I know that the formula for pulling three consecutive clubs from a standard shuffled card deck is (13/52) x (12/51) x (11/50). In the same vein, I know that the odds of being dealt a royal flush in five cards is (20/52) x (4/51) x (3/50) x (2/49) x (1/48).

BUT -- what is the formula for pulling three clubs from the deck if you have 4 chances instead of three? And in the same vein, what is the formula for being dealt a royal flush if you get 6 cards rather than 5?

You can approach the 3-clubs problem it like this: you "win" if you draw 4 clubs or if you draw 3 clubs plus 1 non-club. The probability of four clubs is (13/52)x(12/51)x(11/50)x(10/49). The probability of 3 clubs and a non-club is: (13/52)x(12/51)x(11/50)x(39/49) x 4. The factor 4 is there because the non-club could be either the 1st, 2nd, 3rd, or 4th card drawn. We really don't care which. Add these two probabilities to get the total.

For the royal flush problem first consider the probability of a royal flush in spades, given 6 draws. That probability is (5/52)x(4/51)x(3/50)x(2/49)x(1/48)x(47/47) x6. Here the '6' is because the non-flush card may be appear in any of the 6 positions. Similarly the probability of a royal flush in hearts, or diamonds, or clubs is the same. Hence the total probability is: 4x6 x (5/52)x(4/51)x(3/50x(2/49)x(1/48). It turns out this probability is six times higher than if you only draw 5 cards.

Thank you very much. Very clear explanation!