Probability of Duplicates in Randomly Generated Sequences

[SW VandeCarr]

What is the probability of getting a duplicate of a digit in a randomly generated sequence of four digits?

I would approach this intuitively as follows.

(1.0)(0.1)(0.9)(0.9)=0.081
" (0.1)(0.1)(0.9)=0.009
" (0.1)(0.1)(0.1)=0.001
Sum =0.091

There must be a more efficient and elegant way of doing this but I've not been able to find one. Also, is there a more general formula for a set of n distinct objects each with a probability 1/n of being selected?

 

[tiny-tim]

Hi SW VandeCarr!

What is the probability of getting a duplicate of a digit in a randomly generated sequence of four digits?

hmm … what is the probability of not getting a duplicate?

 

[SW VandeCarr]

Hi SW VandeCarr!


hmm … what is the probability of not getting a duplicate?

That's included in my calculation. I believe the calculation is correct. The probability of not getting a duplicate for randomly generated n distinct objects is (n-1/n)^n, but duplicates are included in triples, quadruples,etc.

 

[tiny-tim]

The probability of not getting a duplicate for randomly generated n distinct objects is (1/n)^n,

sorry, completely wrong

start again

 

[SW VandeCarr]

sorry, completely wrong

start again

You're too fast. You missed my edit.

 

[tiny-tim]

You're too fast. You missed my edit.

The probability of not getting a duplicate for randomly generated n distinct objects is (n-1/n)^n

oooh … still completely wrong …

and why is there only one variable (n)?

 

[SW VandeCarr]

oooh … still completely wrong …

and why is there only one variable (n)?

OK. First, is my calculation correct? If so, is there a general formula for this?

You're correct, there is more than one variable. From a set of N distict objects, a random sequence of n objects is generated. The probability of not getting a duplicate in the sequence of n objects is ((N-1)/N)^n.

Now, let r equal all replications (duplicates, three of a kind, four of a kind, etc) each considered to contain just one duplicate, what is the most efficient way to calculate the probability of a duplicate so defined?

 

[tiny-tim]

The probability of not getting a duplicate in the sequence of n objects is ((N-1)/N)^n.

No, that's the probability of not getting any 0s

 

[John Creighto]

Well, if the number of digits is greater then the base then the probability is equal to one.

 

[SW VandeCarr]

No, that's the probability of not getting any 0s

I don't understand that answer. The objects in the general case aren't necessarily numbers. Each random selection yields one object which may be selected more than once out of n selections. Perhaps the probability is ((N-1)/N)^n-1. Do you know?

My reasoning regarding the probability of getting only a duplicate in four randomly generated digits is:

The first selection can be any digit and therefore has a probability of one.

The second selection has a probability of 0.1 of being the same digit.

The third and fourth selection have a probability 0.9 of not being the same digit.

Since multiplication is commutative, the actual order doesn't matter.

If we consider the probability of getting at least a duplicate, then we need to add the probabilities of getting three or four of a kind.

I don't think the answer is as simple as 1-(0.9)^3.

 

[John Creighto]

The first selection can be any digit and therefore has a probability of one.
The second selection has a probability of 0.1 of being the same digit.

So we're dealing with base 10.

The third and fourth selection have a probability 0.9 of not being the same digit.

Since multiplication is commutative, the actual order doesn't matter.

I Think I get the idea.

If we consider the probability of getting at least a duplicate, then we need to add the probabilities of getting three or four of a kind.

I don't think the answer is as simple as 1-(0.9)^3.

Okay. I agree after two digest:

Probability of a repeat 1/B where B is the base

There are now two digest taken. So when we add another diget:

1/(B)+(1/(B-1))*(1-1/B)

As there is 1/B chance of the third digest matching the fist digit, but if the third digit doesn't match the first that leaves (B-1) numbers in the diget left. So the probability of it matching the second digit is 1/(B-1) multiplied by the probability that it didn't match the first digit.

Let P(N) be the probability of a match after N digest then

$$P(N)=P(N-1)+(1-P(N-1))*{1 \over B-N+2}$$

and

$$P(2)=1/B$$

 

[SW VandeCarr]

Actually the argument is counter-intuitive if one is thinking in terms of cumulative probability. For cumulative probability 1-(0.9)^3 is correct. That is, once the marker digit is selected, the cumulative probability that a duplicate will be selected in the next three selections is about 0.271. However, cumulative probability is not the same as sequential probability. The probability that the marker will be matched in any particular one of the next three independent selections is still 0.1 in frequentist theory.

