I Card Draw Probability with 2 copies of the same card

AI Thread Summary
The discussion centers on calculating the probability of drawing one of two identical cards from a deck in the first seven draws. The formula used is P(q or q') = 2*P(q) - P(q and q'), where P(q and q') is calculated as Binomial[28,5] / Binomial[30,7]. The final probability is derived as P(q or q') = 2*(Binomial[29,6] / Binomial[30,7]) - (Binomial[28,5] / Binomial[30,7]), resulting in a value of 0.4183. An alternative method to arrive at the same probability involves calculating the chance of not drawing the card over seven attempts and subtracting from one. The discussion confirms the accuracy of the calculated probability of 0.4183.
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I am trying to answer my own question of

"If I have a 30 card deck with 28 unique cards and 2 identical cards of interest, what are the odds of drawing one of the 2 identical cards in the first 7?"
I understand that the odds of drawing one of the unique cards in the first 7 is expressed as

29c6 / 30c7

where NcK is "N choose K" or Binomial[N,K] in Mathematica.

Am I correct in using the following to answer my original question?

Let q be the first card of interest and q' be the second:

P(q or q') = P(q) + P(q') - P(q and q')

which in the case of identical cards of interest becomes:

P(q or q') = 2*P(q) - P(q and q')

and the probability of P(q and q') is the following:

Binomial[28,5] / Binomial[30,7]

thereby giving a solution to my original question

P(q or q') = 2* ( Binomial[29,6] / Binomial[30,7] ) - ( Binomial[28,5] / Binomial[30,7] ) = 0.4183
 
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I arrived at the same answer (0.4183) as you did, using this thinking:

Probability of not drawing a card of interest is (28/30) on first draw.
Then for next draw it is (27/29), then (26/28), then (25/27) etc. to (22/24).
Multiply all of those probabilities together to get the probability of not drawing on 7 tries.

Subtract that from 1, to get Probability of drawing.
 
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