How To Calculate Odds of A, K, Q, or J Hitting

[bugeyebug]

I was told recently that the odds of an A, K, Q, and/or J hitting on the flop is 70% or so. This seems high to me, but another knowledgable person said it sounds right to them.

I'd like to learn how to make this calculation, but am getting lost in the math.

I know what two of the 52 cards (mine) are, and let's assume I don't have an A, K, Q, or J in my hand. This leaves 50 unaccounted for cards in the deck, of which there are 4 possible aces, 4 possible kings, 4 possible queens, and 4 possible jacks.

If I deal the first card of the flop, this means I have a (4+4+4+4)/50 chance of hitting an A, K, Q, or J, right? This is 32%.

Now, when I turn over the next card of the flop... here is where I get lost in the math. Is it 16/49? If so, then how do I combine it with the 32% number to get the "and/or" probability.

Then when I add the third card of the flop, is it 16/48?

Help! Can someone help me walk through this?
Cheers,
Bug

[jay_shark]

It's easier to do it this way :

sol - The probability that flop does not contain an A,K,Q or j is 34c3/50c3 , so if we take the complement , we get about 69.46% .

[Top]

Or if you want to do it your way:

In this case you want to do an OR operation. There is no easy way of doing an OR operation in probability, so a reverse AND operation is usually done instead, because an AND in language is equivalent to multiplication in probabilty math.

So instead of doing 16/50 or 16/49 or 16/48 you do:
34/50 and 33/49 and 32/48. This will give you the probabilty of NOT getting a face card or ace on the flop. You then subtract the NOT probability from 1 to get the probability of getting a face card or ace on the flop.

The simplified formula for this problem is then:
1 - (34/50 * 33/49 * 32/48)

That should get around 69.46%.

[bugeyebug]

Wow, that makes perfect sense. Thanks! You guys rock...
Cheers,
-Bug