*Originally posted by cobby*

umph, i am a little confused now.

It's probably better if someone could give me the formula for calculating the probability. That would be kind.

1. Let's say I have K3o. What is the probability that any other player holds at least a A or/and a K? Let's say if there a X players.

2. I have Q3o. What is the probability that any other player holds at least an Ace or a King or a combination of both? Let's say if there a X players.

3. Again I have Q3o and the Flop comes Q-4-5. What is the probability that any other player holds at least a Queen?

It would be very nice if somebody could compile a general formula for these cases. I'm searching the internet for hours now, but wasn't successful, yet. I hope somebody could take the time. Thanks very, very much in advance!

Greetings

cobby

There is not a single general formula, but there *is* a common methodology in calculating these kinds of things. I agree with Rubisilesia that experience will teach you most of these, but IMO calculation can be a valuable addition to get a feel for certain probabilities beforehand, even though you can't perform these calculations during play.

1. The chance that at least one other player holds at least an A or a K is 1 (or 100%) *minus* the chance that *noone* holds either A or K.

What is the chance that your first opponent doesn't hold A or K? You have Kx, so there are 4 Aces and 3 Kings left in a deck of 50 (52 minus your 2) cards. Your first opponent needs one of the remaining 43 (50 minus 7) non-AK cards (chance of 43/50) and then another one of the then-remaining 42 non-AK cards from the remaining 49-card deck (chance of 42/49), for a total chance of (43/50)*(42/49) = 0.74 or 74%

Given the above, what is the chance that the second opponent *also* doesn't hold A or K? There are still 4 Aces and 3 Kings in the deck, but the deck in total now only contains 48 cards (you hold 2 cards, and your first opponent also holds two cards). That makes the chance of the second opponent also not holding an A or a K: (41/48)*(40/47) = 73%

You can repeat this for any additional opponent.

Assuming two opponents in total, the chance that *both are not holding A or K is 74% * 73% = 54%. Therefore the chance that there is *at least* one more A or K in play is 100 - 54 = 46%

Try and calculate the probability yourself for 3, 4, and 5 opponents, then see if you can calculate 2. and 3. yourself. If not, let me know, and I'll do those as well.