Card Draw Probability with 2 copies of the same card

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SUMMARY

The probability of drawing one of two identical cards in the first seven draws from a deck of 30 cards is calculated using the formula P(q or q') = 2*P(q) - P(q and q'). This results in P(q or q') = 2*(Binomial[29,6] / Binomial[30,7]) - (Binomial[28,5] / Binomial[30,7]), yielding a final probability of 0.4183. This calculation aligns with the alternative method of determining the probability of not drawing the card of interest across seven draws and subtracting from one.

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TL;DR
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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