Poker Odds and Math - Explained!
Suppose you’re holding the following pair in your hand: 5♥, 5♣
What are your chances of getting another 5, to make it 3-of-a-kind?
The cards that can improve your hand are called outs … and, in this case, your outs (to improve to 3-of-a-kind) are 5♦ and 5♠.
There are 52 cards in the deck and 2 are in your hand, leaving 50 cards unseen.
The probability of getting one of your 2 “outs” is 2 in 50. … and that’s 4% (since 100 * 2 / 50 = 4%).
>So it’s just (number of outs) / (number of cards unseen) ?
Yes, if there are 50 cards unseen.
However, after the Flop there’ll be another 3 cards on the table, so the number of cards unseen is 47.
Suppose the flop doesn’t have a 5♦ or a 5♠.
Then you need to calculate the poker probability of getting one of these 2 outs from the 47 unseen cards, and …
>I’d say the chances are 2 out of 47. Right?
Right, and that’s about 4.25%.
Some like to note the “odds” rather than the probability.
If there are X ways to win and Y ways to lose, then the odds are Y-to-X or Y/X to 1.
It’s just saying for every Y ways to win there are X ways to lose … and you normally say it as: “The odds are 7-to-1”.
>And that’d mean that for every 7 ways to win there’s 1 way to lose.
Yes. Of 8 possible ways events can occur, 7 will be losses and 1 will be a win.
That means your winning probability is 1 / 8 or 12.5%.
>So if odds are Y / X, then the probability of winning is X / (X+Y) .
Uh … yes, but if somebody says the odds are 7-to-1 it’s easier just to calculate the probability as 1/(7+1) = 0.125.
Normally, you’d multiply by 100 changing it to 12.5%.
In our example, of the 47 unseen cards, X = 2 is the number of cards that’ll give you your 3-of-a-kind.
The number of cards that’ll “miss you” is then 47 – 2 = 45. Then Y = 45.
The odds are then Y-to-X or 45 to 2 or 22.5 to 1.
>Last time you got the probability as 4.25%.
Yes, and that’s 1/(1+odds) = 1/(1+22.5) = 0.0425 or (multiplying by 100, to get a percentage) 4.25%.
After the Flop, there are three cards face up on the table. That means there are now 47 cards unseen.
After the Turn, there are four cards face up on the table. That means there are now 46 cards unseen.
After the River, there are five cards face up on the table. That means there are now 45 cards unseen.
(That’s 52, less the 2 in your Hand, less the cards face up, on the table.)
Suppose you hold 5♥, 5♣ and the Flop doesn’t have either the 5♦ or the 5♠.
Suppose you want to calculate the probability of getting one of the 2 cards that’ll improve your hand on the next two cards laid face up.
The probability of getting either the 5♦ or the 5♠, immediately after the flop (when there are 47 cards unseen) is 4.25% as we’ve noted before in Table 1. (That’s 2/47.)
That’s the probability of getting one of your outs on the Turn and the River.
If there were just one card to go, you’d expect the probability to be half as much.
>That’s still confusing!
Okay, we’ll do it this way:
* If you have 52 cards and you ask: “How many ways can I be dealt 2 cards?”
* The answer is: 52!/(2! 50!) = (52)(51)/(2)(1) = 1326, denoted by 52C2 … read as 52-Choose-2.
Note that 52! = (52)(51)(50)…(3)(2)(1), the product of all integers from 1 to 52.
* In general, if you have N cards and you ask: “How many ways can I be dealt M cards”, the answer is NCM.
>So there are 1326 possible 2-card hands dealt in Texas Hold’em?
Yes. In fact, there are 52 possibilities for the first card and (having dealt that card) 51 for the second so that makes (52)(51) = 2652.
However, among these 2652 possibilities we’d find 2 identical hands: [5♦, 5♠] as well as [5♠, 5♦].
Since one doesn’t distinguish between these 2 hands, we divide 2652 by 2 getting our 1326 … and that’s 52C2. Suppose we ask: “How many ways can I deal a pair?”
* There are 1326 ways to be dealt two cards, how many of these are pairs?
* There are four 2s in the deck, 2♥, 2♦, 2♣, 2♠.
* We want to choose 2 of them. The number of ways is 4C2 = (4)(3)/2 = 6.
* For all 13 cards, from the 2 to the A, that means 13*6 = 78 possible pairs.
* So, of the 1326 possible 2-card hands, 78 will be a pair.
* The probability of being dealt a pair is then: 78/1326 = 0.0588 or 5.88%.
The odds are then: 100/probability – 1 = 100/5.88 – 1 = 16, hence 16-to-1.
>Isn’t there some magic (and simple!) formula?
* You calculate the total number of ways that something can happen (like 1326 ways to deal 2 cards).
* From these, you identify the number of ways that something special can happen (like 78 ways to get a pair).
* You divide the latter by the former (like 78/1326 to get 5.88% probability of something special happening).
>I think there’s a magic formula there.
Yes, there are a couple and they go like this >>>>>>>>>>>>>
Okay, let’s do something more complicated:
* Suppose there are 4 outs and you want 1 of these 4 outs to improve your hand.
* After the Flop, there are 47 card unseen.
* The number of possible combinations of 2 cards for the Turn and the River is 47C2.
* How many of these will have 1 of the 4 outs?
* To calculate this, we remove the 4 outs, leaving 47 – 4 = 43 cards.
* There are 43C2 ways to deal 2 cards from these 43 cards … and NONE will have any of the 4 outs (because we’ve removed them).
* So how many of the 2 cards (from the Turn and the River) WILL have 1 of the 4 outs?
* It’s 47C2 – 43C2 = (total 2-card deals) – (those that do NOT have an out card).
* The probability is then: [ 47C2 – 43C2] / 47C2 = (number with at least 1 of the outs) / (total 2-card deals).
* That’s a probability of 1 – 43C2 / 47C2 = 1 – 0.835 = 0.165 or 16.5%.
Now look carefully at Table 2. Notice that the probabilities, when there are 2 cards to go, is roughly 4x the outs. >>>>>>>>>>>>>>>>>>>>
That’s because of the following (when there are n outs):
* Probability = 1 – (47-n)C2 / 47C2 = 1 – [(47-n)(46-n)]/[(47)(46)] = 1 – (1-n/47)(1-n/46) = n (1/47 + 1/46) – n2/(47)(46)
* To get a percentage, multiply by 100 and get:
Probability = n (100/47 + 100/46) + 0.046 n2 ≈ 4.3 n … when n (hence n2) isn’t too large.
>Earlier, you got 2/47 and now you get …
And now I get 1 – 46C2 / 47C2 = 1 – 45/47 = 2/47.
>When you say the dealer has 47 cards, you’re assuming I’m the only player, right? >>>>>>>>>>>>>>
Uhhh… That’s a good point.
If the dealer deals 2 cards to you, he’ll have 50 left.
However, before the Flop, he’ll deal 2 cards to each of the your opponents.
So, if there are m opponents, how many ways can 2 cards be dealt to each?
There are 50C2 ways to deal the first 2 and, for each of these, there are 48C2 to deal the next 2, then 46C2 etc. etc.
So, for m = 3 opponents, that’ll be (50C2)(48C2)(46C2) = 1,430,163,000.
Suppose the hands dealt to the 3 other players are:
[J♥, 9♠] [6♣, 9♠] [2♥, A♦].
When we counted up to 1,430,163,000, that number included all of the rearrangements of these 3 hands.
It counted [J♥, 9♠] [6♣, 9♠] [2♥, A♦] and [6♣, 9♠] [J♥, 9♠] [2♥, A♦] and [2♥, A♦] [6♣, 9♠] [J♥, 9♠] etc. etc. as distinct.
Clearly, it makes no difference which opponent holds which hand … so they shouldn’t be counted as distinct possibilities.
Since there are 3*2*1 = 3! = 6 possible rearrangements for each set of 3, the number 1,430,163,000 is 6x too large … so we divide by 6.
In fact, for m opponents, we’d divide by m! = 1*2*3*…*m. (In our m=3 example, that’s 1*2*3 = 6.)
Okay, so we have >>>>>>>>>>>>>>>>>>
>Okay, suppose there are 9 opponents and I’d like to know all the possibilities so that …
Answer: Over 20,000,000,000,000,000,000,000,000,000.
>Yeah, but if I had a computer I’d just …
If your computer could calculate the possibilities at the rate of 10,000,000,000,000 per second, it’s take over 600 million years.
In real cases, one wouldn’t bother calculating all the possible combinations that would beat your hand … else you’ll be late for supper.
Instead, one imagines thousands (millions?) of hands being dealt to your opponents and seeing what percentage beat your hand.
>And that’s the probability of being beaten by an opponent?
Approximately. It’s sorta like polling 1000 people to see which prefer chocolate ice cream to vanilla.
Then you use that percentage to gauge the entire population of millions.
Or it’s like doing a few thousand Monte Carlo simulations to see how long your portfolio might last if you withdraw x% each year.
Or it’s like …
>Okay! I get it.