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For example, if I hold two different cards, what's the probability I get a pair at river?

Here is my calculation:

(3/50 + 3/49 + 3/48 + 3/47 + 3/46)*2

Is it correct? How does one calculate the other winning combinations?

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Based on your calculations... If you hold 4 cards (A to 4), you will ALWAYS make a pair?

Estimate with 3/50 * 5 instead of the annoyingly similar fractions: (3/50 * 5) * 4 = 60/50 !?

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The reason you won't get the right answer this way even though I can see the logic in your math is because of "double counting". You think the chance of hitting the right card in the first card on the flop is 3/50 and the chance of hitting the right card in the second card on the flop is 3/49, but you can't add those two together because they are not exclusive. You are forget about the chances that you hit BOTH first card and second card (so you got trips).

So the real straight forward logical math will look like:

6/50 = 0.12 (this is the chance of hitting one of your cards on the FIRST card)

44/50 * 6/49 = 0.1078 (this is the chance of hitting one of your cards on the SECOND card)

44/50 * 43/49 * 6/48 = 0.0956 (3rd card)

44/50 * 43/49 * 42/48 * 6/47 = 0.0863 (4th card)

44/50 * 43/49 * 42/48 * 41/47 * 6/46 = 0.0769 (5th card)

Total: 0.12 + 0.1078 + 0.0956 + 0.0863 + 0.0769 = 0.487 (rounded up to get rid of the least significant digit.

Note: the 5 equations can be combined to this one: 6/50 * (1 + 44/49 + 44/49*43/48 + 44/49*43/48*42/47 + 44/49*43/48*42/47*41/46) = 0.487

You can verify this by calculating 1 - (chance of hitting nothing)

1 - (44/50 * 43/49 * 42/48 * 41/47 * 40/46) = 0.487

It's much easier to calculate the probability of NOT hitting your cards and then subtract that from total (one), however, it's also good to understand the straight forward logical math too (calculating the probabilities of each individual events and then adding them properly)

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For other combination of cards on the river, you will need to first identify how many distinct combo they can be in, and then calculate the exact probability for one variation of that combo... For example, calculating Trips (with two different hole cards):

You want the two cards to match one of your hole cards, so you have this many situations: 11---, 101--, 1001-, 10001, 011--, 0101-, 01001... etc... That's 4+3+2+1 = 10 combinations. (where 1 is success, 0 is fail, - is doesn't matter)

So for 11---, you get:

6/50 * 2/49 * 1 * 1 * 1

for 101--, you get:

6/50 * 47/49 * 2/48 * 1 * 1

for 1001-, you get:

6/50 * 47/49 * 46/48 * 2/47 * 1

... I will not list all 10, but you SHOULD understand the pattern...

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  • I don't disagree with what you're saying but this should be a comment, not an answer. This is not how StackExchange work: you are not supposed to answer "try again", no matter how correct your comment. Answers should be left to actual answer and people shouldn't upvote answers which aren't answers.
    – TacticalCoder
    Dec 15, 2014 at 15:35
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    Thanks, I am just used to helping people find the answer their way as opposed to providing the answer. I updated mine with the actual answer.
    – Ying Li
    Dec 15, 2014 at 18:41
  • i think i understood it :D Good answer Liy. So, for the Trips, i have then to Sum all the 10 combination results?
    – mark
    Dec 21, 2014 at 8:47
  • So, the chances to hit a pair with my higher card at river is 1-(47/50 * 46/49 * 45/48 * 44/47 * 43/46) ?
    – mark
    Dec 21, 2014 at 9:31
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    @mark Yes, for the trip, add all ten combinations and you get the the final odds. Yes, for hitting the higher card by river is exactly what you wrote. You see, UNDERSTANDING the math really helps you create your own equations. You can logically figure out all other combos by analyzing it further :)
    – Ying Li
    Dec 21, 2014 at 23:00
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if I hold two different cards, what's the probability I get a pair at river?

Probabilities computation can quickly get very complicated and you need to be very precise in what you want to exactly compute.

What if the board comes with a pair in it? Do you consider that "getting a pair at river" or do you only want to know the probability of making a pair with one of your two holecards? Do you want the probability to only have exactly one pair at river or to have at least one pair?

At flop, if your holecards do not make a pair, you have:

p = 44/50 * 43/49 * 42/48 = 0.675

to not get a card matching one your holecards (but there's still a small probability you could hit a flush if your two holecards are of the same suit or a straight if your holecards are "close" enough).

So the probability to get, on the flop, at least one of the three cards matching one your holecards is: p = 0.324.

On the river that would be:

p = 1 - (44/50 * 43/49 * 42/48 * 41/47 * 40/46) = 0.487

So, to answer your question, if I'm not mistaken you have 48.7% chance to hit one or more cards matching the rank of one of your two holecards (if your two holecards are different) by the river.

How does one calculate the other winning combinations?

I don't know exactly what you mean by "other winning combinations": do you mean other odds to improve? Because the only "winning combination" you ever get for sure in Texas Hold'em is when you're holding the "nuts" (the absolute best poker hands). For example you can hold a straight flush and still lose to a higher straight flush...

There are several page explaining poker odds, including the Wikipedia one (which is quite advanced):

http://en.wikipedia.org/wiki/Poker_probability_%28Texas_hold_%27em%29#After_the_flop

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