probability of getting three cards with consecutive value marked on them and they are of different suits

Question

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)

Answer 1

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.

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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 $N$.

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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

$$\frac{N}{D} = \frac{264}{\binom{52}{3}}.$$

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Please use MathJax do display your Math.

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Addendum
Responding to the comment question of the OP (i.e. original poster).

ok, can you please mention the condition which I may have extraneously added or I have missed in my approach?

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.

Answer 2

Here is my reasoning, based on a solution I came up with on my own. The answer above verified my solution, and I believe it explains it a little better.

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In a 52-card set, we can assume the following are true:

• There are 4 suits
• There are 13 ranks in each suit (shown in order below)

Now, when Ace (A) counts as either a low and high card, we observe the following orderings:

• A 2 3 4 5 6 7 8 9 10 J Q K <-- Ace counts as low
• 2 3 4 5 6 7 8 9 10 J Q K A <-- Ace counts as high

From the above, we note that a Sequence can have one of 12 possible starting points (it can be anything from A to Q). For example, we can have sequences of (A, 2, 3) and (Q, K, A). So, in the case where A counts as both low/high, our starting point is set to 12.

Once the starting point is determined, the values of the remaining cards are fixed, and each subsequent card must be one of the remaining suits, as one suit is eliminated by each card.

There are 3 cards in the sequence, the first of which can be one of 4 possible suits, the second of 3 possible suits, and the third one of 2 possible suits.

Hence, the number of suite choices (S), is given by:

4 * 3 * 2 = 4! = 24 possible suite choices

Thus, there are a total of: 12 * S = 12 * 24 = 288 possible Sequence hands

The number of different hands in Three-Card Draw is the number of 3-element subsets of a 52-element set, which is:

$\binom{52}{3}$ = 52!/3!(49!) = 52 * 51 * 50 / (3 * 2) = 132,600 / 6 = 22,100

Thus, the probability of drawing 3 cards in a sequence and where each card is of a different suit, is 288/22,100 = 72/5525 = 1.3%

And when Ace can only count as the low card, then the probability instead becomes, 264/22,100 = 1.2%