Originally Posted by

**Archie Meade** Ok,

first examine what the possibilities are...

the first two cards can be

1 and 2

1 and 3

1 and 4

2 and 3

2 and 4

3 and 4

This rules out four of these pairings since the highest number card is 4

The first two cards can only be 1 and 2 or 1 and 3.

If the first two cards are 1 and 2 then the 3rd card must be 3.

If the first two cards are 1 and 3 then the third card must be 4.

Now that the problem is clear, you can use arrangements or selections to calculate the probabilities.

1,2,3 has a $\displaystyle \frac{1}{4}\ \frac{1}{3}\ \frac{1}{2}$ probability

because there is a 1 in 4 chance of picking 1 first,

a 1 in 3 chance of picking 2 second

and a 1 in 2 chance of picking 4 third.

Also, the 2 could come before 1, but the probability is the same.

So the probability of a 1 and 2 followed by 3 is $\displaystyle 2\frac{1}{4}\ \frac{1}{3}\ \frac{1}{2}=\frac{1}{12}$

Working out the probability of 1,3,4 or 3,1,4 is the same procedure.

Adding up the probabilities, you get 1/12+1/12=2/12=1/6

You can list the options also

1,2,3

1,2,4

1,3,2

1,3,4

1,4,2

1,4,3

2,1,3

2,1,4

2,3,1

2,3,4

2,4,1

2,4,3

3,1,2

3,1,4

3,2,1

3,2,4

3,4,1

3,4,2

4,1,2

4,1,3

4,2,1

4,2,3

4,3,1

4,3,2

There are 24 arrangements and 4 of these are in the required order