Everyone gets a 5 card poker hand to start. We frequently play Jacks or better to open. After someone opens everyone that stays in gets an additional 3 cards - no cards are discarded. So basically you have 8 cards to make your best 5 card hand.

My question is - what is the probability that a two pair hand improves to a full house? My gut tells me probably somewhere between 20-25% but not versed enough in statistics to confirm.

All input appreciated.

Quote:linksjunkieNeed a little help with a new 5 card poker variant we’ve recently added to our home game

Everyone gets a 5 card poker hand to start. We frequently play Jacks or better to open. After someone opens everyone that stays in gets an additional 3 cards - no cards are discarded. So basically you have 8 cards to make your best 5 card hand.

My question is - what is the probability that a two pair hand improves to a full house? My gut tells me probably somewhere between 20-25% but not versed enough in statistics to confirm.

All input appreciated.

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Greetings!

This should be a pretty easy question to answer.

You have Two Pair and are specifically looking to improve to a Full House, I'd prefer quads, myself, but Full Houses are also good.

Cards: 52

Known Cards: 5

Unknown Cards: 47

Cards to be Drawn: 3

Since we already have a four-of-a-kind on the brain, let's get that knocked out! We're going to make your hand:

J-J-Q-Q-5

Jacks Remaining: 2

Fives Remaining: 3

Queens Remaining: 2

4OAK

The first thing we should do is look at our easiest ways to make quads. The easiest ways would be to get the other two Jacks or Queens:

https://web2.0calc.com/

When it comes to getting the other two Jacks or Queens, we really don't care what else comes out, so we will treat it all as just the other 45 cards.

nCr(2,2)*nCr(45,1)/nCr(47,3) = 0.0027752081406105

We want to multiply this by two because it can either be Queens or Jacks, so:

0.0027752081406105 * 2 = 0.00555041628

1/0.00555041628 = 180.166666706

Which means that you are about 1 in 180.16666671 to get a Four-of-a-Kind of either Jacks or Queens. You can't get a Four-of-a-Kind in both, because you would then need to be drawing four cards.

The second you of Four-of-a-Kind you can get is fives, which would require that you draw all three other fives:

nCr(3,3)*nCr(44,0)/nCr(47,3) = 0.0000616712920136

As you can see, that is much less likely and only occurs 1/0.0000616712920136 = 1 in 16,215 attempts in this scenario. That's okay; quad jacks and queens are better, anyway.

Let's add that result to our result for Jacks and Queens to get:

1/(0.00555041628 + 0.0000616712920136) = 1 in 178.186813226

Full Houses

Full Houses are slightly trickier than they normally would be because you don't discard anything. For that reason, we will start with a Full House, "Queens Full of Anything," because that is the best ranking Full House possible. The good news is there will always be a best-ranking Full House possible, at least, of cards we already have in hand; though I will admit that the math is going to end up being slightly different if the kicker to the Two Pair is HIGHER than the rank of either of the Two Pair, or both of the Two Pair, but, in this case, Queen is also the highest ranking card.

If you have a hand like AQQJJ, however, Aces become the highest ranking Full House. That won't change things very much (in terms of overall probability) but it does change things a little.

The reason it would change things is because, in this hand, if we get ONE queen, then we could get one or two other fives and it wouldn't make a difference that our best hand is Queens Full of Jacks. In the AQQJJ hand, our best Full House would be Aces Full of Queens, in the event we got two aces.

All of the math in the Four of a Kind section would apply to anything.

Anyway, your Mission, should you choose to accept it, is to do the math for Full Houses on the AQQJJ hand; you should be able to do it by doing something similar to what I am about to perform!

FULL HOUSE-QUEENS FULL OF?

Okay, so what we are going to do here is we want to get a Queen, but not both, and we don't want to get the other two Jacks; getting one Jack doesn't change anything, so let's do the math for both zero jacks and one jack:

ZERO JACKS:

nCr(2,1)*nCr(2,0)nCr(43,2)/nCr(47,3) = 0.1113783533765032

ONE JACK:

nCr(2,1)*nCr(2,1)nCr(43,1)/nCr(47,3) = 0.0106074622263336

With that, we will add these two things together to determine the probability of getting a Full House in Queens. Remember, we have already done the math for Four of a Kind Jacks, so we really don't care if the third card (to the Jacks) is another Queen because we have 4OaK already.

