Dice probability (throw 10 times; find probability I will pick all numbers)

[misgi]

Hello,
I am throwing a dice 10 times.
What is the probability I will pick all numbers?
To be more clear:
After I am throwing a dice 10 times, I will get sequence of 10 numbers.
Each number on this sequence can be 1 up to 6 (6 options).
what is the probability I will see all 6 numbers (1,2,3,4,5,6 )in the sequence?
Thank you

 

[BigBeachBanana]

what is the probability I will see all 6 numbers (1,2,3,4,5,6 )in the sequence?

The question needs to be clarified further. When you say "(1,2,3,4,5,6) in the sequence".

1) Does it have to occur in the exact order or do you want only at least one of the digits to show up once and order not matter?
2) If yes to the former of 1) then can it be something like this (1, 1, 1, 2, 2, 3,4 5, 6, 6 ) or (1,2,3,4,5,6, *,*,*,*) where * can be any digits.

 

[blamocur]

The question needs to be clarified further. When you say "(1,2,3,4,5,6) in the sequence".

1) Do it have to occur in the exact order or do you want only at least one of the digits to show up once and order not matter?
2) If yes to the former of 1) then can it be something like this (1, 1, 1, 2, 2, 3,4 5, 6, 6 ) or (1,2,3,4,5,6, *,*,*,*) where * can be any digits.

Or it can be interpreted that each digit must be seen at least one, with "in the sequence" referring to the sequence of 10 random digits.
Which one is it?

 

[Steven G]

This seems difficult for me at the moment.
I guess the problem reduces to the following:
If I list a string of 10 numbers from the set {1,2,3,4,5,6}, then what is the probability that I use each of the 6 numbers in the set.
I see one way of doing this but it too tedious.
I guess that one can list all the possible 10 digit combinations from smallest to largest and see where that takes you.

1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 2
1 1 1 1 1 1 1 1 1 3
...
1 1 1 1 1 1 1 1 1 6
1 1 1 1 1 1 1 1 2 1
1 1 1 1 1 1 1 1 2 2
...
1 1 1 1 1 1 1 1 2 6
...
1 1 1 1 1 1 1 1 6 6
1 1 1 1 1 1 1 2 1 1
...
1 6 6 6 6 6 6 6 6 6
2 1 1 1 1 1 1 1 1 1
...
6 6 6 6 6 6 6 6 6 6.

The point to the OP is that you have to try something.

Ah, I see the solution now.

 

[pka]

I am throwing a dice 10 times.
What is the probability I will pick all numbers?

Conceptually this is a difficulty question. That is, it is not a regular counting question.
It belongs to to a class known as Occupancy problems.
We ask about the number of ways to place [imath]\bf{n}[/imath] objects into [imath]\bf{k}[/imath] cells.
The objects can be distinguishable or indistinguishable; cells can be distinguishable or indistinguishable
If objects are indistinguishable & the cells are distinguishable
with possible empty cells there are [imath]\dbinom{n+k-1}{n}[/imath] ways to do that.

If objects are indistinguishable & the cells are distinguishable
with no possible empty cells there are [imath]\dbinom{n-1}{k-1}[/imath] ways to do that.

If one on has excess to a mathematics library Discrete Mathematics by J.A. Anderson has a a good discussion of this.
See here
[imath][/imath][imath][/imath]

 

[misgi]

To be more clear,
All numbers must be seen at least one, no matter what is the order.
I got an answer on Youtube for this if someone interested:
"you can use a recursion: Let L(n, k) be the probability that you have got exactly k different numbers after n throws.
Then L(n, k) = (k/6)*L(n-1,k) + ( (7-k)/6)*L(n-1,k-1) because after n-1 throws you must either have already k different numbers, in this case you must throw one of this k numbers once again, or you have exactly k-1 different numbers, in this case you have to throw one of the remaining 6 - (k-1) = 7 - k numbers. Initial values are
L(n, 0) = 0
L(0, k) = 0
L(1,1) = 1
L(n,k)=0 when k>n
"

 

[Steven G]

In how many ways can you choose 6 slots out of 10 to place the numbers 1,2,3,4,5 and 6? Once you choose the 6 slots, in how many ways can you place the numbers 1,2,3,4,5 and 6 into those slots? How many ways can you place numbers in the 4 remaining slots?
Next, in how many ways can you place the numbers in the 10 slots w/o any restrictions?

Hmm, this yields a result greater than 1. Where did I go wrong?

(10 C 6)(6!)(6^4)/6^10>1

 

[pka]

In how many ways can you choose 6 slots out of 10 to place the numbers 1,2,3,4,5 and 6? Once you choose the 6 slots, in how many ways can you place the numbers 1,2,3,4,5 and 6 into those slots? How many ways can you place numbers in the 4 remaining slots? Next, in how many ways can you place the numbers in the 10 slots w/o any restrictions? this yields a result greater than 1. Where did I go wrong?

