What are the chances of someone in a 10 handed game being dealt a higher PP than you in the same hand?

It happened every once n a while on PP but now I'm playing on Inter Poker and I swear to god every time I have a low PP someone has a PP one or two numbers higher. It's getting quite annoying especially when I flop an over pair and these calling stations never raise or bet to define their hands, just call all the way down.

Thanks for the help.

4% for every PP higher than yours

There is a good chance this answer is completely wrong, but i think this may be right.

You have a 220:1 to get any 2 excact cards

So if there are 10 people playing then there is a 22:1 chance that someone has a pocket pair.

Someone confirm for me? BS or truth?

Um, pocket pairs aren't exact two cards.

For example, pocket Aces can be As-Ah, or As-Ac, etc.
But 220:1 are the right odds for a specifice pocket pair, e.g. Aces or Kings.

isnt this roughly 4%?? in which case you would simply add this for every possible higher PP?

[quote=Royal_Tour][quote=dna4ever]There is a good chance this answer is completely wrong, but i think this may be right.

You have a 220:1 to get any 2 excact cards

So if there are 10 people playing then there is a 22:1 chance that someone has a pocket pair.

Someone confirm for me? BS or truth?[/quote]

isnt this roughly 4%?? in which case you would simply add this for every possible higher PP?[/quote][/quote]

so, you have pocket 2's, so there is a 48% chance that someone has a higher pocket pair than you (4 x 12 higher pocket pairs)? If that is right, then all I got is WOW :wink:

so, you have pocket 2's, so there is a 48% chance that someone has a higher pocket pair than you (4 x 12 higher pocket pairs)? If that is right, then all I got is WOW :wink:

Umm. it sounded right at first.. Can you prove othewise?..

I really have no clue, this was an assumption., and is 4% off??

I know 50% seems like much., I guess., basically, you could say., with pocket 2's there is a 4% chance someone else has pocket pairs, and there is a 100% chance that if they do.. they are higher.. LOL..

It sounds like. 60% of the time, they work all the time. lol

Sorry if my post sounded sarcastic, as it was not meant to be. I am just as baffled on this as you may be. I was never very good with figuring out the odds and such (but very good at remembering them if someone tells me). :wink:

i edited

Ahhh ok i figured it out..

chances of hitting a pocket pair. any 2 cards, 17: 1 or 5.8%.

Lets say you have pocket 10's chances that someone has higher pocket pair is 68 : 1. or 1.4%


I'm almost certain this is correct.

all i know is if you're dealt KK the chance that someone has AA is 22:1

Where does the 17:1 come from? I think I follow the rest of your math:

4 (different pairs higher than ten)x 17 (odds of making a pocket pair according to Royal) = 68. 1 divided by 68 = 1.47%.

Seems extremely low, considering how many times we see high pocket pairs against each other. :?

The 17 comes from 13 cards 2->A. Odd of any pockets 220:1. Odds of all pockets 220:13 or 16.92%.

For you math guys out there, I wanted to work this out realistically, not just speculatively, so bear with me. This problem seems fairly straightforward, but is a bit tricky if you want to figure it out accurately.

We know there are 1326 possible 2 card combinations in a 52 card deck. This is derived from the following formula:

The number of combinations is as follows, where number = n, (52 in this case) and number_chosen = k, (2 in this case)

n!
_____

k!(n - k)!

We also know that there are 6 possible ways to make any 1 pocket pair (using the exact same formula as above only substituting n=4 and k=2).

From here, we simply sum the probablilities of each of our opponents having a higher pocket pair. So there are two variables we need:

How many opponents at the table?
How many higher pairs than ours? we'll call this var. "x"

Then the formula becomes (sorry, math doesn't really work in the forum. you get the idea.)

6x 6x
______ + ______ + ...... for each player at the table (*100)

1325 1324


(We start with 1325 in the denominator and count down because we hold 1 combination, and each additional opponent holds one combination.)

So running this through a few extreme examples:

We find that at a full, 10 player table, if we hold pocket deuces, the probablility that at least one player holds a higher pocket pair is 48.9962%. Now, if we're heads-up with those same pocket deuces, the probability of our opponent having a higher pocket pair goes down to 5.43396%.

We also find that at a full table, if we hold pocket queens, the probability that at least one player holds a higher pocket pair is 8.16603%. With one less opponent at the table, that probability becomes 7.25625%. Now, with QQ heads-up, the probability that our opponent holds KK or AA is a mere 0.90566%.

Hope that helps!

so, you have pocket 2's, so there is a 48% chance that someone has a higher pocket pair than you (4 x 12 higher pocket pairs)? If that is right, then all I got is WOW :wink:

This is a decent rough estimate, but it only applies to a full table. If you're playing short-handed, the probablility goes way, way down.

Thank you MK.. I assumed i was waaaaay off., but i guess i had the right idea.