0.1113783533765032 + 0.0106074622263336 = 0.1219858156 or 1/0.1219858156 = 1 in 8.1976744188

Of course, we could also do better than the Jacks and get a Full House Queens full of Kings, or Aces, but we really don't care for these purposes. We're just going to do all Full Houses Queens Full of Anything, which this covers. The Four of a Kind in Fives is irrelevant because, if we got three fives, then we did not get a Queen anyway.

FULL HOUSE IN JACKS:

For the Full House in Jacks, once again, the fives are immaterial as Jacks Full of Queens (that we already have) is better anyway. However, we can't draw a Queen as one of the other two cards, otherwise, we have a Full House Queens Full of Jacks, Kings or Aces, depending how the other two cards go.

Okay, so we want no Queens whatsoever:

nCr(2,1)*nCr(2,0)*nCr(43,2)/nCr(47,3) = 0.1113783533765032 or 1/0.1113783533765032 = 1 in 8.97840531561

As you can see, this will be the only probability we have to deal with. We can't possibly get Quad Fives because we have a Jack and two cards is fewer than three. We could get a Jack with two Kings, or Aces, but that's immaterial to us getting a Jack, but not a Queen.

FULL HOUSE IN FIVES:

Finally, we could get Fives Full of something, but that would only happen if we got Two Fives, No Jacks and No Queens, so let's do that:

nCr(3,2)*nCr(2,0)*nCr(2,0)*nCr(40,1)/nCr(47,3) = 0.0074005550416281 or 1/0.0074005550416281 = 1 in 135.125

The left hand numbers inside of all parenthesis on the left side of the divisor add up to 47, and add up to three on the right, so I think we have this right.

Once again, this is the probability of getting two fives, but not three fives and the third card being neither a Jack or a Queen.

FULL HOUSE IN ANYTHING ELSE:

What, you thought we were done? No way! We have ten ranks remaining and, from all ten of them, we could conceivably draw three of the four available cards. Fortunately, since that is what would be required, no card can be anything else, which makes the math easy:

nCr(4,3)*nCr(44,0)/nCr(47,3) = 0.0002466851680543 * 10 = 0.002466851680543 or 1/0.002466851680543 = 1 in 405.375

I think we're about ready to wrap it up, so here we go:

FINAL HAND PROBABILITIES

Four of a Kind (Fives): 0.0000616712920136 or 1 in 16,215---->(0.0061671292%)

Four of a Kind (Queens): 0.0027752081406105 or 1 in 360.3333333333---->0.27752081406%

Four of a Kind (Jacks): 0.0027752081406105 or 1 in 360.333333---->0.27752081406%

Four of a Kind (ANY): 1/(0.00555041628 + 0.0000616712920136) or 1 in 178.186813226 (Approximate, remember, we rounded off a little on Queens + Jacks)---->0.561208757%

Full House (Queens Full): 0.1219858156 or 1 in 8.1976744188---->12.19858156%

Full House (Jacks Full): 0.1113783533765032 or 1 in 8.97840531561---->11.13783533765032%

Full House (Fives Full): 0.0074005550416281 = 1 in 135.125---->0.74005550416281%

Full House (A/K/10/9/8/7/6/4/3/2 Full) = 0.002466851680543 or 1 in 405.375---->0.2466851680543%

With that, we must sum up the potential Full Houses:

0.1219858156+0.1113783533765032+0.0074005550416281+0.002466851680543 = 0.24323157569 or 1 in 4.11130831662 or 24.323157569%

Full House ANY: 0.24323157569 or 1 in 4.11130831662---->24.323157569%

ANY Full House OR Quads: (0.00561208757+0.24323157569) = 0.24884366326 or 1 in 4.01858736083 or 24.88436626%

With that, you're close enough to 1 in 4 to get either a Full House or Quads that you'll likely not play enough hands to notice the difference.

You probably tried to catch me forgetting to include the other ten Full Houses; I almost did---you're a sly one!

Quote:odiousgambitif I was in this game, I'd figure the expected best hand would be similar to 7 card stud, only somewhat better of course. So if there were 6-7 players, two pair wouldn't be worth a hoot. 3 of a kind would seldom be worth playing, and a straight could be in trouble. Of course if there are fewer players, things change.

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High three of a kinds would be strong plays for the Full House, not to mention the potential for quads, which would always have a chance at winning.

