Mathematical Question

Suppose we have a card game with a reduced deck (7's to aces's) with a total of 32 cards. Each player is dealt two cards face down and then one card is dealt face up. 12 or 13 players may play this game at any one time. What is the probability of a player being dealt a pair of Aces. What about being dealt trips (such as QQ and the first upcard being a Queen).. I also have more question to come later, but for now any help would greatly be appreciated and these questions are very important so i hope some of the advice is accurate.


Thanks again,

MJ

"Suppose we have a card game with a reduced deck (7's to aces's) with a total of 32 cards. Each player is dealt two cards face down and then one card is dealt face up. 12 or 13 players may play this game at any one time."


If each of twelve players is dealt three cards, wouldn't there be a total of 36 cards dealt? Wouldn't you run out of cards before you were finished dealing three cards to everyone?


I think ten is the maximum number of players, if there are only 32 cards in the deck, and if each player is dealt 3 cards.


1. Your probability of being dealt a pair of aces and one other card is 6*28*6/(32*31*30) = 0.03387. Regardless of the number of opponents, you should expect a pair of aces once in every 29 or 30 hands.


2. If you have nine opponents, if you do not hold an ace yourself, if the opponent on your left has an ace showing, and if no other ace is showing, the probability of the opponent sitting on your left holding one ace (plus another card which is not an ace) in the hole is 3*17*2/(20*19) = 0.268.


3. If you have nine opponents, if you do not hold an ace yourself, and if no aces are showing, the probability of the opponent sitting on your left holding a pair of aces in the hole is 6*2/(20*19) = 0.03158.


4. If you have nine opponents, if you do not hold an ace yourself, and if no aces are showing, one or more of your opponents will have a pair of aces in the hole slightly more often than one hand out of four.


5. If you have nine opponents, and if you do not have an ace yourself, at least one of your nine opponents will have a pair of aces (either with one ace up and the other down or with both aces in the hole) approximately one third of the time.


I'm not a mathematician and there is never a guarantee my math is correct, but numbers 1, 2, and 3 above, for which I have shown the set-ups, are fairly simple - and someone is usually kind enough to correct me if I have made a mistake.


Numbers 4 and 5 above are quite a bit more difficult to solve rigorously and I have only shown the approximate answers.


Hope this helps.


Buzz

I think MJ means that the third card is a community card. Also if the deck contains Aces through 7's that is 4*7=28 cards not 32.


Getting a pair of Aces in the hole:

The number of ways to choose 2 cards from a set of 28 is


28!/(26! * 2!) = 28*27/2 = 378


The number of ways to choose 2 out of the 4 Aces is 6, so the probability of being dealt two Aces is 6/378 = 0.016 or 62 to 1.


Getting trips:

In a deck of Ace's through 7's you can't get trip Queens!


The number of ways to choose 3 cards from a set of 28 is


28!/(25! * 3!) = 28*27*26/6 = 3276


There are 7 ranks for the trips and 4 ways to choose the trips for each rank. 7*4 = 28 ways to have trips. So the probability is 28/3276 = 0.009 or 116 to 1.


If you need accurate answers you will need to formulate more precise questions! Hope this helps.

Another way to look at the trips problem:


The third card is a community card so the probability of having a pair that matches it is calculated as follows. You are now choosing 2 cards from the remaining 27. This can be done in 27*26/2 = 351 ways. The number of ways to choose 2 cards from the remaining three that match the rank of the upturned card is 3. So the probasbility of having a pair that matches the upturned card is 3/351 or 116 to 1.


Same answer, different method.

"I think MJ means that the third card is a community card."


Chaos - That would make it possible to have more than ten players, and would make all the values I calculated meaningless. After re-reading MJ's post starting this thread, I think your interpretation on this point is better than mine was, although I think you will agree it's not crystal clear.


"Also if the deck contains Aces through 7's that is 4*7=28 cards not 32."


Depends if you discard high cards or low cards. A deck with AKQJT987 would have 32 cards.


Buzz

Buzz, I had not thought of A down though seven. The trip Queens should have tipped me off.


Redoing my math:


Getting a pair of Aces in the hole:

The number of ways to choose 2 cards from a set of 32 is


32!/(30! * 2!) = 32*31/2 = 496


The probability of being dealt two Aces is 6/496 = 0.012 or 81.66 to 1.


