Need to find the probability when three cards are chosen from a deck of 52 cards, the three cards have consecutive values marked on them and they are from different suit (Ace – 1, Jack – 11, Queen – 12 and king – 13)
My approach is:
Lets say that the set of numbers we got is 1,2,3 –
probability of choosing 1 is 4/52
probability of choosing 2 is 3/51 (ignoring the card with value 2 from the 1st suit from which 1 was picked)
probability of choosing 3 is 2/51 (ignoring the card with value 3 from the first two suits from where card with value 1 & 2 were picked)
there are total 11 such cases -> (1,2,3), (2,3,4), (3,4,5),…(11,12,13)
IMO the probability should be: 11*(4/52)(3/51)(2/50)
Best Answer
The problem is much more complicated than that.
For one thing, if the $3$ card hand turns out to be $(1)$, $(2)$, $(3)$, these $3$ cards can be selected in any order.
For another, if the first card selected is (for example) a $(3)$, there are $3$ different sequences possible. Namely, $[(1,2,3), (2,3,4), (3,4,5)].$
Further, the Ace may count high or low.
So, although $[1,2,13=king]$ does not count, $[1,12=queen,13=king]$ does count.
So, the approach that I prefer is
$$\frac{N\text{(umerator)}}{D\text{(enominator)}},$$
where $~\displaystyle D = \binom{52}{3}$.
So, the problem has been reduced to computing $N$.
For any given sequence of $[n,n+1,n+2]$ there are $4$ choices for the suit of $(n)$, and then $3$ choices for the suit of $(n+1)$ and then $(2)$ choices for the suit of $(n+2)$.
Then, $(4 \times 3 \times 2) = 24$ suit choices.
Forther, the low card can range anywhere from $(1=Ace)$ through $(12=Queen)$, so you have $12$ choices for the ranks involved.
Therefore, $N = 12 \times 24 = 288.$
Final computation:
$$\frac{N}{D} = \frac{288}{\binom{52}{3}}.$$
Edit
If $(1=Ace)$ is not permitted to count high, then you have to deduct $24$ from $N$. So the computation would be
$$\frac{N}{D} = \frac{264}{\binom{52}{3}}.$$
Please use MathJax do display your Math.
Addendum
Responding to the comment question of the OP (i.e. original poster).
Assuming that $(1=Ace)$ only counts low, your approach has $2$ major flaws.
The 1st flaw is that if the first card involved is a $(2)$, then you have the possibility of $2$ different sequences. Either $[1,2,3]$ or $[2,3,4].$ Similarly, if a $3$ is involved, then there are potentially $3$ different sequences: $[(1),(2),(3)],[(2),(3),(4)]$ or $[(3),(4),(5)].$
It is true that this flaw may be bypassed by considering the events of (for example) $[(1),(2),(3)]$ and $[(2),(3),(4)]$ as disjoint events. So, assuming that Ace only counts low, you have $(11)$ disjoint events to deal with. So, you ask yourself, what is the probability of each of these events.
The event of $[(1),(2),(3)]$ of $3$ different suits should be computed as follows:
$$\frac{12}{52} \times \frac{6}{51} \times \frac{2}{50}. \tag1 $$
In (1) above, there are $(12)$ different cards that may be chosen first. Then, the 2nd card chosen must be one of the $8$ cards of a different rank. Of these $8$ cards, you have to eliminate the $2$ cards of matching suits.
So, you end with a computation of
$$\frac{11 \times 12 \times 6 \times 2}{52 \times 51 \times 50}$$
This agrees with the computation that I gave of
$$\frac{264}{\binom{52}{3}}.$$
However, your approach was complicated, and left more room for error.