I know that this problem can be correctly calculated using inclusion-exclusion principle. However, I thought solving for the complement (i.e. the probability that it contains all suits) would be the more intuitive approach, though my answer turned out to be wrong. $P =1- \frac{{13\choose 1}{13\choose 1}{13\choose 1}{13\choose 1}{48\choose 9}}{52 \choose 13}$ which is negative. Could anyone point out what's wrong with this expression? thanks!
2026-03-26 16:10:18.1774541418
Calculate the probability that a poker hand of 13 cards does not contain at least one suit
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You are picking one card of each suit with the $13 \choose 1$s then picking the other $9$ cards. You count every hand many times. If it only has the A and 2 of spades you will count it once when the A is the first spade and the 2 is among the other 9 cards, then again when the 2 is the first spade and the A is among the other 9 cards. That multiple counting makes your subtraction much too large. A hand with $4-4-3-2$ distribution will get counted $4 \cdot 4 \cdot 3 \cdot 2=96$ times.