# Probablity a randomised four digit number does not have two specific consecutive numbers

I am trying to work out the probability a four digit number does not have two consecutive numbers, for example two consecutive 5's, not starting with a 0 is assumed.

Now I could work out how many numbers in this range contain two consecutive numbers manually, but that seems like a terrible method and impractical if I get it on an exam.

My attempt:

$9×10^3=$ total ways to arrange a four-digit number

$10^2=$ ways to arrange a four-digit number starting with two consecutive numbers(55) (e.g. 5545)

$2=$ ways to arrange two consecutive numbers along the last 3 digits.

$9*10*2=$ ways to arrange a four-digit number with two consecutive 5's placed in the last three digits (e.g. 6550, 6155)

$$(9*10^3) - (10^2) - (9*10*2)$$ $$=8720$$ So I have a probability of $$8720/9000$$ or $$.96888\text{%}$$

Is this correct or have I made mistakes?

You say you have $9000$ four-digit numbers.

To count numbers with no consecutive repeat digits is quite easy: you say you have $9$ choices for the first digit; given the first you have $10-1=9$ choices for the second; given the second you have $9$ choices for the third; given the third you have $9$ choices for the fourth.

$$\dfrac{9^4}{9 \times 10^3} = 0.729$$ so if this is what you were trying to do then you have made a mistake.

(Added) If alternatively you are looking for the numbers which do not have any consecutive $5$s, the easiest way to count is to look at those which do, as you attempted.

There are three possible patterns of the forms 55AA, B55A or CD55 where A is any digit , B is any digit except $0$ or $5$, C is any digit except $0$, and D is any digit except $5$. So there are $10\times 10 + 8 \times 10 + 9\times 9 = 261$ four-digit numbers which have consecutive $5$s and so $9000-261 = 8739$ which do not.

$$\dfrac{8739}{9000} = 0.971$$ so again you have an error, but much closer this time. You might try to spot how you have over counted the second two patterns

• Is this for a specific number? I mean for a consecutive number of specific choice, probability that there isn't two 7's in a row, but 6's and 5's etc are all fine, or two 5's but the rest are fine. – Display Name Oct 24 '13 at 9:15
• @DisplayName: I have given an alternative answer – Henry Oct 24 '13 at 21:10
• Hello Henry, sorry for not being more clear initially, and your alternative answer definitely led me to the understanding I was looking for. Great work! – Display Name Oct 24 '13 at 23:19

The consecutive two digits can take three places, the first two, the middle two, and the last two.

The two consecutive numbers could range from 11 to 99 and they could take the first, the middle and the last. Let us analyze the first two with this example. 99_ _ The third digit could be from (0-8) giving us 9, the fourth digit could be from 0-9. Thus the total number is 9*9*10.

Now, let us analyze the middle two _ 99 _, The first digit could be anything from 1-9(giving us 9 choices), the last digit could be anything from 0-8 ( giving us 9 choices) Thus the total number is 9*9*9 and lastly the last two _ _ 99, the second digit could be anything from 0-8 (9 choices), given that the first digit could be any from 1-9 (9 choices). Thus the total number is 9*9*9.

Coming to 00, for the pattern _ 00 _, the choices for the first digit is 1-9 ( 9 choices), the last digit should be 0-9 choices, Thus 9*10 = 90. For the pattern __00, first digit could be anything from 1-9 (9 choices), the second digit could be anything from 1-9 = 9*9 = 81.

Adding up all , 9*9*10 + 9*9*9 + 9*9*9 + 90+81 = 2439 = (9000 - 2439)/9000 = 0.729

I ran a check on the computer and made sure the answer is right!