In a set of 25 cards where two cards are only the same, what is the probability of picking those two same cards if I can only pick twice?

What I only know is that at first pick it is 2/25 equivalent to 8% chance. For the second pick it would be 1/24 (only if the first pick was a success) which is equivalent to 4.17% chance.

How can I compute for the probability having these results? I'm just asking this out of curiosity. Thanks a lot.


Your first steps are correct. Let $A$ be the event in which we pick one of the two cards in the first turn, and $B$ the event in which we pick the other card in the second turn. Using the definition of conditional probability, we have:

$$P(A, B) = P(A) P(B | A) = \frac{2}{25} \frac{1}{24} = \frac{1}{300}$$

The answer by cognitive uses another approach, but unfortunately is wrong. We can use the concept of combinatorics, by considering the amount of ways we can draw two cards out of 25. Since each combination is equally likely to be drawn, the probability of ending up with the two similar cards, equals:

$$P(A, B) = \frac{1}{25 \choose 2} = \frac{1}{\frac{25!}{23!2!}} = \frac{1}{\frac{25 \cdot 24}{2}} = \frac{1}{300}$$

  • $\begingroup$ Thank you so much for the explanation $\endgroup$ – Moon Feb 12 at 9:04

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