12 replies to this topic

[runthemover]

If you are not a math/probability person RUN AWAY. run far far away.

I am somewhat ashamed to have to ask this. I can't figure out for the life of me what I am missing here.

Probability of seeing all 6 faces of a fair die in 6 independent rolls. This answer I know

6! / 6^6 = .01543

run a simulation of 100000 repetitions and I get that approximate answer

What I can't figure out.

Probability of seeing all 6 faces of a fair die in 7 independent rolls. I have

7! / 6^7 = .018 (approx)

6! ways of getting all six faces seen multiplied by C(7, 6) aka 7 choose 6 or just 7

divide that by 6^7 ways of rolling 7 dice.

when I run a simulation of this I get .05411. This is about 3 times my combinatoric answer. I am drawing a blank for some reason. I know my simulation is correct. The combinatoric answer doesn't look right to me. Any ideas?

[dna4ever]

you forgot to carry the 1

your welcome

[booyaga]

you forgot to carry the 1

your welcome

you're....js

[dna4ever]

you're....js

SOAB! that one gets me everytime

[LongLiveYorke]

You made your mistake when you considered (or forgot to consider) the different possabilities for what the 7th one could be and where it could go. In other words, with 7 die, we have to get one repeat. Let's work from that fact:

There are seven slots:

_ _ _ _ _ _ _

We're going to get exactly one repeated if we're going to have all 6 faces show up in 7 rolls. So, the name of the game is to count how many ways we can get 6 faces out of 7 rolls and divide that by the total amount of combinations (ie 6^7).

There are 6 ways that we can get a double (11,22,33,44,55,66). These have to go somewhere. There are 7 Choose 2 ways that we can arrange the doubles on the board. Here are examples:

6 _ _ _ 6 _ _

_ _ _ 5 5 _ _

_ _ 2 _ _ _ 2


etc.

So, starting from this, we get 6 *(7C2) ways we can make the above formations. We now need to count the ways to fill in the rest. But that's easy now. It's just:

5*4*3*2*1 ways to fill in the remaining 5 slots. So, our total is:

6*7C2*5*4*3*2*1 / 6^7

=.054012

I really hope this is right. It seems to somewhat match the answer you got. The problem with combinatorics is that you can really be convinced that you're right, you can be so sure of it and have everything make so much sense, and then totally get the wrong answer. Let's hope I didn't just do that.

Cheers.

[dna4ever]

You made your mistake when you considered (or forgot to consider) the different possabilities for what the 7th one could be and where it could go. In other words, with 7 die, we have to get one repeat. Let's work from that fact:

There are seven slots:1

_ _ _ _ _ _ _

We're going to get exactly one repeated if we're going to have all 6 faces show up in 7 rolls. So, the name of the game is to count how many ways we can get 6 faces out of 7 rolls and divide that by the total amount of combinations (ie 6^7).

There are 6 ways that we can get a double (11,22,33,44,55,66). These have to go somewhere. There are 7 Choose 2 ways that we can arrange the doubles on the board. Here are examples:

6 _ _ _ 6 _ _

_ _ _ 5 5 _ _

_ _ 2 _ _ _ 2
etc.

So, starting from this, we get 6 *(7C2) ways we can make the above formations. We now need to count the ways to fill in the rest. But that's easy now. It's just:

5*4*3*2*1 ways to fill in the remaining 5 slots. So, our total is:

6*7C2*5*4*3*2*1 / 6^7

=.054012

I really hope this is right. It seems to somewhat match the answer you got. The problem with combinatorics is that you can really be convinced that you're right, you can be so sure of it and have everything make so much sense, and then totally get the wrong answer. Let's hope I didn't just do that.

Cheers.

yea thats what i meant

[Petoria]

You made your mistake when you considered (or forgot to consider) the different possabilities for what the 7th one could be and where it could go. In other words, with 7 die, we have to get one repeat. Let's work from that fact:

There are seven slots:

_ _ _ _ _ _ _

We're going to get exactly one repeated if we're going to have all 6 faces show up in 7 rolls. So, the name of the game is to count how many ways we can get 6 faces out of 7 rolls and divide that by the total amount of combinations (ie 6^7).

There are 6 ways that we can get a double (11,22,33,44,55,66). These have to go somewhere. There are 7 Choose 2 ways that we can arrange the doubles on the board. Here are examples:

6 _ _ _ 6 _ _

_ _ _ 5 5 _ _

_ _ 2 _ _ _ 2
etc.

So, starting from this, we get 6 *(7C2) ways we can make the above formations. We now need to count the ways to fill in the rest. But that's easy now. It's just:

5*4*3*2*1 ways to fill in the remaining 5 slots. So, our total is:

6*7C2*5*4*3*2*1 / 6^7

=.054012

I really hope this is right. It seems to somewhat match the answer you got. The problem with combinatorics is that you can really be convinced that you're right, you can be so sure of it and have everything make so much sense, and then totally get the wrong answer. Let's hope I didn't just do that.

Cheers.

You're too damn quick for me. I come into gen. OT and I see Combinatorics and think "ship it", but then I open the thread and see that LLY has already posted the correct answer (not so surprisingly).

[runthemover]

thank you for the explanation. sometimes you get stuck into one way of thinking. very much appreciate your help. your explanation was very good I have to say.

[dna4ever]

thank you for the explanation. sometimes you get stuck into one way of thinking. very much appreciate your help. your explanation was very good I have to say.

any time

[runthemover]

any time

you probably should give me your cell phone # just in case. well you know what, just in case give me your credit card # so I can fly out to ask you in person.

[LongLiveYorke]

thank you for the explanation. sometimes you get stuck into one way of thinking. very much appreciate your help. your explanation was very good I have to say.

Using the above way of computing it, it is very easy to generalize the problem to many rolls (more than 7). One needs to only sum the different ways that one can get repeats (ie a triple, two doubles, etc).

[tobytobey]

You made your mistake when you considered (or forgot to consider) the different possabilities for what the 7th one could be and where it could go. In other words, with 7 die, we have to get one repeat. Let's work from that fact:

There are seven slots:

_ _ _ _ _ _ _

We're going to get exactly one repeated if we're going to have all 6 faces show up in 7 rolls. So, the name of the game is to count how many ways we can get 6 faces out of 7 rolls and divide that by the total amount of combinations (ie 6^7).

There are 6 ways that we can get a double (11,22,33,44,55,66). These have to go somewhere. There are 7 Choose 2 ways that we can arrange the doubles on the board. Here are examples:

6 _ _ _ 6 _ _

_ _ _ 5 5 _ _

_ _ 2 _ _ _ 2
etc.

So, starting from this, we get 6 *(7C2) ways we can make the above formations. We now need to count the ways to fill in the rest. But that's easy now. It's just:

5*4*3*2*1 ways to fill in the remaining 5 slots. So, our total is:

6*7C2*5*4*3*2*1 / 6^7

=.054012

I really hope this is right. It seems to somewhat match the answer you got. The problem with combinatorics is that you can really be convinced that you're right, you can be so sure of it and have everything make so much sense, and then totally get the wrong answer. Let's hope I didn't just do that.

Cheers.

Why didn't I choose the blue pill?

By the way, that sound you just heard was my head exploding.

[_Great_Dane_]

SOAB! that one gets me everytime

One would think that he'd eventually get it right.

Why didn't I choose the blue pill?

By the way, that sound you just heard was my head exploding.