Quick Probability Question

[SlyR]

Can someone help settle a bet we have here at work?
A cell phone number contains eleven digits.
One recent ESN came up as 05101555151.

Excluding the exact order and the fact that some digits recur more than others, what is the P that an ESN will contain only three specific digits?

My very unqualified opinion is that the solution would have something to do with (10!)/(3!). Is that correct?

Please respond to xscott.raymond@xverizonwireless.com (remove x'es).

[MrBlini]

If we know the three specific digit values, the probability that an 11-digit phone number will contain only these three digit values is:

P = (3/10)^11, approximately 0.00000177,

simply the product of the probability that each of eleven digits will be one of the three specified values.

Note that this does include combinations that utilize only one or two of the three digit values.

If the item of interest is that an 11-digit phone number will contain only three digit values or fewer, but these can be ANY three digit values, we need to multiply the probability corresponding to any specific 3 digit
values by the number of such combinations (http://mathworld.wolfram.com/Combination.html) of digit values. That does involve factorials:

P = (3/10)^11 * 10!/(7!3!) , approximately 0.000213

[M.B.E.]

[ QUOTE ]
If the item of interest is that an 11-digit phone number will contain only three digit values or fewer, but these can be ANY three digit values, we need to multiply the probability corresponding to any specific 3 digit
values by the number of such combinations (http://mathworld.wolfram.com/Combination.html) of digit values. That does involve factorials:

P = (3/10)^11 * 10!/(7!3!) , approximately 0.000213

[/ QUOTE ]
Actually that's incorrect; it's too high because you are counting ESNs like 33433334343 eight times. That ESN would get counted for the three-digit sets {0,3,4}, {1,3,4}, {2,3,4}, {3,4,5}, {3,4,6} etc.

Also, ESNs like 66666666666 are being counted 36 times. It gets counted for the three digit sets {0,1,6}, {0,2,6} etc.

The correct answer is 0.000206128. That is the exact probability that a randomly chosen ESN contains three or fewer distinct digits (assuming all ESNs from 00000000000 through 99999999999 are equally likely).

The probability equals approximately 1/4851.

[M.B.E.]

[ QUOTE ]
(assuming all ESNs from 00000000000 through 99999999999 are equally likely)

[/ QUOTE ]
Incidentally, this assumption is almost certainly wrong. For one thing, I would expect that one of the digits is a check digit, determined by the other ten. Maybe someone could confirm that? For another, I would expect that ESNs are allocated in some kind of sequence.

[M.B.E.]

If you have only the digits 0, 1, and 5 to work with, you can form 177,147 eleven-digit numbers. That is just 3^11. For each of the eleven digits you have three choices (0, 1, or 5), so you just multiply together the eleven 3s.

If you assume that every ESN from 00000000000 to 99999999999 is equally likely, then the probability that an ESN contains no digits other than 0, 1, and 5 is

177,147/(10^11) = 0.00000177147 = 1/564,503

That's equal to (3/10)^11, which is what MrBlini posted.

[MrBlini]

Oops. You're right about the multiple counting.

[bigpooch]

The following gives a count of all 3 number combinations of
11 digit sequences.

10 sequences with exactly 1 digit represented

C(10,2) x (2^11 - 2) = 92070 sequences with exactly 2 digits
represented

C(10,3) x (3^11 - 3(2^11-2) - 3) = 20520720 sequences with
exactly 3 digits represented

the total is 20612800 which verifies the answer given by
M.B.E. in a previous post.