0
$\begingroup$

In a deck of cards with 52 cards we change clubs 8 with spades 10 and clubs 9 with spades jack from another deck of cards. What is the probability of getting two pairs if you draw 5 cards?

In a standard deck with 52 cards, the probability of getting two pairs could be calculated as follows:

We can determine the sample space by choosing 2 ranks of 13 existing ranks and then 2 suits of 4 existing suits for each pair of cards. Finally choose the last one among the remaining 11 ranks which can be done in 44 different ways. On the other hand the space is determined by choosing 5 cards among 52 cards.

Help me to find out how to determine the total number of possibilities in this case with a non-standard deck of cards.

$\endgroup$
2
  • 1
    $\begingroup$ It's still just a matter of counting cases. Note that the suits are irrelevant, you only care about rank. One case would be $AABBC$ where none of the three ranks are $8,9,10,J$. And so on. $\endgroup$
    – lulu
    Commented Feb 15, 2022 at 15:05
  • 2
    $\begingroup$ To organize the cases, I suggest dividing the ranks into three types: Ordinary, Strong, Weak. The Ordinary ranks, like $2$ or $K$, have $4$ cards each. Strong ones, $10,J$, have $5$ and Weak ones, $8,9$, have only $3$. $\endgroup$
    – lulu
    Commented Feb 15, 2022 at 15:09

1 Answer 1

1
$\begingroup$

In a deck of cards with 52 cards we change clubs 8 with spades 10 and clubs 9 with spades jack from another deck of cards.

The modified deck consists of $3$ eight, $3$ nine, $5$ ten, $5$ jack, and $4$ each of the remaining $9$ ranks. @Lulu suggests dubbing these Weak, Strong, and Ordinary ranks: so you have $2$ Weak ranks, $2$ Strong ranks, and $9$ Ordinary ranks.

What is the probability of getting two pairs if you draw 5 cards?

The six cases for two pairs and a singleton are:

  • Two Weak pairs:
  • Two Strong pairs:
  • Two Ordinary pairs: $\left.\binom 92\binom 42^2\cdotp\binom{3\cdot 2+5\cdot 2+4\cdot 7}{1}\middle/\binom{52}5\right.$
  • A Weak and Strong pair: $\left.\binom 21\binom 32\cdot\binom 21\binom 52\cdot\binom{3+5+4\cdot 9}{1}\middle/\binom{52}5\right.$
  • A Weak and Ordinary pair:
  • A Strong and Ordinary pair:

I've included probabilities for two among these cases, you can surely do the rest.

$\endgroup$
1
  • $\begingroup$ Actually I found out too that there must be 6 cases as you mentioned in your comment. I think I managed to calculate the probibility and it is approximately 0.049. Does it sound correct? $\endgroup$
    – Marx
    Commented Feb 17, 2022 at 18:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .