# Your Questions About Texas Holdem Poker Odds

Donald asks…

## what are the poker odds of getting beat by Aces full of Jacks or better in a hand?

what are the **texas** **holdem** **poker** **odds** of getting beat by Aces full of Jacks or better in a hand, while using both cards in your hand aka THE BAD BEAT? ( quads over quads , straight flush over quads etc.)

### poker holdem answers:

That depends on what 2 cards you have your hand. Please provide more details.

Chris asks…

## what is a formula for poker odds in Texas hold ’em?

i need a formula for **poker** **odds** in **holdem** that takes into account suit and card type (2-ace)

also (if you can) how i could tweak the formula to change the amount of cards in your own hand and the “community hand”. I made up my own **poker** game similar to **Texas** hold ’em and i have added some new hands to it and i need to know which are harder to get so i can rank them higher

### poker holdem answers:

*Edit* I forgot, there’s a simpler formula that does the same thing for certain calculations.

N_a * N_b * …(for # different types needed) * (#total picks Permute(#total picks) )

divided by

D * D-1 *…(for # total picks)

For the sake of demonstration, suppose you’re drawing 3, and you need 3 types of cards to hit your hand. Suppose one type has 8 remaining cards in the deck, another has 5, and the other has 3. Suppose the deck has 47 cards remaining. The math would look like this:

[(8 * 5 * 3) / (47*46*45)] * 3p3 = 0.0074

Now suppose we were drawing 4 cards, and we needed 2 of the card with 8 in the deck, instead of just one. There would be a slight change in the formula:

[(8*7 * 5 * 3) / (47*46*45*44)] * 4p4

/ 2

= 0.00235

The division by 2 is because, needing two of one type of card isn’t the same as needing one each of two types of cards. We’re dealing with permutations. With two types of cards, we’d have more permutations. Consider: (8/52)*(7/51) and (7/52)*(8/51). But with two of one type of card, there is no (7/52), it wouldn’t make sense. It so happens that there are exactly half as many permutations, so we divide the whole thing by 2.

Here’s a simpler example where the formula is again different. Find the probability of getting 4 aces when drawing 5 cards. In this example, there’s only 1 type of card needed, so order doesn’t matter at all, therefore there’s no permutation used. There will be a combination used, because there are more cards drawn than needed.

[4! / (52*51*50*49) ] * (5 c 4) = 1.847 E-5

You don’t have to memorize the variations of the formula, just understand the simple logic of the formula and you’ll know how to tailor it to each situation.

Below is my original answer. It’s a one-size-fits-all formula, you’ll never have to tweak it for a specific problem, it can solve all the above examples plus more. However, it is more complicated, so it’s perhaps better to use it only for more complicated problems.

{ (x C a) * [(z-x) C (n-a)] / (z C n_total) } * { (d C a_s)^r / (n_total C a_ss) } = probability of getting exactly A of X object in N draws, for dependent events

z = total number of objects to choose from (in hold’em this will be 52)

x = number of objects of the type of thing you need (for instance, there are 13 spades)

a = how many of those x objects do you need (for instance, you need 5 spades for a spade flush)

n = how many picks to try to get the A objects (for instance, if you see the river you get 7 cards)

There will be more than one value for X, A and N if you need more than one type of card. You would have two separate numerators, and multiply them together before dividing by (z C n_total). N_total would be the sum of the individual n-values.

In most cases, d=1 and r=1. But some problems require the calculation to be divided into what I call “rounds”. For instance, someone recently asked the probability of getting dealt blackjack 8 times in a row. For that there are 8 rounds (r = 8), because order somewhat matters: getting dealt double-aces 4 times and double-tens 4 times doesn’t count as blackjack, you need the aces and tens to happen together in eight two-card clusters. The d-value is the number of draws per round. In this example, d=2 because a blackjack hand consists of two cards.

Furthermore, some problems require “sub-rounds”. This is when not all rounds are equivalent (or put another way, when order matters on a micro-level as well). This kind of problem requires an additional (d C a)^r to be multiplied into the numerator before dividing by (n C a). (However, my formula might have a flaw for problems with sub-rounds, I’m looking into it.)

It’s a powerful formula, but undoubtedly I’ve made it confusing. So I’ll provide some demonstrations of how it can be used:

“An urn has 9 balls that are identical except that 5 are white and 4 are red. A sample of 6 is selected randomly without replacement.

What is the probability that exactly 4 are white…?”

(5 c 4) * [(9-5) c (6-4)] / (9 c 6) = 0.357

Bingo: 75 numbers to draw from, 5×5 card, 1 free space in the middle

Calculate: the probability of marking all 24 spaces in 47 picks or less.

(24 c 24) * [(75-24) c (47-24)] / (75 c 47) = 6.3 E-7

Blackjack: calculate the probability that the dealer gets 8 blackjacks in a row, using a 4-deck shoe.

(16 c 8) * [(208-16) c (8-8)] * (64 c 8) * [(208-64) c (8-8)] = (16 c 8) * (64 c 8)

(16 c 8)*(64 c 8) / (208 c 16) = 1.7548 E-10

(1.7548 E-10) * (2 c (8/8))^8 / (16 c 8) = 3.490556 E -12

Now try this harder formula on the problems I solved using the easier formula (4 aces in 5 cards, etc.). If you get the same answers I did, then you have the basic idea of this formula.

It would be better to have a program that crunches this formula for you, so all you have to do is input the variables and it outputs the answer. I already have such a program, so if you wanted I could email the code for a TI calculator, or attach my java version of the program.

Mark asks…

## poker odds, wats the odds of winning the pot in texas holdem with jack 3 verses jackjack and ace ace?

### poker holdem answers:

I deal poker for a living and can answer this for ya,

J3 vs AA:

The overpair aces are roughly an 85% favourite. This number will vary slightly depending on if the J3 is suited and if one of the aces is that suit.

J3 vs JJ:

This is even worse for the J3. The JJ will win approx 85 – 90% of the time, again fluctuation depending on suits.

J3 vs JJ vs AA:

I assume the question meant that the hands are all played in a pot together and they’re all on their backs before the flop, right?

In this case the rate of wins for the J3 highly varies depending on whether they’re suited or not and if that suit appears in the other hands. Best case scenario for the J3 is that it’s suited (say spades) and no other spades are in play yet. Then the J3 wins roughly 10% of the time. Worst case is that it’s unsuited and can’t win with a flush due to higher cards of that suit being present in the opponents hands. In this case the J3 wins only about 3 or 4% of the time.

Short answer = slim to fu*k all.

Linda asks…

## If you know every single poker odd in Texas Holdem, does that mean you definitely know when your going to win?

### poker holdem answers:

Knowing the odds does count, but knowing what kind of hand your opponent has and the pot odds are also will help you out. You have to gage your live outs against against what you think you may be up against. For example if the board is paired and you are on a straight draw with over-cards then you have to think is would my my king be any good or if my straight hit would that be any good. Also you don’t have to have a hand to win in no limit anyways so knowing your odds is good, but it is not everything.

Mary asks…

## Is this a bad call in texas holdem poker?

Is this a bad call in **texas** **holdem** **poker**?

i had Ace of diomands and Ace of hearts

flop dealt

Ace Spades, 4 spades, K clubs,

one person bet $1000, i had $25,000 to play with

i called because i had trip aces

turn card was Ace Clubs(yeah i know what are the **odds**) so i betted $1,000 trying to lure other player in with low bets to trick him

then river was 2 spades

leabing house cards with

Ace spades, 4 spades, K clubs, Ace clubs and 2 spades

so i went all in because i knew quad aces are hard to beat.

but he had 3 and 5 spades giving him straight flush.

should i have seen that coming and not have gone all in with quad aces or should i have gone all in seein as straight flushes are really hard.

so the Q is did i make a bad call.

BTW dont worry i got most me money back. it wasnt real money it was on PKR fake money. but still was it a good call knowing the 600,000 to 1 odd risk

yes i did raise preflop

i always raise preflop if i have pocket royal cards

such as QQ,KK,JJ,AA

lol i still cant get over the fact he got straight flush. it is so unbelevable.lol it was a good game anyway

### poker holdem answers:

When you have “the nuts” you go all in…and you didn’t…end of story…you let the person get there

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