Getting trips:

The number of ways to choose 3 cards from a set of 32 is


32!/(29! * 3!) = 32*31*30/6 = 4960


There are 8 ranks for the trips and 4 ways to choose the trips for each rank. 8*4 = 32 ways to have trips. So the probability is 32/4960 = 0.0065 or 154 to 1.

Sorry guys, i made a mistake. Each player receives two hole cards and one community card is dealt. A deck (7's throug to Aces) meaning only 28 cards.


I want understand these probabilities, also i would like to know what the probability of someone holding a pair bigger than your's is. To make it clear i will give an example -


Suppose I hold JJ and the first community card is a 7. What is the probability that someone holds QQ, KK, or AA and how do calculate these probabiliies in different situations???


THANK YOU ALL ALOT, i greatly appreciate it.

MJ

Another Question: I hope i am not being a jerk with all these questions, but as i mentioned they are very important since i will be playing in a huge no limit high stakes game in a month and i want to learn everything i can (the buy in is $10000 ) and its not a tourney lol.


Suppose you hold J9 and the first community card is a 9. What is the probability of catching a Jack to make two pair. I know you have 3 outs so how do you calculate this.


-MJ

WAIT A SECOND. After rereading the previous responses i am sure i am right (7, 8, 9 ,10, J, Q, K, A) that is 8 * 4 = 32 cards.


??????

-MJ


(Sorry for being so stupid)

Chaos - You based your calculation on not knowing the whereabouts of 32 cards. However because you have two cards in your hand, before the one-card flop there are 30 unknown cards, and after the one-card flop, there are 29 unknown cards.


Your value is correct; it just doesn't consider all the information available to MJ when he has to decide how to bet. MJ will get a more useful value if he considers all the information at his disposal, which, of course, includes consideration of the cards he can see. Don't you agree?


MJ should be more interested in knowing the chances of an opponent having a pair of aces when he, himself, has a pair of kings or queens, or perhaps a single ace himself.


The number of 2 card combinations, when you have 30 cards from which to choose, is 435 (rather than 496). Similarly, the number of 2 card combinations, when you have 29 cards from which to choose, is 406 (also rather than 496).


However, none of these values should be of much interest to MJ. Why? Because it is not useful to know the probability of any particular opponent holding a pair of aces. Why? Because in a twelve handed game, MJ has eleven opponents. You can't simply find the probability of one particular opponent holding something and then multiply that probability by the number of opponents (although if the probability is small enough, as in this particular case, the value you get by doing so is very close to the result you would get by using a more rigorous method, such as "semi-deals").


What MJ really should want to know is how scary the possibility of a pair of aces is, when he, himself, has been dealt a particular hand, say a pair of kings.


Thus MJ should want to know how likely his opponents collectively are to have had a pair of aces distributed to at least one of them.


To cut the long story which I have started short, when MJ has a pair of kings (or anything else that does not include an ace, before he sees the flop, he should expect at least one of his eleven opponents has been dealt a pair of aces roughly one hand in seven. Once MJ sees the flop card, (and it is not an ace), the probability one of his eleven opponents was dealt a pair of aces is greater than it was before the flop, but not by much, approximately two hands in thirteen.


What should this all mean to MJ? Well, before the one-card flop, he can expect his pair of kings to get cracked by a pair of aces in the hole about one time out of seven, but the other six times out of seven, none of his eleven opponents will have been dealt a pair of aces in the hole. Thirteen times out of fifteen, when MJ sees an eight on the flop, he can figure none of his opponents was dealt a pair of aces. Although an opponent is likely to have been dealt a pair of aces more often in this game than in Texas hold 'em, a pair of kings in this game is even stronger than in Texas hold 'em, because there is only one card on the board instead of five.


What if the one-card flop is an ace? If MJ has a pair of kings, it doesn't matter whether an opponent was dealt one ace or two; an opponent with an ace has him beaten. If the flop is an ace, the probability that one or more of his eleven opponents was originally dealt either one or two aces is 0.9904.


Here's the calculation:

26 choose 22 = 14950 (no aces amongst MJ's foes)

29 choose 22 = 1560780 (total number of 22 card combinations taken from a pack of 29 cards)

14950/1560780 = 0.0096 (probability no ace has been distributed to MJ's foes)

1 - 0.0096 = 0.9904


In other words, roughly ninety nine per cent of the time, when the flop is an ace, at least one of MJ's opponents was dealt an ace. Since at least one of his opponents would probably see the flop with an ace, calling a bet with a pair of kings when the flop is an ace is very scary.


