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

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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



:cry: 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.
 
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Hi Arcana! :smile:

I'd say this problem is a little too annoying to tackle immediately.
Can you solve first the following for me:

In how many way can you roll 2 dice such that 2 numbers are the same?

Solve this first, then we'll analyze what you did.
 
Hi ArcanaNoir! :smile:

Let's start with the first.
You need to consider the negation of the outcome to make the problem a lot easier.

What is the negation of: "at least 2 numbers are the same"?
 
micromass said:
Hi Arcana! :smile:

I'd say this problem is a little too annoying to tackle immediately.
Can you solve first the following for me:

In how many way can you roll 2 dice such that 2 numbers are the same?

Solve this first, then we'll analyze what you did.

6/36
because 1,1 or 2,2, etc.
 
ArcanaNoir said:
6/36
because 1,1 or 2,2, etc.

OK, so you're finding tuples with entries between 1 and 6.

So for the first one, you need to find 6 tuples (a,b,c,d,e,f) such that we have at least 2 numbers that are the same.

Following ILS's lead: what is the negation of that.
 
I like Serena said:
Hi ArcanaNoir! :smile:

Let's start with the first.
You need to consider the negation of the outcome to make the problem a lot easier.

What is the negation of: "at least 2 numbers are the same"?

All numbers are different? Which is nPr(6,6) = 6!
Right? So, okay, for at least 2 are the same it's 6^6-6! and then over 6^6 for the probability.

I'm really getting messed up for at least 3 and at least 4.
 
micromass said:
OK, so you're finding tuples with entries between 1 and 6.

So for the first one, you need to find 6 tuples (a,b,c,d,e,f) such that we have at least 2 numbers that are the same.

Following ILS's lead: what is the negation of that.

I'm confused.
 
ArcanaNoir said:
I'm confused.

Did my tuples analogy confuse you? :frown: I thought it made it easier...
I'll probably leave the thread to ILS, then. He can explain things much better than me :biggrin:
 
Well yes, the tuples sort of confused me. Can you be more specific and less vague? lol... I need a shove more than a nudge here.
 
  • #10
For the next one, first you need to find what the chance is for exactly 2 numbers that are the same.
After that you can find the answer again with negation.

So going back to micromass's example, suppose you have 3 dice. :wink:
How many possibilities for exactly 2 numbers to be the same?
 
  • #11
I like Serena said:
So going back to micromass's example, suppose you have 3 dice. :wink:
How many possibilities for exactly 2 numbers to be the same?

I don't know how to do that without tabling the possibilities.
 
  • #12
You might find it easier to start at the other end. How many ways are there to roll 6 dice so all 6 numbers are the same?

To figure out "exactly 5 numbers the same", you have to figure out how many ways you can roll 5 dice so all 5 numbers are the same, the number of ways you can roll the die that's different, and then how they can be arranged, e.g. yxxxxxx, xyxxxx, etc.

Then "at least 5" is "5 all the same" plus "6 all the same".
 
  • #13
All right, zooming in.
Suppose we roll the dice one by one.

What is the chance for the first 2 dice to be the same, and the 3rd die to be different?
 
  • #14
ArcanaNoir said:
I don't know how to do that without tabling the possibilities.
That might not be a bad idea. If you write out the combinations in an orderly way, you should be able to identify what factors contribute to the total, and then generalize from there.
 
  • #15
vela said:
You might find it easier to start at the other end. How many ways are there to roll 6 dice so all 6 numbers are the same?

To figure out "exactly 5 numbers the same", you have to figure out how many ways you can roll 5 dice so all 5 numbers are the same, the number of ways you can roll the die that's different, and then how they can be arranged, e.g. yxxxxxx, xyxxxx, etc.

Then "at least 5" is "5 all the same" plus "6 all the same".

Exactly 5 is 6 times 5= 30

It's really 3 and 4 that I can't get, and it's because I don't know what formulas to use. The other numbers I did by tabling.
 
  • #16
I like Serena said:
All right, zooming in.
Suppose we roll the dice one by one.

What is the chance for the first 2 dice to be the same, and the 3rd die to be different?

:frown: I don't know.
 
  • #17
ArcanaNoir said:
Exactly 5 is 6 times 5= 30
Does the order matter? I'm assuming it does, in which case you're missing a factor of 6.
It's really 3 and 4 that I can't get, and it's because I don't know what formulas to use. The other numbers I did by tabling.
Well, the factor of 5 above is the total number of ways to roll one die so that it doesn't match the value of the other dice. In the case where you have 4 matching dice and two different, how many combinations are there for the two dice? You know that each can take on one of five possible values.
 
  • #18
There's a formula where you find the number of ways you can arrange the letters in PEPPER,
and you do 6!/(1!2!3!).

So, for exactly 4 numbers the same, would I do 6!/(4!1!1!)?

[edit] never mind I see that's not going to work.
 
  • #19
All right.
I'll make the table for you (first 2 dice the same, 3rd different).

Code: 
                chance
1 1  2,3,4,5,6  (1/6)(1/6)(5/6)
2 2  any not 2  (1/6)(1/6)(5/6)
3 3  any not 3  (1/6)(1/6)(5/6)
4 4  any not 4  ...
5 5  any not 5
6 6  any not 6

Total chance is: 6 x (1/6)(1/6)(5/6)

I might also say:
First die: any number - chance is 6/6
Second die: the same number - chance 1/6
Third die: any different number - chance 5/6

The product is (6/6)(1/6)(5/6).

Does that make sense?

If so, then what would the chance be for 4 dice with the first 2 dice to be the same and the other 2 different?
 
