How to calculate % of getting called by "x" range of hands

[jkstay]

se2schul, this was a very good explanation. However, I don't think the last part of your post (where you explain how to find the probability of getting dealt the 7 of clubs or a pocket pair larger than Jacks) is correct.

If you follow your logic, then the probability of being dealt a pocket pair greater than fives would be as follows:
Number of ways to get dealt a pocket pair = 221
Number of ways to get dealt a pocket pair larger than fives = 9 * 221 = 1989
1989/1326 = 150%

What you should do instead is add the individual probabilities together (going back to your original example of 7 of clubs or pocket pair larger than jacks):
Probability of 7 of clubs = 52C1/52C2 = 52/1326
Probability of AA = 4C2/52C2 = 6/1326 (=1/221)
Probability of KK = 4C2/52C2 = 6/1326
Probability of QQ = 4C2/52C2 = 6/1326

Get the sum of the individual probabilities:
52/1326 + 6/1326 + 6/1326 + 6/1326 = 70/1326 = 35/663 = 5.3%

Hopefully this doesn't confuse things...

[se2schul]

Good catch. Thanks. That's what I get for posting quickly after too many Friday work beers... I really should've checked my numbers over.

ss

[DRD66]

[ QUOTE ]
But do you see why there are 6, 4, and 12 ways to be dealt those hands?

[/ QUOTE ]
See if I'm thinking right:
6 pp - 4 instances of first card x 3 remaining to pair = 12 ways. HOWEVER, order doesn't matter (eg 2d2h is same as 2h2d), so half these combos eliminated 12/2=6
4 specific suited - 4 of 1st card (A here) x 1 of second card of same suit (K) = 4
12 specific os - 4 of 1st card (say its Ad) x 3 of 2nd card diff suit (Kc,Ks,Kh) = 12

[ QUOTE ]
you got 1 chance out of 52 to hit the first card, and then 1 chance out of 51 to hit the second card. 1/52 * 1/51 = 1/2652

[/ QUOTE ]

>>>>DISCLAIMER THE FOLLOWING RAMBLE IS NOT EXPERT OPINION, JUST LOGIC BASED AND SIMPLE MATH. PEER REVIEW SURE TO FOLLOW. <<<<<<<

This is correct, but 1/2652 is the ODDS of drawing a specific hand, and we are talking about WAYS to draw a hand. The odds calculation treats AdKh, KhAd, and AdKc as three different hands, when for your purpose they're the same. My thought is to use the number of "unique" hands - this number ignores the order you recieve the cards and only considers suit if your hand is suited. So, for non-pairs there are 13 of 1st card rank x 12 of 2nd card rank = 156 combinations. Again, since order doesn't matter (AK is the same as KA) so divide by 2 to get 78 unique starting non-pair combinations disregarding suit. Therefore, there are 78 unsuited and 78 suited non-pairs for a total of 156 unique non-pairs. Add the 13 pairs for a total of 169 "unique" starting hands.
Another way to think of it is that there are 13 Ax combinations (A2-AA) + 12 Kx combinations (K2-KK, we've already counted KA as AK) + 11 Qx + 10 Jx + ... + 1 2x (22) = 91. (there's a name for this type of sequence I think. simple solution is n(x1+xn)/2) Of these 91, all but the pairs (91-13=78) could also be suited combinations, so 91+78=169 total unique combinations of starting hands.
The easiest way to get this number is probably to open Poker Tracker to the general tab, click "filters" then "select specific hand" and count the boxes.
Since I've started getting serious about poker, I need to find a way to think about this stuff that makes sense to me. Seems like there are several ways to arrive at the same conclusion, but you must be VERY CAREFUL that you are comparing apples to apples.

I see other people posting with some more detailed probabilty math (which I'm trying to learn), and I also know a few people who can just memorize enormous ammounts of data and recall it instantly. The stuff above is just how I work through it. Any suggestions or corrections would be appreciated.

[DRD66]

Excellent post. Would you have reccomendations for further resources? I'm OK with any math short of calc. I'd read "Lady Luck: The Theory of Probablilty" by Warren Weaver a few years back, is this a decent text?
Also, for the lazy and/or geeky among us, the Texas Instuments TI-30XIIS calculator has an nCr function and all kinds of other cool buttons. $10-15 at Circuit City, Staples, etc. You'll save that much cutting back on Advil.

[Allinlife]

[ QUOTE ]
So, this is how to do "choosing".
To workout 4C2, start counting down from the first number multiplying each term until it's one bigger than the bottom number. Then divide that by the number you get when you multiply the bottom number being counted down to 1. That sounds complicated, but it's really not. I'll do a couple of examples:

4C2 = (4 * 3)/(2 *1) = 6
5C2 = (5 * 4 * 3)/(2 * 1) = 15
6C3 = (6 * 5 * 4)/(3 * 2 * 1) = 20
Do you see the pattern??


[/ QUOTE ]
sorry, but I still don't get why you do
52*51/2 when doing 52C2 , I get why you divide it by 2*1 but not why you simply stop at 52*51

you said you start counint down from first number, multiplying each term until it's one bigger than the bottom number.
so wouldn't that make it 52*51* so on to *3?
am I wrong to assume this bottom number is 2 in "52C2"

once again, thank you very much for detailed lesson (sorry for my think head)

[eastbay]

[ QUOTE ]

sorry, but I still don't get why you do
52*51/2 when doing 52C2 , I get why you divide it by 2*1 but not why you simply stop at 52*51


[/ QUOTE ]

Because half the combinations are the same hand, so you only count them once.