But when the flop is not an ace, a pair of kings is even stronger than in Texas hold 'em, in spite of the greater chance (in this game compared to THE) of an opponent having been dealt a pair of aces. That's because there is only one board card instead of five. The increased chance (compared to THE) of an opponent being dealt a pair of aces is more than offset by there being only one board card.


Similarly, when when MJ is dealt a pair of queens and the flop is either an ace or a king, calling a bet may be very scary, depending on how likely various opponents are to play AX or KX. Similarly, when MJ is dealt a pair of jacks, when the flop is an ace or a king or a queen, calling a bet may be very scary, depending on how likely various opponents are to play AX, KX, or QX.


Although I started this as a reply to Chaos, I'm sure you're reading it, MJ. Hope it helps, you find what you are seeking. I'll answer some of your other questions tomorrow. It's 4:20 a.m. here in L.A. and sleep beckons me.


Buzz

How many more cards are there to come? Is this game basically HE but with a betting round after each community card is revealed?


Some of Buzz's comments are based on their being only a single community card. Your pocket pairs will be vulnerable to someone pairing their overcard in the hole just as in HE. The difference hear is that you can try and bet them off their hand for each of the first three comminty cards.


If your opponents play this game regularly, they may already have most of the probabilities calculated. They probably also have some strategy and tactics specific to this form of poker alsready in their arsenal. You are probably a big underdog. And playing no limit will quickly punish your errors.

Buzz I agree the information you calculated is more useful to the play of the game.

It is also known as Asian holdem and Greek holdem.


It has only a small blind structure, with a single blind being posted by the dealer, who acts last on the first round, and first for all susequent rounds. This actually makes no difference to the game, as the player on the right of the dealer, always has the same positional advantage as the dealer in holdem.


Each player gets 2 cards (dealt both at the same time) from a 32 card deck with 6's through 2's removed.


A single community card is turned up and a betting round ensues. Then a 2nd, 3rd, 4th, and 5th card are turned one at a time, with betting at each stage.


Limits vary depending on the game. Assume the blind has a base value of 1. Common betting limits for the 5 betting rounds are as follows.


22224 22228 22448


Players most use exactly 3 cards from the board and both hole cards to make the best possible hand.


If you require strategy tips MJ feel free to ask. I won over $2,000 at this game in the last 4 sessions i have played, where the blinds were either $5 or $10 (dealers choice).

MJ - How many community cards in all are dealt? Are you writing about the same game that Pete has described (somewhere else in this thread)? (Pete has written that the game is called two card Manilla and there are 5 community cards dealt in all, but they are dealt one at a time).


"Each player receives two hole cards and one community card is dealt."


From the above quote, I get the impression that only one community card is dealt.


"Suppose I hold JJ and the first community card is a 7."


However, from the above quote, I get the impression that more than one community card is dealt.


Which is it? Makes a big difference in terms of reasoning.


However, to directly (but only approximately) answer your question, If you have a pair of jacks in the hole, and if there is a seven on the board, the approximate probability one of your eleven opponents has a pair of aces, kings, or queens is 0.42. In other words, after you have seen the first common card, and it is below a jack, you can figure that roughly two hands in every five an opponent will have been dealt a higher pair than your jacks. You had better see Pete about the strategy tips, but that would be often enough for me to feel very insecure about a pair of jacks in the hole, especially if this is a pot-limit or no-limit game. With a pair of jacks you're a slight (roughly three to two) starting favorite, but in pot limit that can come back to bite you.


You can't, IMHO, play poker well when you play scared, but in this game, one of your opponents is much more likely to have been dealt a pair higher than jacks than in the full deck game. In addition, if there are five community cards, one of your opponents is much more likely than in the full deck game to end up with a pair higher than jacks.


Let me restate that I'm not a mathematician. I only took up probability so that I could figure out odds for the game favored by the great Greek God, Zeus-aha-8.


/images/glasses.gif

Buzz

Good analysis Buzz. MJ you might want to read the below strategy advice.


In fact a pair of Jacks is playable only under special conditions. From any position if the first community card is a J (well duh), however even now expect to lose a lot of the time unless you can really thin the field. Other than that you can only play them for a limp with passive players still to act in late position if the upcard is a T,9,8 or 7.


