Probability of 10th Number Being 3rd Unique Number in a Die Throw

[blondii]

Homework Statement

Consider a die. What is the probability that the 10th number that comes up is the 3rd unique number of the entire throw.

Eg: a throw that could be considered a success is the throw which is represented by the set
{1,2,1,1,1,1,1,1,1,3} Where the 3 unique numbers would be: {1,2,3}

Let
N = Number of attempts (in this case number of throws)
K = Number of items to select from (in this case 6 because its a die).
S = Number of similar or repetitive items (In the eg case the repetitive number is 1).
R = Number of unique items to choose from (In the eg case it is 3, i.e {1,2,3})
NcS = N combination S (Ways of selecting S things from N different things)

Homework Equations

The equations below I just quickly thought up. Don't know if they hold for this case or not but please do correct me.

S = N - R + 1

The Attempt at a Solution

Note: Some formulas below I just quickly thought up. Don't know if I am correct or not but please do let me know otherwise.

I have included explanations of how i am using the formulas in my answers. But the actual numerical answer is at the bottom so please scroll down if you just want to view and confirm if my answer is correct or not.

Number of ways of selecting a sample of N items which contains R unique items:

(NcS)x(K-1)xK
--------------------------------
Explanation for my formula:
NcS - The number of ways of selecting S (repetitive) and N-S (unique) combinations from N things

(NcS)x(K-1) - From the NcS different arrangements a repetitive item can be arranged, there are (K-1) things that are unique. Thus (NcS)x(K-1) gives the total number of ways a particular repetitive item can be arranged with the different (K-1) item combinations.

(NcS)x(K-1)xK - From the (NcS)x(K-1) different ways (which holds for a single case), there is a possibility of K different combinations of items that can be repetitive. Thus (NcS)x(K-1)xK takes into account the K different items.
------------------------------------------

Number of ways of selecting a sample where the Nth item is the R'th Unique item:
-------------------------------------
Variable Declaration
S = (N - 1) - (R - 1) + 1
-------------------------------------
thus
((N - 1)cS)x(K-1)x(K)x(K+1-R)

-------------------------------------
Explanation of Variable S and Formula
S = (N - 1) - (R - 1) + 1 - Number of repetitive items in the (N-1) item list. (N-1) because the list excludes the N'th item.

((N - 1)cS)x(K-1)x(k)x(K+1-R) - Because the N'th item selected is the R'th unique item, we multiply ((N - 1)cS)x(K-1)x(k) by the remaining (K+1-R) different possible unique items that are left to be selected as the N'th item (thus added to the end of the already selected (N-1) items.
-------------------------------------

Solutions
N = 10
K = 6
R = 3
S = (N - 1) - (R - 1) + 1 = 7
((N - 1)cS) = 9c7 = 36
K^N = 6^10 = 60466176
((N - 1)cS)x(K-1)x(K)x(K+1-R) = (36)(5)(6)(4) = 4320

Therefore the probability that the 10th number that comes up is the 3rd unique number of the entire throw is:

((N - 1)cS)x(K-1)x(K)x(K+1-R) / K^N = 4320 / 60466176 = 7.144 x 10^-5

Please can someone confirm this answer or tell me if I'm going in the right direction. I would appreciate any suggestions to another possible solutions if mine is not correct.

Cheers

 

[PAllen]

The answer is not correct. I will pose a few questions to think about to get you on the right track:

1) For the first 9 throws, how many distinct die choices are allowed?

2) What is the probability that 9 throws are drawn from a set of size <answer to (1)>?

3) Can some of these throws cause the overall throw to fail to meet the requirements of the problem? How would you account for this?

4) How many ways are there to choose sets of die numbers meeting (1)?

 

[blondii]

Thanks PAllen. I will attempt again at the solution and try repost. I think I can see where I went wrong. Thanks also for the suggestions.

 

[Ray Vickson]

Homework Statement

Consider a die. What is the probability that the 10th number that comes up is the 3rd unique number of the entire throw.

Eg: a throw that could be considered a success is the throw which is represented by the set
{1,2,1,1,1,1,1,1,1,3} Where the 3 unique numbers would be: {1,2,3}

Let
N = Number of attempts (in this case number of throws)
K = Number of items to select from (in this case 6 because its a die).
S = Number of similar or repetitive items (In the eg case the repetitive number is 1).
R = Number of unique items to choose from (In the eg case it is 3, i.e {1,2,3})
NcS = N combination S (Ways of selecting S things from N different things)

Homework Equations

The equations below I just quickly thought up. Don't know if they hold for this case or not but please do correct me.