Example: One of the 52*51 possibilities is Ac Th. Another one of the 52*51 possibilities is Th Ac. But that's the same hand. So only half the combinations matter. Hence the divide by 2.

eastbay

[Allinlife]

thanks for that reply eastbay, but do you know what he means by [ QUOTE ]
start counting down from the first number multiplying each term until it's one bigger than the bottom number.

[/ QUOTE ]

[eastbay]

[ QUOTE ]
thanks for that reply eastbay, but do you know what he means by [ QUOTE ]
start counting down from the first number multiplying each term until it's one bigger than the bottom number.

[/ QUOTE ]

[/ QUOTE ]

He's trying to write out the formula for combinations ("choosing") in words. I don't bother trying to remember this formula. It's on any decent handheld calculator, or you can find tons of applets on the web that do this, like here:

Combinations Calculator

The formula doesn't really lend much insight into the concept, so I don't see any advantage in memorizing it.

eastbay

[DRD66]

[ QUOTE ]
sorry, but I still don't get why you do
52*51/2 when doing 52C2 , I get why you divide it by 2*1 but not why you simply stop at 52*51
you said you start counint down from first number, multiplying each term until it's one bigger than the bottom number.
so wouldn't that make it 52*51* so on to *3?

[/ QUOTE ]
<font color="red">***WARNING! Scary looking math early, but you don't need to follow it. It's simplified further down**</font>

Allinlife's question stumped me, so I dug out the aforementioned Weaver book for clarification. All this info is from that text. The formula for nCr is n!/r!(n-r)!. "n!" (pronounced "n factorial") is n*(n-1)*(n-2)*...*2*1. So for 52C2, we have 52!/2!(52-2)! = 52!/2!*50!. Longhand this is 52*51*50*49*48*...*1 / (2*1)*(50*49*48*...*1). Since 50*49*48*...*1 (or 50!) is in both top and bottom, they cancel out, leaving 52*51/2*1 = 2652/2 = 1326 (lorinda and eastbay - the AcKd=KdAc is built into the equation by multiplying the bottom by r!. If order matters, it called a permutaion, nPr = n!/(n-r)!. There are other cases where you divde by 2 for the reason you state, see my post earlier in this thread)

[ QUOTE ]
To workout 4C2, start counting down from the first number multiplying each term until it's one bigger than the bottom number. Then divide that by the number you get when you multiply the bottom number being counted down to 1.

[/ QUOTE ]
What you do is start counting down from the first number multiplying each term UNTIL THE NUMBER OF TERMS IS THE SAME AS the bottom number. Then divide by the number of terms. It just happens that you get the same result if you do 4C2 the way you wrote it or the way I wrote it.
[ QUOTE ]
I'll do a couple of examples:

4C2 = (4 * 3)/(2 *1) = 6
5C2 = (5 * 4 * 3)/(2 * 1) = 15
6C3 = (6 * 5 * 4)/(3 * 2 * 1) = 20

[/ QUOTE ]
These actually work out as:
4C2 = (4 * 3)/(2 * 1) = 6
5C2 = (5 * 4)/(2 * 1) = 10
6C3 = (6 * 5 * 4)/(3 * 2 *1) = 20
Again, the fact that 6C3 worked out your way is coincidence due to the fact your first term is exactly double your second. Let's look at 6C2. Is it 6*5*4*3/2*1=360/2=180 or 6*5/2*1=30/2=15? There should be FEWER combinations of two things than of three (bottoming out at 6C1=6, eg. roll one die there are 6 different results). In poker terms, think of 5C2 as how many hole combos will draw to a wheel - A2, A3, A4, A5, 23, 24, 25, 34, 35, 45. 10 ways to start an A-5 straight, not 15.

Dam, this gets confusing, huh? Maybe this should be in the Probability forum. Bottom line is, there are ways to do the calculations but you gotta do the right math for the problem, and that's difficult. Deciding WHICH formula to apply to WHAT problem is a huge labryinth before you even get to the math. There are many software resources available. There are also other easy-math ways to think about some of this (see my earlier reply to Ilya, which I'm still not sure is valid). If your just curious and want to play with the numbers, go for it and have fun! If you're basing a monetary descision on a probability or calculation, assure yourself you've got the right info. As always, the player is responisble for how and why he bets.

Still an excellent post se2schul! You're good with words, maybe you could modify your original post and put it up on the probability or theroy forum. Cheers!

[se2schul]

Oh Crap. It seems everyone is right except me. I've messed up my post in so many ways it's embarrasing. That's what I get for trying to do shortcuts and explain something that I haven't reviewed in years while drinking many rye and gingers.

I just saw AllInLife's problems as fairly trivial since back in school I used to these in my sleep.

I wasn't going to just work through them, but now it seems that I could use the practice.

If someone could double check these for me, it would be much appreciated.

Prob AA-99:
Probability of a pocket pair is 1/221 (since we've worked this out MANY times).
Probability of AA-99 is 6 * 1/221 = 6/221

Prob AKs:
There are 4 ways to get AKs (AsKs, AdKd, AhKh, AcKc)
So, the probability is 4/(52C2) = 4/1326

Note: The prob of AQs is the same as for AKs = 4/1326

Prob AKo:
There are 4*4 = 16 ways to get AK, but 4 of those ways are suited, so 16-4 = 12 ways to get AKo.
Prob AKo = 12/1326

Now, for the culmination of me demonstrating my lack of math skills:
The probability that someone gets dealt AA-99, AKs, AQs, or AKo is the sum of each individual probability:
=6/221 + 4/1326 + 4/1326 + 12/1326
=56/1326 = 4.2%

Hopefully there are less errors and this is better explained. I feel a little sheepish for some of the earlier mistakes, but it actually isn't all that surprising. There used to be a running joke at school with my friends. They used to say "If it's an EASY math problem, Steve's answer is surely wrong".