  • #20
vela said:
Does the order matter? I'm assuming it does, in which case you're missing a factor of 6.
crap

Well, the factor of 5 above is the total number of ways to roll one die so that it doesn't match the value of the other dice. In the case where you have 4 matching dice and two different, how many combinations are there for the two dice? You know that each can take on one of five possible values.

halp.

I really want to know how to do this without tabling, but every question people ask, I can only answer by tabling.
 
  • #21
I like Serena said:
All right.


Total chance is: 6 x (1/6)(1/6)(5/6)

do you mean 6 x (6/6)(1/6)(5/6) ?

If so, then what would the chance be for 4 dice with the first 2 dice to be the same and the other 2 different?

(6/6)(1/6)(5/6)(4/6)?
 
  • #22
To get exactly four dice the same, you need something like XXXX YZ. Let's ignore the order for now. How many values can X, Y, and Z each take?

Similarly, for exactly three dice the same, you need XXX YZW. How many values can X, Y, Z, and W assume?

Do you see a pattern?
 
  • #23
vela said:
To get exactly four dice the same, you need something like XXXX YZ. Let's ignore the order for now. How many values can X, Y, and Z each take?
x can take 6, y can take 5, and z can take 4.

Similarly, for exactly three dice the same, you need XXX YZW. How many values can X, Y, Z, and W assume?
6, 5, 4, 3
Do you see a pattern?
maybe?

is it (6/6)(1/6)(1/6)(5/6)(4/6)(3/6) for exactly 3 numbers the same?
 
  • #24
ArcanaNoir said:
do you mean 6 x (6/6)(1/6)(5/6) ?

Err... no.
I meant there are 6 (disjunct) lines in the table with each a chance of (1/6)(1/6)(5/6).
So the chance on any of the lines in the table is the sum of the chances.
In this case that is 6 x (1/6)(1/6)(5/6).


ArcanaNoir said:
(6/6)(1/6)(5/6)(4/6)?

Yep! :smile:

And with 6 dies, where the first 2 dices are the same and the rest different?

Next we'll consider the possible rearrangements, like you did with PEPPER.
 
  • #25
i like serena said:
and with 6 dies, where the first 2 dices are the same and the rest different?

(6/6)(1/6)(5/6)(5/6)(5/6)(5/6)?
 
  • #26
ArcanaNoir said:
x can take 6, y can take 5, and z can take 4.
No, Z can take 5 values, right? We don't care if it matches Y.
 
  • #27
vela said:
No, Z can take 5 values, right? We don't care if it matches Y.

okay
is it (6/6)(1/6)(1/6)(5/6)(5/6)(5/6) for exactly 3 numbers the same?
 
  • #28
To use vela's terminology, we're looking at the pattern XX YZWV here.
 
  • #29
I like Serena said:
To use vela's terminology, we're looking at the pattern XX YZWV here.

so what do we do with it?
 
  • #30
ArcanaNoir said:
and with 6 dies, where the first 2 dices are the same and the rest different?

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

No, you allow the 4th die to be the same as the 3rd die here (you didn't before!)

And also for the 5th and 6th die to be the same as one of the other dies.

That's not right.
 
  • #31
I like Serena said:
No, you allow the 4th die to be the same as the 3rd die here (you didn't before!)

And also for the 5th and 6th die to be the same as one of the other dies.

That's not right.

But vela said it was wrong to say the fourth could NOT the same as the fifth or sixth. :cry:
 
  • #32
XX YZWV

X any: 6/6
X the same: 1/6
Y different from X: 5/6
Z different from X and Y: 4/6
W ...
V
 
  • #33
I like Serena said:
No, you allow the 4th die to be the same as the 3rd die here (you didn't before!)

And also for the 5th and 6th die to be the same as one of the other dies.

That's not right.

Okay I see that for exactly 2 to be the same If I let the last four be whatever they might be more than 2 the same among themselves. so what do I do?
 
  • #34
what if I have 2,2,3,3,4,5,6? Is that allowed as "exactly 2 numbers the same"?
 
  • #35
ArcanaNoir said:
But vela said it was wrong to say the fourth could NOT the same as the fifth or sixth. :cry:

Ah, I see, vela was using it differently than I was.
Sorry. :blushing:
 
  • #36
I like Serena said:
Ah, I see, vela was using it differently than I was.
Sorry. :blushing:

So now what?
 
  • #37
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?
 
  • #38
(6/6)(1/6)(1/6)(1/6)(5/6)(5/6)
 
  • #39
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.
 
  • #40
6!/4! ?
 
  • #41
ArcanaNoir said:
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).
 
Last edited:
  • #42
(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}
 
  • #43
You've got it! :smile:

Now you can find the chance of "at least 4 the same" by summing this with "exactly 5 the same" and "all the same".
 
  • #44
Thanks.
Just to be sure,
for exactly 3 the same would it be:
(shuffle XXXYYZ)(PXXXYYZ)+(Shuffle XXXYYY)(PXXXYYY)+(Shuffle XXXYZW)(PXXXYZW)?
 
  • #45
(Going out to fix the mailbox with Dad now, don't wait for any more replies from me for a lil while)
 
  • #46
ArcanaNoir said:
Thanks.
Just to be sure,
for exactly 3 the same would it be:
(shuffle XXXYYZ)(PXXXYYZ)+(Shuffle XXXYYY)(PXXXYYY)+(Shuffle XXXYZW)(PXXXYZW)?

Yep! :smile:


ArcanaNoir said:
(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. :approve:
 
  • #47
Great, thanks for all the help!

I still hate probability. :devil:
 
  • #48
ArcanaNoir said:
Great, thanks for all the help!

I still hate probability. :devil:

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. :wink:
 
  • #49
I like Serena said:
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. :wink:

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!
 
  • #50
ArcanaNoir said:
sometimes PF makes me want to bang my head against the wall!
Our work here is done.
 

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