Question

In game 77, a 7-digit winning number consisting of the digits 0 to 9 is drawn one after the other (they put the numbers back in the bag before drawing another number) and their digits are arranged in the order of the draw. What is the probability that the numbers drawn in non-descending sequence, i.e. the number of a drawing is at least as large as the number of the previous drawing? Also calculate this probability for the general case of a n-digit number.

Answer 1

We need to choose the number of non-descending sequences possible. Now note that given any 7 numbers there is only one permutation of those numbers which will be non-descending. For example if the seven numbers are: 5,7,9,3,3,6,1 then 1-3-3-5-6-7-9 is the only non-descending permutation of these numbers.

So number of non-descending sequences possible = number of ways of choosing 7 numbers when repetition is allowed.

This can be done as follows: Since replacement is done so at each step we can draw any number from 0 to 9. Consider the urn to contain 7 0's ; 7 1's and so on till 7 9's then we have a total of 7 x 10 = 70 numbers in the urn and we have to choose 7 of them which can be done in 70C7 ways. Note that we put 7 copies of each number in the urn because we want to consider the number of ways of selecting a 7 digit sequence and each digit of the sequence can be any number from 0 to 9 (including repetitions) If we just do 10C7 then those repetions are not considered as it will not take into account drawing the numbers 1,1,1,1,1,1,1 or any other sequence with two or more identical numbers.

The total number of sequences possible is 10^7 as for each place in the sequence we have 10 options.

Hence the required probability is : 70C7 / 10^7

In general if it is a n-digit sequence we can do the same thing and put n copies of each number in the urn taking the tital objects in the urn to be 10n and select n from these in (10n)Cn ways

Hence required probabilty = (10n)Cn / 10^n

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