ITH Poker Forum

[dazzammm]

hi, just got the odds and probability book - good read.

I play alot of 6-max limit hold'em and always asking myself "ive hit top pair, whats the chance of someone else having top pair too"

ive done the following calculation - can some let me know if ive got the calc wrong. its all very rough and not 100% accurate - i would just like to know if the theory is right.

(ignoring a player is more likely to play,say, a king, than a 2)

first i worked out the probability of 1 player not having e.g Ace as 47/50 * 46/49 (considering 50 and 49 unknown and there are 3 aces left in the deck) = 0.88.

then worked out for 2 players, then 3 players, then 4 players etc upto 5.

it works out on average 0.87 that a player does not have an ace if you do

therefore
to work out probability of player having an ace is 1 - (0.87 to the power of "players") (*100 to get %age)

Sooooo.

chance of being headsup and youre opponent has also hit top pair is 13%.
2 players and either hitting top pair is 25%.
......
so finally if its a family pot (all 6 players) its 51% certain that if you have top pair so has someone else

does this sound right ???

[nsidestrate]

does this sound right ???

I don't have time to do the math right now, but it doesn't instinctively sound right.

[Fenris78]

hi, just got the odds and probability book - good read.

I play alot of 6-max limit hold'em and always asking myself "ive hit top pair, whats the chance of someone else having top pair too"

ive done the following calculation - can some let me know if ive got the calc wrong. its all very rough and not 100% accurate - i would just like to know if the theory is right.

(ignoring a player is more likely to play,say, a king, than a 2)

first i worked out the probability of 1 player not having e.g Ace as 47/50 * 46/49 (considering 50 and 49 unknown and there are 3 aces left in the deck) = 0.88.

I think this is a mistake. If you are asking "What's the chance of someone else having top pair too" that means that there are 5 cards already known, namely your hole cards plus the flop. Also, there should be only 2 cards remaining in the deck that make top pair, not 3, since you have one card for top pair in your hand while the other one is on the board.

So if you hold Ax and the flop is Axx, the chance for a random hand to not contain an A is 45/47*44/46 = 91.6%

Thus against 5 random starting hands the chance of no one having top pair as well is 64.4%

[dazzammm]

thanks fenris.

as soon as i came back to this post after a few days my mistake dawned on me. it becasue i was working on two examples at the same time and got the math confused (i was also looking at flopping 2nd pair with 1 overcard and chance of someone having top pair)

[Clint]

Is it all about the math though? I mean if you raise (a proper amount) and get called, you're up against another good Ace, a decent pair, a LAG player looking to break you, or a bad player who loves pretty painted cards. From a math perspective you need to know what's the likelihood that its another A or a Pair. But even that ignores the bizarre 2-pairs, hidden sets, or A-rag (that hits the rag) you'll encounter from LAG and BAD players.

The way I see it, either way, when you are debating your course of action and all you have is top pair, proceed with caution unless you're lucky enough to be playing against bad players. Top Pair only = a pot control hand. Manage the pot, don't BUILD the pot, again, unless bad players.

Post #1, out!

[mchilger]

As Clint says, this isn't so much an exercise in math, but an exercise in range analysis. Basically, take any starting hand chart and put your opponent on a range. For example, you raise in the cutoff and the button reraises, assign a 3-bet range to him - then determine based on combinations how many of those hands contain an ace.

One estimate to arrive at your answer is to use PokerStove (however, I don't think it subtracts the fact that you have an ace in your hand). Enter your hand and the range of hands you expect your opponent to be holding.

For example above, say you put the button on the following range: 55+, A9, A-7s, Kq, KJs, QJs. That range represents 13% of all hands. Of that range, take out all the hands without an ace, and you find that 7.1% include an ace. So 45% of the time he doesn't hit an ace. Now, once you check-raise the flop and he calls or 3-bets, of course you have new information.

Matthew

[dazzammm]

thanks for the comments.

i didnt make it clear but the original maths was based on the oppenents holding random hands (not ideal but made the maths easier). i just wanted a ROUGH idea how the number of players affected my chances if i hit top pair (the levels i was playing at oppenents holding random hands was very likley ).

[chrisjp]

Rough idea is fine as long as you consider it very rough.

Some players, Snyder calls them Acemasters, just love to play a hand with an Ace in it. Most others are more selective obviously.

The playing tendencies grossly affect the odds since it's not random anymore.

Still it's interesting. Like the probably that an Ace flops is tremendously affected by whether or not you hold an Ace (or two) in your hand.

Chris