PLO8 Math Question

wagon596 · Mar 2, 2015

I play a PLO8 Tournament almost every day. Was wondering if there are any math wizards that can answer my question.

Say at a 8 player table I have an Ace and a Deuce in my hand. What are the odds someone else is holding another Ace-Deuce? It seems to happen fairly often, was just wondering if someone could let me know how often?

I'm a percentage player, but have no idea how to figure that question. I hate getting quartered or worse.

Thanks

enolan · Mar 2, 2015

Probability that a single opponent has the same pre-flop non-pocket pair hand as you is ...
6/50 * 3/49 = 0.007347 i.e. 0.7347%
(in your example this would be 3 Aces OR 3 Twos from 50 unknown cards times 3 from 49 for the 2nd card to match yours)

For multiple players at the table:
Take the complimentary probability (i.e. 1-0.007347) and raise it to the power of the number of players. Then take the compliment of that value as your answer (and multiply by 100 to get the percentage value you looking for).

So for 5 other players
(1-0.007347) * (1-0.007347) * (1-0.007347) * (1-0.007347) * (1-0.007347)
= 0.9638
Then 1-0.9638 = 0.0362 which as a percentage is 3.62%

For a full ring (8 other players)
Probability at least one opponent has the same pre-flop non-pocket pair hand = 1-0.9427 = 0.0573 or 5.73%

(This is a little higher than I would have guessed but it reminds me of the surprising statistic that on a football(soccer) field, if you include the referee, it's more likely than not that two players will share the same birthday)

wagon596 · Mar 2, 2015

enolan said:
Probability that a single opponent has the same pre-flop non-pocket pair hand as you is ...
6/50 * 3/49 = 0.007347 i.e. 0.7347%
(in your example this would be 3 Aces OR 3 Twos from 50 unknown cards times 3 from 49 for the 2nd card to match yours)

For multiple players at the table:
Take the complimentary probability (i.e. 1-0.007347) and raise it to the power of the number of players. Then take the compliment of that value as your answer (and multiply by 100 to get the percentage value you looking for).

So for 5 other players
(1-0.007347) * (1-0.007347) * (1-0.007347) * (1-0.007347) * (1-0.007347)
= 0.9638
Then 1-0.9638 = 0.0362 which as a percentage is 3.62%

For a full ring (8 other players)
Probability at least one opponent has the same pre-flop non-pocket pair hand = 1-0.9427 = 0.0573 or 5.73%

(This is a little higher than I would have guessed but it reminds me of the surprising statistic that on a football(soccer) field, if you include the referee, it's more likely than not that two players will share the same birthday)

Thanks, Are we talking Omaha Hi/Lo, with 4 hole cards ? I told you I'm not the brightest light on the Christmas Tree...

enolan · Mar 2, 2015

My bad! I switched to HE as soon as I started thinking about it.
So a slight change to the working ...

We have 0.007347 (from above) as the probability that two random cards are the specific ones you're thinking of (not a pair though).

For an Omaha hand of 4 cards there are 6 way to make a 2 two card pairing so the probability of an opponent having two specific non-paired cards is 6 * 0.007347 = 0.044082 (or 4.4%) (I think this is a bit of a cheat way to do this bit and a more accurate answer is 0.04328 or 4.3%}

For an 8 player game (7 other players)
The probability that at least one opponent has two specific non-pocket pair cards is (1- (1-0.044082)^7 )= 1-0.7294 = 0.2706 or 27%

Yip! you would see it quite often.

jj20002 · Mar 2, 2015

wagon596 said:
Thanks, Are we talking Omaha Hi/Lo, with 4 hole cards ? I told you I'm not the brightest light on the Christmas Tree...

for omaha, 4 cards are dealt to each player instead of two so the number of unknown cards are 48 instead of 50, so follow the formulae above considering 48 and 47 instead of 50 and 49,

if holding more than an ace or a deuce then the chances are reduced substracting each to the the same formulae

duggs · Mar 2, 2015

I think this is the wrong way to go about it, i think we should bayesian probability it by estimating a range for opponents given an action and see how heavy A2 is in that range given blockers.

enolan · Mar 3, 2015

jj20002 said:
.... so follow the formulae above considering 48 and 47 instead of 50 and 49 ...

oops ..
6/48 * 3/47 = 0.007979 i.e. 0.7979%

so the 8 handed Omaha figure should be
(1- (1-(6*0.007979))^7 )= 1-0.7094 = 0.2906 or 29%

duggs idea would give a more realistic statistic and be more valuable for decision making but also more complex (any takers?). My Omaha game is barely at novice level.

BearPlay · Mar 3, 2015

I'm not a math wizard but on the other side of the situation, I would just offer that having A2xx is not reason enough to pursue every street, or even past the flop, if your hand is not balanced (two-way) and/or with a backup to the nut low in case you are counterfeited, for the reason you stated: being quartered.

In many cases, folding A2xx, especially when you foresee the pot growing too large, is the better choice post-flop, when you consider that at most, you are aiming for 1/2 of the pot.

Our goal in PLO8 should always be to scoop, or at least win 3/4 of, the pot.

Best success to you as you move forward, Larry.

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