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View Full Version : A, K or Q on flop.

EvilivE
04-10-2005, 04:21 PM
If your holding JJ what is the probability of an A or K or Q appearing in the flop?

I came up with approximately 57% of the time an overcard will come on the flop when you are holding JJ. 57% for some reason just seems a little high. Could somebody verify if that approximation is correct.

Here are my calculations, I solved the problem two different ways.

Method 1
=========
12/50 + (38/50 * 12/49) + (38/50 * 37/49 * 12/48)

Method 2
=========
1 - (38/50 * 37/49 * 36/48)

Thanks

niin
04-10-2005, 04:39 PM

PygmyHero
04-10-2005, 05:20 PM
I agree with your solution. I did it a little bit differently and I thought I'd share my solution since it simplifies the problem to some extent.

There are 38 cards in the deck you are okay with seeing on the flop (52 minus your hole cards minus all the aces, kings, and queens. The 38 include all the undercards as well as the last two jacks in the deck). So the equation for a 'safe' flop is:

C(38,3) = 8436

Total number of flops is C(50,3) = 19600

8436/19600 = 43% or 1.3:1 against.

So it is 43% to catch a flop with no A, K, or Q, and thus it is a 57% chance you will see one or more of them.

One last note - it gets worse...The 43% chance includes some unfavorable flops (three suited boards, three sequenced boards, paired boards, etc.). You may want to take that into consideration if you're trying to get a more accurate read on the value of JJ.

TomBrooks
04-12-2005, 09:38 PM
I think the OP would only be worried about a Q, K, or A coming on the flop if one of his opponents held a Q, K, or A. If you assume this, I think you would change the equation slightly.

and I wish I knew how to calculate those "C" equations. I just asked how in another thread though.

i wanna be me
04-13-2005, 01:00 AM
[ QUOTE ]
and I wish I knew how to calculate those "C" equations. I just asked how in another thread though.

[/ QUOTE ]

isn't it just like 38 "choose" 3 - or 38C3 like in high school?

TomBrooks
04-13-2005, 02:06 AM
I never heard that mathematical term before reading it here. Not even in 4 semesters of College Engineering Calculus.

BruceZ
04-13-2005, 03:51 AM
[ QUOTE ]
I never heard that mathematical term before reading it here. Not even in 4 semesters of College Engineering Calculus.

[/ QUOTE ]

But you used them when you did the binomial expansion of (a + b)^n. Perhaps you called them the "binomial coefficients" and used the notation (n k).

(a + b)^n = a^n + C(n,1)*b*a^(n-1) + C(n,2)*b^2*a^(n-2) + C(n,3)*b^3*a^(n-3) + ... + C(n,n)*b^n

You can get the C(n,k) from the nth row of Pascal's triangle (starting with n=0). Each number is the sum of the two numbers above it.

<font class="small">Code:</font><hr /><pre>
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
....
</pre><hr />

EvilivE
04-13-2005, 01:11 PM
[ QUOTE ]

and I wish I knew how to calculate those "C" equations. I just asked how in another thread though.

[/ QUOTE ]

Another way of writing that is nCk. Where n = the entire set and k = the subset you seek. So you get 52C2, which is essentially asking "How many unordered 2 element subsets can I create from a 52 element set?"

The formula looks like this:

n!
--------
k!(n-k)!

using example 52C2:

52!
---------
2!(52-2)!

TomBrooks
04-13-2005, 01:37 PM
[ QUOTE ]
nCk
n!
--------
k!(n-k)!

[/ QUOTE ]
Great. Thanks. I found a function for it on my scientific calculator. I didn't even realize what it was before. I remember binomial coeficients, but only vaguely. It's been a long time since I studied math. I don't remember the use of k. But I've forgotton a lot of that stuff.