JJ with a J up is a dangerous holding if the board does not pair, or if the board pairs high. Consider the following layout. You have JcJs with Jh up in early position but do not raise to trim the field, as you wish to build a large pot.


The board unfolds as so and 3 players are there at the end. JhKhAsAdTh. You are third if there is any significant action. Most likely you are losing to Aces full which is losing to the hearts flush. (Flush beats full in short deck try making a flush, you'll see why).


However if the board pairs lower, you are in big time action. Last night i won a sizeable pot in a $10 blind hand where i held TT with T up. Board finished T889K. This is not a dangerous layout without a flush card fallin 2nd or 3rd, and then completing on the end, as it is unlikely a player would be in with 88 and a T up, and it is also unlikely that a pair of Kings can handle the heat from all the raising between Fulls and straights and flush draws.

Chaos - I agree with you.


Some of MJ's statements lead me to believe there is more than one community card. For example, "Suppose you hold J9 and the first community card is a 9. What is the probability of catching a Jack to make two pair. I know you have 3 outs so how do you calculate this."


After MJ, holding jack-nine has seen the first community card, a nine, the location of 29 cards is still unknown. Assuming five community cards in all, there are 29*28*27*26/24 = 23751 possible 4-card combos for the remaining four community cards. (This is the same as "29 choose 4" if that makes more sense to you).


(1) There are 3*26*25*24/6 = 7800 two-card combos with exactly one jack.


(2) There are 3*26*25/2 = 975 two-card combos with exactly two jacks.


(3) There are 1*26 = 26 two-card combos with exactly three jacks.


(4) Since (1), (2), and (3) are mutually exclusive, there are 7800 + 975 + 26 = 8801 ways to end up with at least one jack on the board.


(5) The probability of at least one jack is 8801/23751 = 0.37.


(6) The probability of only one jack is 7800/23751 = 0.33.


I think MJ is really interested in the probability of at least one jack, rather than exactly one jack, and I think MJ is also really interested in the probability of another nine or two, as well.


One thing I have learned about probability is that the questions (as to what you really should want to know) are often originally unclear. Stating a question in a slightly different way leads to a different result. We may be able to answer MJ's probability questions, but I'm not sure the answers will be as helpful as some sage advice from Pete as to how to play the game.


When I first started playing Omaha-8, I wondered how often a low was possible. I knocked myself out figuring the answer, even invented some new kind of math to get the result. Took me about twenty written pages of calculations. I finally figured out how to do it a different, more conventional, way expressing the calculations in about one page. Turns out it doesn't make much difference. Why? Because what you really want to know is the probability of making low after you look at your cards, and then after you see the flop, and then after you see the turn. (My efforts weren't wasted because now I can easily do those more pertinent calculations). However, reading Ray Zee's Omaha-8 book was much more helpful, in terms of winning, than knowing the answer to my question about how often low was likely.


If you're reading this, MJ, I think you want to ask Mr. Peterson about general strategy tips.


I know nothing about strategy in the game you are planning to play, but off the top of my head I suspect playing jack-nine is a big money loser, regardless of how often you can improve when the first community card is a nine. I'd fold jack-nine before the single-card flop rather than put any additional money into the pot, and I'd also fold jack-nine to a bet after the single-card flop was a nine. And that's in a limit game. In pot limit or no limit, playing jack-nine will get you second place often enough to be financially disastrous, regardless of the probability of improving. That's just my opinion, offered in the spirit of helping you be successful.


Hope this helps.


Buzz

Pete - Thanks. I was interested in what you would have to say, even though it's doubtful I'll ever play the game. But if I ever do encounter the game, thanks to your tips I'll have a better idea of how to proceed.


Buzz

HERE IT IS THE ENTIRE GAME --------


Each players receives two hole cards from a deck of 32 card (aces through to sevens), then right after everyone is dealt there cards a community card is deat (there is no betting preflop).

The betting now goes from the left of blinds as in holdem but it is no limit, players can fold call or raise the bet up to the amount of their stack. After betting round is complete remaining players see the next community card, again another round of betting and then the third community up card and again another round of betting then there is a showdown.


Thats how the game is played,

Any strategies....

-MJ

Each player antes $5 and the blinds are $5 - $10 therefore a pot size raises thins the field dramatically and keeps the game either headsup or three way action.


-MJ