You are using the wrong model. Suppose there there are six boxes numbered [imath]1\text{ to }6[/imath].
We have ten identical white balls and one die. We toss the die ten times to get ten outcomes: [imath]1,~2,~3,~4,~5\text{ or }6[/imath].
After each toss of the die we put one ball into the box with the number corresponding to the number on the die.
Reread reply #5. After ten tosses the boxes can be filled in [imath]\dbinom{10+6-1}{10}[/imath] ways with some boxes may be empty.
With no boxes left empty that can be done in [imath]\dbinom{10-1}{5-1}[/imath] ways. SEE HERE

 

[Dr.Peterson]

You are using the wrong model. Suppose there there are six boxes numbered [imath]1\text{ to }6[/imath].
We have ten identical white balls and one die. We toss the die ten times to get ten outcomes: [imath]1,~2,~3,~4,~5\text{ or }6[/imath].
After each toss of the die we put one ball into the box with the number corresponding to the number on the die.
Reread reply #5. After ten tosses the boxes can be filled in [imath]\dbinom{10+6-1}{10}[/imath] ways with some boxes may be empty.
With no boxes left empty that can be done in [imath]\dbinom{10-1}{5-1}[/imath] ways. SEE HERE

My difficulty with this is that the problem is not just about counting, but about probability. The results you are counting aren't equally likely, are they? For that, I think we'd want the balls to be numbered (first die, second die, ...).

 

[pka]

My difficulty with this is that the problem is not just about counting, but about probability. The results you are counting aren't equally likely, are they? For that, I think we'd want the balls to be numbered (first die, second die, ...).

Please tell me why you think, tossing a die ten times, the outcomes are not equally likely.
Tossing a die ten times [imath]6,1,1,3,6,5,6,6,1,1[/imath] is equally likely as [imath]6,1,2,3,6,5,4,6,1,1[/imath].
But the first does not contain all possible outcomes but the second does.
Order has nothing to do with this question. Among the ten outcomes does each of the six numbers appear at least once.
Again, one die is tossed ten times. The probability: the number of ways in which each of the six numbers shows at least once
divided by the number of ways we can randomly get ten of [imath]1,2,3,4,5,6[/imath]
How many ways can we put ten identical balls into six distinct cells.
If done on a random (but equality probable) way, what is the probability no cell will be empty?

 

[Dr.Peterson]

Please tell me why you think, tossing a die ten times, the outcomes are not equally likely.
Tossing a die ten times [imath]6,1,1,3,6,5,6,6,1,1[/imath] is equally likely as [imath]6,1,2,3,6,5,4,6,1,1[/imath].
But the first does not contain all possible outcomes but the second does.
Order has nothing to do with this question. Among the ten outcomes does each of the six numbers appear at least once.
Again, one die is tossed ten times. The probability: the number of ways in which each of the six numbers shows at least once
divided by the number of ways we can randomly get ten of [imath]1,2,3,4,5,6[/imath]
How many ways can we put ten identical balls into six distinct cells.
If done on a random (but equality probable) way, what is the probability no cell will be empty?

The problem is that there can be more ways to obtain one outcome (arrangement of balls in boxes) by rolling dice, than to obtain another.

Take a small example: flip a coin (2-sided "die") 4 times; what is the probability that you get all possible "numbers" (both H and T) among the four?

Your calculation, using [imath]n=4[/imath] and [imath]k=2[/imath], gives [math]\frac{\dbinom{n-1}{k-1}}{\dbinom{n+k-1}{n}}=\frac{\dbinom{4-1}{2-1}}{\dbinom{4+2-1}{4}}=\frac{\dbinom{3}{1}}{\dbinom{5}{4}}=\frac{3}{5}[/math]
In reality, there are [imath]2^4=16[/imath] outcomes from the 4 flips, of which only 2 are all heads or all tails, so the probability of getting both H and T is

[math]\frac{\text{some H, some T}}{\text{all sets of 4 tosses}}=\frac{14}{16}=\frac{7}{8},[/math]
which is different from your result, unless I've badly misinterpreted something.

How did this happen? There are clearly 5 outcomes from putting the balls in the bins, namely (H,T) = (4,0), (3,1), (2,2), (1,3), and (0,4). But there is only one way to get (4,0), namely HHHH, whereas there are 5 ways to get (3,1), namely (HHHT, HHTH, HTHH, and THHH. And so on, with each outcome having its own probability. They are not equally likely.

I am very cautious when modeling an event, to make sure that the process I use in counting behaves the same as the process specified in the problem. In the process of tossing coins to decide where balls should go, the dice (or rolls) are distinguishable. In counting only how many balls (rather than which specific rolls) are in each bin, you have flattened out the process. In effect (though not in egregiousness!) what you are doing is the equivalent of saying the probability of tossing heads is 1/3, because there are 3 possible outcomes (heads, tails, and edge).