If I pick 4 cards from 52 and put them back, and pick 8 again, what is the possibility I will get back the 4 cards successively?

Question

My question:

I first pick 4 outta 52 cards and put them back. I pick 8 cards again. What is the possbility that in those 8 picks, I will get the previously picked 4 in successive order.

My approach is kinda tedious and not using a formula. I am not sure if it's right. Hope someone can check it. Thanks Can anyone think of another way?

So each arrow represents a possbility. You add all those 5 possibilities up.

$$(\frac{4}{52})(\frac{3}{51})(\frac{2}{50})(\frac{1}{49})+(\frac{4}{51})(\frac{3}{50})(\frac{2}{49})(\frac{1}{48})+(\frac{4}{59})(\frac{3}{48})(\frac{2}{47})(\frac{1}{46})+(\frac{4}{48})(\frac{3}{47})(\frac{2}{46})(\frac{1}{45})$$

Which is $=21.89\times 10^{-6}$

Edit: if you want to answer this question, can you explain it more thoroughly, so far every answer is different. And to me, Trần Thúc Minh Trí's answer makes the most sense. Combinatorics is kinda confusing.

Answer 1

I changed my answer for the second time, this time more certain:

• The number of ways to pick $8$ cards is $52 \times 51 \times 50 \times 49 \times 48 \times 47 \times 46 \times 45 = 30342338208000$, the order DOES MATTER, because you have to pick $4$ original cards in a row, but not in any specific order.

• Note that for any $4$ cards, there are $4!=24$ ways of swapping them.

• Assume that your new eight cards have the first four cards all correctly picked from the original four cards, the remaining four cards should be $4$ in $48$ cards remaining. The number of possible ways for this case is $24 \times 48 \times 47 \times 46 \times 45 = 112078080$ ways, again the order does matter.

• Assume that your new eight cards have the $1$st card is not from the original four cards, the $2$nd to $5$th card are all correctly picked from the original four cards, the remaining three cards should be $3$ in $47$ cards remaining. The number of possible ways for this case is $48 \times 24 \times 47 \times 46 \times 45 = 112078080$ ways, again the order does matter.

• Assume that your new eight cards have the $1$st and $2$nd cards are not from the original four cards, the $3$rd to $6$th cards are all correctly picked from the original four cards, the remaining three cards should be $2$ in $46$ cards remaining. The number of possible ways for this case is $48 \times 47 \times 24 \times 46 \times 45 = 112078080$ ways, again the order does matter.

• Assume that your new eight cards have the $1$st, $2$nd, $3$rd cards are not from the original four cards, the $4$th to $7$th cards are all correctly picked from the original four cards, the remaining three cards should be $1$ in $45$ cards remaining. The number of possible ways for this case is $48 \times 47 \times 46 \times 24 \times 45 = 112078080$ ways, again the order does matter.

• Assume that your new eight cards have the $1$st, $2$nd, $3$rd, $4$th cards are not from the original four cards, the $4$th to $8$th cards are all correctly picked from the original four cards. The number of possible ways for this case is $48 \times 47 \times 46 \times 45 \times 24 = 112078080$ ways, again the order does matter.

So the probability of having picked back $4$ cards successively is:

$$\frac{112078080 \times 5}{30342338208000}=\frac{1}{54145}$$

Answer 2

Again, you have a total of $\binom{48}{4}$ ways.

As you said, there are five ways in which the cards can come in right order. But the other $4$ cards can be arranged in $4!$ ways.

So total candidates = $5 * 4! \implies$ prob $ = \frac{5 * 4!}{\binom{48}{4}}$

Answer 3

I have two ways-

1) For each pick of $4$ cards in the $1^{st}$ round, number of ways for the pick of $8$ cards in the $2^{nd}$ round -

Total ways = $_{52}$P$_8$

Required ways $= _{52-4}$P$_{8-4}(8-4+1)=_{48}$P$_4(5)$ ,($5$ for the $4$ cards taken in the $1^{st}$ round and $_{48}$P$_4$ for the other $4$ cards)

$\therefore$ Required probability $= \frac{_{48}P_4(5)}{_{52}P_8}=\frac{5(48)!}{52!}$

2) For each pick of $8$ cards in the $2^{nd}$ round, number of ways for the pick of $4$ cards in the $1^{st}$ round -

Total ways = $_{52}$P$_4$

Required ways $= 8-4+1=5$

$\therefore$ Required probability $= \frac{5}{_{52}P_4}=\frac{5(48)!}{52!}$

In general, if we have n cards, we have to pick k cards then put them back then after picking x cards, then the probability of getting the k cards successively $= \frac{(x+1-k)(n-k)!}{n!}$