S = N - R + 1

The Attempt at a Solution

Note: Some formulas below I just quickly thought up. Don't know if I am correct or not but please do let me know otherwise.

I have included explanations of how i am using the formulas in my answers. But the actual numerical answer is at the bottom so please scroll down if you just want to view and confirm if my answer is correct or not.

Number of ways of selecting a sample of N items which contains R unique items:

(NcS)x(K-1)xK
--------------------------------
Explanation for my formula:
NcS - The number of ways of selecting S (repetitive) and N-S (unique) combinations from N things

(NcS)x(K-1) - From the NcS different arrangements a repetitive item can be arranged, there are (K-1) things that are unique. Thus (NcS)x(K-1) gives the total number of ways a particular repetitive item can be arranged with the different (K-1) item combinations.

(NcS)x(K-1)xK - From the (NcS)x(K-1) different ways (which holds for a single case), there is a possibility of K different combinations of items that can be repetitive. Thus (NcS)x(K-1)xK takes into account the K different items.
------------------------------------------

Number of ways of selecting a sample where the Nth item is the R'th Unique item:
-------------------------------------
Variable Declaration
S = (N - 1) - (R - 1) + 1
-------------------------------------
thus
((N - 1)cS)x(K-1)x(K)x(K+1-R)

-------------------------------------
Explanation of Variable S and Formula
S = (N - 1) - (R - 1) + 1 - Number of repetitive items in the (N-1) item list. (N-1) because the list excludes the N'th item.

((N - 1)cS)x(K-1)x(k)x(K+1-R) - Because the N'th item selected is the R'th unique item, we multiply ((N - 1)cS)x(K-1)x(k) by the remaining (K+1-R) different possible unique items that are left to be selected as the N'th item (thus added to the end of the already selected (N-1) items.
-------------------------------------

Solutions
N = 10
K = 6
R = 3
S = (N - 1) - (R - 1) + 1 = 7
((N - 1)cS) = 9c7 = 36
K^N = 6^10 = 60466176
((N - 1)cS)x(K-1)x(K)x(K+1-R) = (36)(5)(6)(4) = 4320

Therefore the probability that the 10th number that comes up is the 3rd unique number of the entire throw is:

((N - 1)cS)x(K-1)x(K)x(K+1-R) / K^N = 4320 / 60466176 = 7.144 x 10^-5

Please can someone confirm this answer or tell me if I'm going in the right direction. I would appreciate any suggestions to another possible solutions if mine is not correct.

Cheers

The probability that the third unique number comes on toss n is related to the probability that the sum of two independent but non-identically-distributed geometric random variables is equal to n-1. Can you see why?

RGV

 

[blondii]

Ok this is just to improve on my answer above. I have been pulling my hair out for almost a day now so I hope I'm close to the answer this time.

For the first 9 throws, the acceptable sets would be ones that contain 2 unique numbers with their repetitions.

Thus the set {1,2,2,2,1,2,2,1,1} or {1,3,3,3,1,1,1,3,3} would be an acceptable set where the unique numbers are {1,2} and {1,3} respectively.

Now the different combination of times the two numbers can be repetitive can be represented by:

Say L{X,Y} would mean the first number has X repetitions (if any) and the the second number has Y repetitions (if any).
Where X+Y = 9.
Each of these combinations of repetitions can be arranged in 9cX = 9cY.

In the case of this problem, the possible L{X,Y}'s are:
L{1,8}, L{2,7}, L{3,6}, L{4,5}

Eg: L{1,8} stands for the set where there are 8 repetitions of one number and 1 occurrence of the other number. Eg an acceptable subset could be the set {1,2,2,2,2,2,2,2,2} or {2,2,2,2,4,2,2,2,2} etc.

Ways of arranging:
L{1,8} = 9c8 = 9
L{2,7} = 9c7 = 36
L{3,6} = 9c6 = 84
L{4,5} = 9c5 = 126

Total number of combinations with repetitions with two unique numbers is:
Let N = number of unique numbers = 2.

Total of permutations = N(9 + 36 + 84 + 126) = 510.
This answer holds for only one pair of combination of numbers.

To take into account all the possible pair combinations:
510*(6c2) = 7650.

Since K (as mentioned earlier) represents the number of items to choose from = 6.