I used the old rule of thumb for Texas Hold’em to guesstimate the odds. The rule I’m referencing is to multiply your outs by 4x after the flop to estimate your chances of catching let’s say a flush if one card needed - 9 x 4 = 36%. After the river then by 2x. So my thinking was 6x might be appropriate with three more cards added to my hand.

This may be way off base but your math seemed to validate my short cut.

Thanks again!!

Straights? Flushes? Those odds are heavily dependent on what the two pairs are, what the odd single card rank is, what suits are in the 5 initial cards, etc…. There will be overlaps…. A full house hand could be with a flush. Or a straight.

Draw AJJ

Two 4OAK

On the simplest level, and one I see newbies fail at, a player is evaluating their final hand and what the chances are that they have the best hand. So that was where I was at with my comments.Quote:Mission146Quote:odiousgambitif I was in this game, I'd figure the expected best hand would be similar to 7 card stud, only somewhat better of course. So if there were 6-7 players, two pair wouldn't be worth a hoot. 3 of a kind would seldom be worth playing, and a straight could be in trouble. Of course if there are fewer players, things change.

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High three of a kinds would be strong plays for the Full House, not to mention the potential for quads, which would always have a chance at winning.

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Your mathematical approach reminds me of a couple of things:

* every once in a while you see someone claim he takes 'pot odds' into consideration for his decisions. I never see that shot down as impractical, beyond what I can do, which is to notice if the pot is small, big, or in between. But a 'pot odds' guy will claim that he sees the pot paying 5:1 while he has a 1 in 4 chance of hitting a flush ... I guess 5:1 is good enough for those times he still loses with a flush when he hits. I can't do this except by gut, but I think you need that in your evaluation if you're going the math route

* in the movie 'Cincinnati Kid' , during the big poker scene, there is a player who has a pencil and notepad and he's calculating the odds and probabilities openly. At one point he protests to the other players he is losing even though he's playing mathematically perfect. It's kind of overdone, who wouldn't want to be more discrete? but gets the point across that poker isn't just about who's good at that ... after all a very good player may be pretty mediocre at that aspect of the game.

>>>

The showdown and coming in second best hand, yep, when I lose big that's how it happens, coming in second all the time.

- playing position

- playing the player

- Pot odds

- Playing the stack sizes (including ICM: Integrated Chip Modeling)

- playing the tells (ex: a player's hands)

- mixing up your level of aggression, your ranges (range shanking)

- bet sizing

- even acting, giving false tells

- playing GTO (Game Theory Optimal) versus exploitive

But the games I play in involve semi-professionals and a high level of quality

as for my abilities, I just take more chances when it is a huge pot and for some reason the betting went back down to smaller bets ... I mean, you don't have to be a good poker player to realize several things improved for you, like 'somebody didn't hit I guess' . And you have the factor of low cost to stay in meanwhile you are presumably about to lose a wad by folding.

Our home games ? we play for peanuts for most of them I can get into [potentially 4 different games depending on if they need a player] Am a regular at 2, those are the small stakes games

Quote:linksjunkieThanks Mission. Very thorough response Greatly appreciated.

I used the old rule of thumb for Texas Hold’em to guesstimate the odds. The rule I’m referencing is to multiply your outs by 4x after the flop to estimate your chances of catching let’s say a flush if one card needed - 9 x 4 = 36%. After the river then by 2x. So my thinking was 6x might be appropriate with three more cards added to my hand.

This may be way off base but your math seemed to validate my short cut.

Thanks again!!

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I also want to mention that the overall probability of a Full House would be the same regardless of whether the fifth card is an over card, under card or in-between the Two Pair.

Comparing:

QQJJ5

to

QQJJA

All that happens is that a draw of 55J or 55Q becomes either Queens or Jacks full of Queens or Jacks, so the probability of the Q or J FH increases accordingly. With AAQ or AAJ draws, it does the opposite as the Full House becomes Aces full, but the probability of the Aces Full decreases the probability of the Jacks/Queens full to the same extent that the probability of Queens or Jacks full increases as a result of draws like 55J or 55Q just become Jacks or Queens full.

Thus, the overall probability of a Full House is unchanged; it just gets redistributed.

Quote:SOOPOOIt’s WAY more complicated than Missions post implies.

Straights? Flushes? Those odds are heavily dependent on what the two pairs are, what the odd single card rank is, what suits are in the 5 initial cards, etc…. There will be overlaps…. A full house hand could be with a flush. Or a straight.

