir]

Thanks.
Just to be sure,
for exactly 3 the same would it be:
(shuffle XXXYYZ)(PXXXYYZ)+(Shuffle XXXYYY)(PXXXYYY)+(Shuffle XXXYZW)(PXXXYZW)?

 

[ArcanaNoir]

(Going out to fix the mailbox with Dad now, don't wait for any more replies from me for a lil while)

 

[I like Serena]

Thanks.
Just to be sure,
for exactly 3 the same would it be:
(shuffle XXXYYZ)(PXXXYYZ)+(Shuffle XXXYYY)(PXXXYYY)+(Shuffle XXXYZW)(PXXXYZW)?

Yep!

(Going out to fix the mailbox with Dad now, don't wait for any more replies from me for a lil while)

Thanks for telling me so I won't wait unnecessarily for another post.

 

[ArcanaNoir]

Great, thanks for all the help!

I still hate probability.

 

[I like Serena]

Great, thanks for all the help!

I still hate probability.

In retrospect, we might as well have done the "exactly 2" case, seeing how easily you finished the exercise, once you knew what to do.

 

[ArcanaNoir]

In retrospect, we might as well have done the "exactly 2" case, seeing how easily you finished the exercise, once you knew what to do.

Seeing how easily I finished the exercise, once I knew what to do, I wish you could have just told me what to do in the first place. Maybe I'll remember better since we did it this way, but sometimes PF makes me want to bang my head against the wall!

 

[vela]

sometimes PF makes me want to bang my head against the wall!

Our work here is done.

 

[I like Serena]

Seeing how easily I finished the exercise, once I knew what to do, I wish you could have just told me what to do in the first place. Maybe I'll remember better since we did it this way, but sometimes PF makes me want to bang my head against the wall!

Any suggestions?

Is there something that might better have been done differently?
I'm still trying to learn as well how to do these things. :shy:

Although I do secretly derive some pleasure of the image seeing you banging your head.
I guess it compensates for the times I was banging my head myself.

 

[ArcanaNoir]

Something's wrong. :( When I applied the technique of (shuffle)(Prob)+... to all the dice are different, I got:

(Shuffle abcdef)(Probabcdef)= (6!)(6/6)(5/6)(4/6)(3/6)(2/6)(1/6) = 100/9

Which is clearly not right. What's up?

 

[I like Serena]

Something's wrong. :( When I applied the technique of (shuffle)(Prob)+... to all the dice are different, I got:

(Shuffle abcdef)(Probabcdef)= (6!)(6/6)(5/6)(4/6)(3/6)(2/6)(1/6) = 100/9

Which is clearly not right. What's up?

Ah, yes, I'd forgotten about that. Sorry.

Your formula for PEPPER was not complete yet.
It only works because the number of P's is different from the E's and the R.

In your current case you already have all the possible combinations, so the shuffle factor should be "1".
To understand this, look only at the first 2 dice.
The combinations (3,4) and (4,3) are both covered by your calculation.
So there are no extra combinations to be found by shuffling.

The same holds true in the pattern XXXXYZ.
I gave you the wrong number before. Sorry.
The proper formula for XXXXYZ is: 6! / [(4!1!1!)(2!)]
The extra 2! in the denominator is because Y and Z can be exchanged, finding a pattern that you have already counted.
The formula for XXXXYY is still: 6! / (4!2!).

In the case XXXWYZ it is: 6! / [(3!1!1!1!)(3!)]
In the case XXXYYZ it is: 6! / [(3!2!1!)]
In the case XXXYYY it is: 6! / [(3!3!)(2!)], because you can interchange X and Y and find the same pattern.

 

[ArcanaNoir]

So the shuffle formula is (number of spaces!)/(number of identical objects! x number of "unique" objects!)?

For example, Shuffle TENNESSEE would be (9!)/(1!4!2!2! x 4!)?

That's not right I think. Can you explain further the shuffling?

 

[I like Serena]

TENNESSEE would be (9!)/(1!4!2!2! x 2!)
Since there are two objects (NN and SS) with equal length.

AAABBBCCCDDEEFF would be (15!)/(3!3!3!2!2!2! x 3!3!),
since there are 3 objects of length AAA and 3 objects of length DD.

 

[ArcanaNoir]

Okay, thanks. :)

 

[ArcanaNoir]

So.. I still can't get the right answer for "all dice different" using the shuffling. It's supposed to be (720)/(46656)

I did: (shuffle ABCDEF)(P ABCDEF) = [(6!)/(1!1!1!1!1!1!6!)](6/6)(5/6)(4/6)(3/6)(2/6)(1/6) = 5/324

 

[I like Serena]

Errr... how are they different?

 

[ArcanaNoir]

OMG I don't understand how that could have happened. My calculator ALWAYS reduces. Thank you. I'm going to go cry now...

[edit] I know how it happened. This was actually part "a" of the same problem and I wrote it down un-reduced because a number of the solutions in the back of the book are written unreduced. This was yesterday, by today, I forgot I did it that way.

BTW, thank you VERY much for the shuffling explanations, that's going to stick with me a long time, and it wasn't in the book or the notes like that. Just PEPPER. Totally useless to have such a specific example and not even know it shouldn't be generalized without extra information.

 

[I like Serena]

I'm afraid the formula is usually not included in course material.
And actually it's pretty hard to write it down properly.
Afterward it's a puzzle again to decipher it.
If you keep it, make sure you also list a lot of examples on how to use it (more than just PEPPER).

And please don't hold your breath before you forget it again.
I did!

I appreciate the thanks. :)

 

[ArcanaNoir]

Shouldn't the probability that all numbers are different plus exaclt 2 are alike and the rest different plus exactly three are alike plus...up to they are all the same be no greater than one?

 

[I like Serena]

Shouldn't the probability that all numbers are different plus exaclt 2 are alike and the rest different plus exactly three are alike plus...up to they are all the same be no greater than one?

Correct, they should sum up to exactly one (or less than one depending on what you calculate exactly).

 

[ArcanaNoir]

Okay, I think I have this problem all done. I've solved it probably more than one way by now, and I got consistent answers. *runs away*

 

[Ray Vickson]

Homework Statement

How many ways can you roll six dice so that at least 2 numbers are the same? At least 3? At least 4? At least 5?

Homework Equations

The Attempt at a Solution

I've used every equation in the chapter and filled page after page with numbers all moved around. I'm drowning! Please help. I hate probability.

{at least two the same} = E2 + E3 + E4 + E5 + E6, where Ej = {exactly j the same} (and "+" denotes set union). E2 = {2 1s, others all different}+{2 2s, others all different}+... Clearly, all these have the same number of elements, so the number of elements in E2 is 6*|{2 1s, all others different}| (|{.}| = set cardinality). Now think of 6 bins and 6 balls tossed at random into them; we want the number of outcomes in which bin 1 has 2 balls and the others all have 0 or 1 ball each. This the type of thing for which the _multinomial_ distribution was designed.

RGV

 

[I like Serena]

*runs away*

Hey! Come back!
There's more...