The idea of cumulative probability fits with experience, but an it's entirely different concept than the probability of a single independent event. I don't know the answer to this problem, but it is a problem. Randomized scientific trials use threshold probabilities to determine statistical significance but rarely adjust the significance level for the number of similar trials conducted. If you believe in cumulative probability, scientific trials should adjust these levels or be liable for false positive results.

 

[John Creighto]

Which part of my last post do you disagree with and how does this apply to randomized scientific trials. Is there an area of research were these statistics are particularly important? 1-(0.9)^3 is probably an okay approximation if the base is large relative to the number of digits. Since the base is 10 and the number of digest is four, it might be an okay approximation for this problem.

Your formula clearly fails when the number of digits N=10 since it gives a finite probability of no repeats which is impossible with 10 digits.

 

[tiny-tim]

What is the probability of getting a duplicate of a digit in a randomly generated sequence of four digits?

My reasoning regarding the probability of getting only a duplicate in four randomly generated digits is:

uhh? what is the question … getting exactly one duplicate, or getting one or more duplicates?

i assumed from your original post that it was the latter

 

[SW VandeCarr]

Which part of my last post do you disagree with and how does this apply to randomized scientific trials. Is there an area of research were these statistics are particularly important? 1-(0.9)^3 is probably an okay approximation if the base is large relative to the number of digits. Since the base is 10 and the number of digest is four, it might be an okay approximation for this problem.

Your formula clearly fails when the number of digits N=10 since it gives a finite probability of no repeats which is impossible with 10 digits.

The cumulative probability of getting at least one duplicate or at most one duplicate obviously must increase as the number of randomly generated digits increase. For four digits the questions boil down to how many sequences between 0000 and 9999 have at least one duplicate, and how many have at most one duplicate?

For sequential probability, the question is: given an existing digit sequence of n digits with no duplicates, what is the probability that the next digit selected will be a duplicate of one of the previously generated digits? For a generated sequence that already has just one duplicate, the probability that an expanded sequence (by adding more randomly generated digits) will have just one duplicate decreases.

 

[John Creighto]

The cumulative probability of getting at least one duplicate or at most one duplicate obviously must increase as the number of randomly generated digits increase. For four digits the questions boil down to how many sequences between 0000 and 9999 have at least one duplicate, and how many have at most one duplicate?

I'm not sure it's a good idea to ask more then one question in the same post. Anyway, you still haven't critiqued the steps in my solution above for the probability of getting at least one duplicate.

For sequential probability, the question is: given an existing digit sequence of n digits with no duplicates, what is the probability that the next digit selected will be a duplicate of one of the previously generated digits?

Again, I thought I solved this problem.

For a generated sequence that already has just one duplicate, the probability that an expanded sequence (by adding more randomly generated digits) will have just one duplicate decreases.

I'll give this problem a shot later. I think though we should do it in a separate thread, and I think the moderate should make the title of this thread more clear.

 

[SW VandeCarr]

I'm not sure it's a good idea to ask more then one question in the same post. Anyway, you still haven't critiqued the steps in my solution above for the probability of getting at least one duplicate.

You're using different notation than I am. For digits B=N=10 and n is the length of the randomly generated (rg) string. Your formula seems correct for n=2. For n=10 with no duplicates, its clear that the next rg number is a duplicate with p=1.

Again, I thought I solved this problem.

If your N=n and your B=10 then 1/B-N+2 = 0.5 for N=n=10. I haven't done the full expansion, but the 11th rd digit must be a duplicate (assuming no duplicates in the string n=10) as you have said.

I'll give this problem a shot later. I think though we should do it in a separate thread, and I think the moderate should make the title of this thread more clear.

For cumulative probability there are two related questions. I'm including cumulative probability because tiny tim raised the issue. One is the number of 4 digit sequences that contain at least one identical pair (1213, 3333, 4433, 0051, 0110, ect.) The other is the number of 4 digit sequences that contain at most one identical pair excluding triplets and quadruplets.

The title is "probabilities of duplicates" and I'm not giving a tutorial. I'm asking questions because I don't know the answers.

 

[SW VandeCarr]

Hi SW VandeCarr!


hmm … what is the probability of not getting a duplicate?

Both John Creighto and I have shown you're wrong if the question is at least one duplicate in n trials with a finite base B as for digits 0-9. I suggest the correct formula is quite simple based on the uniform distribution:

P(k(i)=(k(j)) = (n-1)/B where n is the number of trials and k is the kth trial within n and (n-1) less than or equal to B.
(
For for the first trial P=0 and for the eleventh trial P=1 as it should..

For at most one duplicate in n trials, I suggest my first post is correct.

Any comments tiny tim or others?