Thus at anyone time when there is a pair of numbers that make up a set, there are K-N remaining numbers which can be the possible 3rd unique number from the 10th throw.

Hence 7650x(K-N) = 7650x4 = 30600.

Thus the probability that the 10th thrown die is the 3rd unique number is:
30600 / K^N = 30600 / 60466176 = 5.0607 x 10^-4

Can someone please confirm if I am on the right track or if this answer is correct. Any suggestion and comments will be much appreciated.

Thanks.

 

[PAllen]

Your answer is correct. However, there was a much easier way to arrive at it based on the hints I was giving.

For a choice of two numbers for the first nine throws, you have:

(1/3)^9

probability of 9 throws from two given die numbers. But these include the possibility of throwing all one number, or all the other number. You can correct for these simply:

(1/3)^9 - 2 * (1/6)^9

Then, for the last throw, any of 4 numbers are good:


(2/3)*[(1/3)^9 - 2 * (1/6)^9]

Finally, there are 15 possible choices for the sets of two numbers drawn from 6, so the final expression is simply:

15*(2/3)*[(1/3)^9 - 2 * (1/6)^9]

which you can verify matches your far more laborious computation.

 

[blondii]

Thanks PAllen. I think your method is far better than mine. Your suggestions earlier did actually help me along of which I admit my attempts to arrive at the same solution were quite long.

Thanks again for your help and time.

Cheers.

 

[Joffan]

Late to the party...

I ran a little spreadsheet calc to see what the numbers looked like (n different numbers after m throws). More like engineering than mathematics, I know. Depending on the interpretation of the question, the answer came out differently; if the 10th throw is the first occurrence of the third number, I get 0.000506 (your answer above) or if it is simply the third coming up (perhaps again, but only three different numbers have been thrown) it's about 0.00651.

On the other hand, it strikes me that the stress could be on "unique" - it could mean that, over the throw of ten dice, some numbers are repeats, some numbers are absent, and three are unique, one of which is the tenth to be thrown. But that is significantly harder.

 

[Ray Vickson]

The probability that the 3rd unique occurs on toss (n+1) is P(X1 + X2 = n), where X1 = number of tosses to get a number different from the first one, and X2 = number of tosses to get a number different from the first two. X1 is a geometric RV with success probability p1 = 5/6, while X2 is geom with p2 = 4/6. X1 and X2 are independent.

RGV

 

[PAllen]

The probability that the 3rd unique occurs on toss (n+1) is P(X1 + X2 = n), where X1 = number of tosses to get a number different from the first one, and X2 = number of tosses to get a number different from the first two. X1 is a geometric RV with success probability p1 = 5/6, while X2 is geom with p2 = 4/6. X1 and X2 are independent.

RGV

There is no indication the OP knows about random variables. Further, I don't see how this saves any computation over my method. Or do you think my method is mistaken?

 

[PAllen]

Random variables agrees with my answer, but is much, much more computation.

[edit: at least as far as I know how to use RGV's approach, it is about 10 times the computation to get the same answer].

 

[blondii]

Late to the party...

I ran a little spreadsheet calc to see what the numbers looked like (n different numbers after m throws). More like engineering than mathematics, I know. Depending on the interpretation of the question, the answer came out differently; if the 10th throw is the first occurrence of the third number, I get 0.000506 (your answer above) or if it is simply the third coming up (perhaps again, but only three different numbers have been thrown) it's about 0.00651.

On the other hand, it strikes me that the stress could be on "unique" - it could mean that, over the throw of ten dice, some numbers are repeats, some numbers are absent, and three are unique, one of which is the tenth to be thrown. But that is significantly harder.

I think depending on the different interpretations we could have:

1) Probability that AFTER 10 throws, you will have EXACTLY 3 unique numbers
In this case there is exactly 3 unique numbers (with their repetitions) in the acceptable set whether it is the 10th thrown or not.

P(X=3) ≈ 0.0135

2) Probability that AFTER 10 throws, you have AT LEAST 3 unique numbers
Which means an acceptable set could have 3-6 unique numbers (with their repetitions) after 10 throws.

P(X≥3) ≈ 0.32052

Those are just my rough probability calculations I just scribbled down.

 

[blondii]

The probability that the 3rd unique occurs on toss (n+1) is P(X1 + X2 = n), where X1 = number of tosses to get a number different from the first one, and X2 = number of tosses to get a number different from the first two. X1 is a geometric RV with success probability p1 = 5/6, while X2 is geom with p2 = 4/6. X1 and X2 are independent.

