Probability of pairs

Question

Let's imagine to have 2 decks of regular 52 cards (104 cards, ace to king, no jokers)
I'm stuck on probabilities of finding exact pair (e.g. 2 aces of hearts) in different number of drawn cards.

The probability of drawing two cards and have one exact pair is 1/103 (first card is free, the second has only one choice to be the matching one. it is 0,97%.

First question:
How about to have one (only) exact pair drawing 4 cards? I'd say it is 1 x (1/103) x (1) x (100/101) = 0,96% no matter what is the extraction order*. Shouldn't be more than 0,97% having more possibilities? Where is my error?

Note: 1 x (102/103) x (1/102) x (100/101) or 1 x (102/103) x (100/101) x (1/102) also is equal to 0,96%

Second question:
What is the probability to have only one exact pair when drawing 13 cards instead of four? How to analitically have the calculation for N cards drawn (N = 13) ?

Third question:
How to reason to calculate the probability to have at least one exact pair when drawing 13 cards? I'd say that is the counter-probability to have 0 exact pairs. That is to say:

1 - [1 x (102/103) x (100/102) x (98/101) x ... x (80/92)] =  1 - 0,425 = 0,574 = 57,4%

Is the right way to reason?

Fourth question:
How to reason to calculate the probability to have only 2 (or 3...) exact pair when drawing 13 cards? Should the sum of the probabilities of having only 1 to 6 exact pairs give the probability to have at least 1 pair as calculated above?

Thank you in advance for all the answers!

Answer 1

To begin, simplify the problem by having each deck of cards have a different colored back so all 104 cards are distinguishable.

For the first problem, finding the probability of exactly one match in four cards, doing so via combinations treating order as irrelevant:

• Pick the card used for the match. $52$ choices.
• Pick the remaining two card fronts. $\binom{51}{2}$ choices.
• Pick the color of the card-backs for each of these other two cards: $2\times 2$ choices.

This is out of the $\binom{104}{4}$ ways of picking cards. This gives the probability as

$$\dfrac{52\times \binom{51}{2}\times 2^2}{\binom{104}{4}}=\dfrac{600}{10403}\approx 0.05768$$

Doing this where order is relevant, choose whether the matching cards occurred in the first and second slot, in the first and third slot, in the first and fourth slot, in the second and third slot, in the second and fourth slot, or in the third and fourth slot... that is to say, from the four positions we choose two of them to be occupied by the match. There are $\binom{4}{2}=6$ ways to make this choice.

Given this choice, from the left to right, given the earlier positions (if any) have been successfully filled in the designated way, find the probability of continuing to fill positions in the designated way and multiply. Recognize that for each of the six different scenarios, they all have the same respective probabilities, so rather than adding them together we can simply multiply the result by $6$.

That gives a probability then of

$$6\times 1\times \dfrac{1}{103}\times 1\times \dfrac{100}{101}=\dfrac{600}{10403}$$

Exactly the same as before.

The punchline here is that it does not make a difference in problems like these whether we treated order of cards in the hand as relevant or irrelevant so long as we are consistent with the choice and use the same understanding in both the numerator and denominator simultaneously, and also so long as the events we are interested in make no reference to exact order of cards either.

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For the second problem, we could do it with treating order of cards as relevant, but it will be in my opinion far too much arithmetic. If you insist, go ahead and work it out as before, noting the $\binom{13}{2}$ ways in which you pick the positions of the match.

Instead, doing it order irrelevant we have the exact same method as before to calculate.

• Pick the card used for the match. $52$ choices.

• Pick the remaining 11 card fronts. $\binom{51}{11}$ choices.

• Pick the color of the card-backs for each of the other $11$ cards. $2^{11}$ choices

This is out of the $\binom{104}{13}$ ways of picking cards, giving the probability as:

$$\dfrac{52\times \binom{51}{11}\times 2^{11}}{\binom{104}{13}}$$

Compare this to an answer of:

$$\binom{13}{2}\times 1\times \dfrac{1}{103}\times 1\times \dfrac{100}{101}\times\dfrac{98}{100}\times\dfrac{96}{99}\times\dfrac{94}{98}\times\dfrac{92}{96}\times\dfrac{90}{95}\times\dfrac{88}{94}\times\dfrac{86}{93}\times\dfrac{84}{92}\times\dfrac{82}{91}$$

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For the third question, yes, this is fine. You could similarly have found the probability of zero pairs directly using the same approach as for the next section using $0$ for $k$.

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For the final question

• choose which $k$ card types were used for matches in $\binom{52}{k}$ ways.

• Choose which $13-2k$ card types were used for the remaining cards in $\binom{52-k}{13-2k}$ ways

• For each of these, choose which color card-backs were used in $2^{13-2k}$ ways.

This is out of the $\binom{104}{13}$ ways of picking $13$ cards.

Answer 2

Your working for two cards is correct.

For the first question - As you are trying to find probability of only one exact pair in $4$ drawings, it should be

$\displaystyle \small 52 \cdot {51 \choose 2} \cdot 2^2 \big / {104 \choose 13}$

Explanation: You choose one of the $52$ cards that you will get exact pair of (both cards) and you then choose $2$ remaining cards out of $51$ that you will get one each of (out of two cards for each chosen rank, suit and color).

For the second question - it is again similar to first

$\displaystyle \small 52 \cdot {51 \choose 11} \cdot 2^{11} \big / {104 \choose 13}$

For third question, please note that $\displaystyle {52 \choose 13} \cdot 2^{13}$ will be the number of ways of choosing cards such that you get no exact pair. So the probability that you get at least one exact pair,

$\displaystyle \small \big[{104 \choose 13} - {52 \choose 13} \cdot 2^{13}\big] \big/ {104 \choose 13}$

Please follow similar approach for fourth.