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Why? He asked about Full Houses; I just went into quads as a bonus.

Also, no offense meant, but if you plan on betting straights in the second betting stage of this game, I'll play it with you all day.

Quote:Mission146Quote:SOOPOOIt’s WAY more complicated than Missions post implies.

Straights? Flushes? Those odds are heavily dependent on what the two pairs are, what the odd single card rank is, what suits are in the 5 initial cards, etc…. There will be overlaps…. A full house hand could be with a flush. Or a straight.

link to original post

Why? He asked about Full Houses; I just went into quads as a bonus.

Also, no offense meant, but if you plan on betting straights in the second betting stage of this game, I'll play it with you all day.

link to original post

I laughed out loud.

Quote:odiousgambitOn the simplest level, and one I see newbies fail at, a player is evaluating their final hand and what the chances are that they have the best hand. So that was where I was at with my comments.Quote:Mission146Quote:odiousgambitif I was in this game, I'd figure the expected best hand would be similar to 7 card stud, only somewhat better of course. So if there were 6-7 players, two pair wouldn't be worth a hoot. 3 of a kind would seldom be worth playing, and a straight could be in trouble. Of course if there are fewer players, things change.

link to original post

High three of a kinds would be strong plays for the Full House, not to mention the potential for quads, which would always have a chance at winning.

link to original post

Your mathematical approach reminds me of a couple of things:

* every once in a while you see someone claim he takes 'pot odds' into consideration for his decisions. I never see that shot down as impractical, beyond what I can do, which is to notice if the pot is small, big, or in between. But a 'pot odds' guy will claim that he sees the pot paying 5:1 while he has a 1 in 4 chance of hitting a flush ... I guess 5:1 is good enough for those times he still loses with a flush when he hits. I can't do this except by gut, but I think you need that in your evaluation if you're going the math route

* in the movie 'Cincinnati Kid' , during the big poker scene, there is a player who has a pencil and notepad and he's calculating the odds and probabilities openly. At one point he protests to the other players he is losing even though he's playing mathematically perfect. It's kind of overdone, who wouldn't want to be more discrete? but gets the point across that poker isn't just about who's good at that ... after all a very good player may be pretty mediocre at that aspect of the game.

>>>

The showdown and coming in second best hand, yep, when I lose big that's how it happens, coming in second all the time.

link to original post

As I understood it, the player draws three additional cards, so they aren't community cards. Because of that, assuming I'm correct, you would have no idea what the other player is holding or what they could be holding. That's not to say that, "Pot Odds," aren't useful in this situation, but I don't think they would be as much of a practical consideration as Texas Hold 'Em, where as the streets come out, you can narrow what your opponent might have.

With this, the only thing you know is that your opponent has eight total cards and stayed in with five total cards. I guess really high flushes could be good in this game, especially if it's a heads-up or three-way sort of game, but I'd be pretty hesitant to ever bet all but Ace-High flushes unless I was semi-bluffing.

The thing you have to look at is, suppose a player is starting with:

888JQ

What you get into with this is the opponent has three ways to make quads, which are any eight, three queens or three jacks.

nCr(1,1)*nCr(46,2)/nCr(47,3) = 0.0638297872340426

nCr(3,3)*nCr(44,0)/nCr(47,3) = 0.0000616712920136 * 2 = 0.0001233425840272

Of course, the real concern in this scenario is the Full House. Holding Two Pair, we've already established that the opponent is roughly 25% to get either quads or a Full House, of course, most of that comes from the Full House. In this scenario, the opponent is more than 6% just to get the quads because he is holding trips.

We also have no idea if we are blocking quads.

The Full House is the real problem here; there are a lot of ways to get one. The first of these is that any single Jack or Queen leads to eights full. Before we get into that, however, we have to look at Full Houses with two jacks or two queens.

nCr(3,2)*nCr(1,0)*nCr(43,1)/nCr(47,3) = 0.0079555966697502 * 2 = 0.0159111933395004*

*The nCr(1,0) is because the eight makes quads.

In the case above, a Full House in Jacks is not foreclosed by a Full House in Queens because we are only drawing three, so you can't have two more of both.