RGV

Ray can you post a solution using your method above. I would very much like to compare with PAllen's solution to help me see the different (and simpler) ways of tackling this problem compared to mine.

 

[PAllen]

The way I know to use Ray's method is to compute a probability for each possible value of X + Y =9, X >=1, Y >=1. For example, P(X=3,Y=6) becomes:

(1/6)^2 * 5/6 * (1/3)^5 * (2/3)

There are 8 such terms. If you do the computation you get exactly the same answer as my method; you've just done a lot more work to get there.

 

[blondii]

Thanks PAllen, i think from your explanation I can now see how Ray would have arrived at the solution. Much appreciated.

 

[Ray Vickson]

Ray can you post a solution using your method above. I would very much like to compare with PAllen's solution to help me see the different (and simpler) ways of tackling this problem compared to mine.

X1 ~ geom(5/6) has distrib. P(X1=k) = (5/6)*(1/6)^(k-1), while X2 ~ geom(4/6) has distrib. P(X2=k) = (4/6)*(2/6)^(k-1) for k = 1,2,... . We have P{X1 + X2 = n} = sum(p1*q1^(k-1)*p2*q2^(n-k-1): k=1..n-1) = p1*p2*[q1^(n-1)-q2^(n-1)]/(q1-q2). Substitute n = 9, p1 = 5/6, q1 = 1/6, p2 = 4/6, q2 = 2/6.

RGV

 

[PAllen]

X1 ~ geom(5/6) has distrib. P(X1=k) = (5/6)*(1/6)^(k-1), while X2 ~ geom(4/6) has distrib. P(X2=k) = (4/6)*(2/6)^(k-1) for k = 1,2,... . We have P{X1 + X2 = n} = sum(p1*q1^(k-1)*p2*q2^(n-k-1): k=1..n-1) = p1*p2*[q1^(n-1)-q2^(n-1)]/(q1-q2). Substitute n = 9, p1 = 5/6, q1 = 1/6, p2 = 4/6, q2 = 2/6.

RGV

Nice, and with that simplification it breaks down to essentially the identical computation as my method.

 

[Ray Vickson]

There is no indication the OP knows about random variables. Further, I don't see how this saves any computation over my method. Or do you think my method is mistaken?

If the OP knows about the geometric distribution, and about sums of independent random variables, then he/she may find the random variable method easier (or at least, more direct); if not, he/she can just ignore it. As for easier: I get the same formula as you, just using simple high-school summations; it does not involve 10 times as much computation! However, the random variable method generalizes quite easily, so one can ask for the probability that the 5th unique number occurs on the 49th toss, etc., using essentially just one common approach (although the computations get a bit more involved). Whether or not that is desirable is a matter of personal choice.

RGV

 

[PAllen]

If the OP knows about the geometric distribution, and about sums of independent random variables, then he/she may find the random variable method easier (or at least, more direct); if not, he/she can just ignore it. As for easier: I get the same formula as you, just using simple high-school summations; it does not involve 10 times as much computation! However, the random variable method generalizes quite easily, so one can ask for the probability that the 5th unique number occurs on the 49th toss, etc., using essentially just one common approach (although the computations get a bit more involved). Whether or not that is desirable is a matter of personal choice.

RGV

I'm embarrassed I missed the geometric simplification, but would also point out that for this class of problems, my approach is general and leads directly to a computationally efficient formula. It very much is a matter of taste. For other types of problems, I might only consider random variables, whereas I, personally, would not bother for problems like this one. For example, your 5th unique number on 49th toss adds minimal extra work:

1) Given a throw that after 48 tosses has exactly 4 distinct values (for which I call P(N=4) as the probability of such a throw for 4 chosen die numbers), then there is 1/3 chance that the 49th throw makes if of interest. There are 15 possible sets of 4. Thus, I have for the overall probability:

15*(1/3)*P(N=4)

2) The method I gave for arriving at something like P(N=4) is general and useful to know: alternately combining simple cases to exclude what you don't want. So I directly write (with experience, and knowing binomial coefficients):

P(N=4) = P(N<=4) - 4 P(N<=3) + 6 P(N<=2) - 4 P(N=1)

3) Each of these last is trivial: (2/3)^48, (1/2)^48, (1/3)^48, (1/6)^48

I don't even have to think about summing additive partitions of 48.