The next type of Full House to look at is to draw two of any of the other ten ranks, which is superseded only by drawing an eight, as that would make quads.

nCr(4,2)*nCr(1,0)*nCr(42,1)/nCr(47,3) = 0.0155411655874191 * 10 = 0.155411655874191

The next thing that could happen is we could draw a single queen or jack, which would give us eights full of Queens or Jacks. The first thing that we have to do is come up with the probability for that, but we must subtract from that probability the probability of drawing a Queen or Jack, while also drawing two of the same of some other card, as we have already included all of those draws above.

nCr(3,1)*nCr(1,0)*nCr(43,2)/nCr(47,3) = 0.1670675300647549 * 2 = 0.3341350601295098 (Ignores Getting Two of Something Else)

Next step is to subtract from those times we get a Queen or Jack, but also get two of something else:

nCr(3,1)*nCr(1,0)*nCr(4,2)*nCr(39,0)/nCr(47,3) = 0.0011100832562442 * 10 = 0.011100832562442

There are two ways this can happen, Queen + Pair or Jack + Pair, so: 0.011100832562442 * 2 = 0.022201665124884

0.3341350601295098 - 0.022201665124884 = 0.3119333950046258

In total, if someone holds trips, the can get:

Quads: 0.0638297872340426 + 0.0001233425840272 = 0.0639531298180698

Full Houses: 0.3119333950046258 + 0.0159111933395004 + 0.155411655874191 = 0.4832562442183172

Full House or Quads: 0.4832562442183172 + 0.0639531298180698 = 0.547209374036387

Conclusion

With that, we establish that an opponent holding Two Pair is effectively 25% to get either Quads or a Full House. In the meantime, the player holding trips is more than 54.7% to end up with either Quads or a Full House.

It's important to remember that, depending on the composition of the 2P or Trips being held, the opponent could also get a better flush than you end up with.

We also know nothing about the opponent's hand as it sounds like there are no community cards and the opponent does not discard.

My tendency is to think that Straights are absolutely unplayable in the final stage and you should not stay in if your most likely best hope is probably a straight. I would also suggest that all but the best flushes, as a finishing hand, are borderline unplayable and I would almost certainly not draw to a Flush (without Straight Flush possibilities) or call in the final stage with even a high flush (except maybe Ace) against multiple players. That said, some high flushes might be good for a semi-bluff if you can get a read on the opponent.

In my opinion, High Flush is the cutoff in this game for what constitutes a betting hand in the final round. I would expect most pots of any meaningful size to be taken down by FH, or better.

Quote:Mission146Quote:SOOPOOIt’s WAY more complicated than Missions post implies.

Straights? Flushes? Those odds are heavily dependent on what the two pairs are, what the odd single card rank is, what suits are in the 5 initial cards, etc…. There will be overlaps…. A full house hand could be with a flush. Or a straight.

link to original post

Why? He asked about Full Houses; I just went into quads as a bonus.

Also, no offense meant, but if you plan on betting straights in the second betting stage of this game, I'll play it with you all day.

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Your answer went far beyond just telling him the odds of hitting a full house. Had you simply answered that, I wouldn’t have added my comment on top of it. Someone even mentioned 3 of a kind, which although beating the two pairs, would frequently lose to straights and flushes. If you have 8s8c9s9c7s I think you are doing a disservice by not figuring in the straight flushes as well, as a simple example.

The bottom line is two pair sucks!

Quote:SOOPOOQuote:Mission146Quote:SOOPOO

Straights? Flushes? Those odds are heavily dependent on what the two pairs are, what the odd single card rank is, what suits are in the 5 initial cards, etc…. There will be overlaps…. A full house hand could be with a flush. Or a straight.

link to original post

Why? He asked about Full Houses; I just went into quads as a bonus.

Also, no offense meant, but if you plan on betting straights in the second betting stage of this game, I'll play it with you all day.

link to original post

Your answer went far beyond just telling him the odds of hitting a full house. Had you simply answered that, I wouldn’t have added my comment on top of it. Someone even mentioned 3 of a kind, which although beating the two pairs, would frequently lose to straights and flushes. If you have 8s8c9s9c7s I think you are doing a disservice by not figuring in the straight flushes as well, as a simple example.

The bottom line is two pair sucks!

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It does suck! I actually did a little experiment with this:

https://www.random.org/playing-cards/

The first experiment was to simply draw 100 hands of eight cards, not particularly relevant, but here are the results:

Nothing: 6

Pair: 25

Two Pair: 36

Trips: 10

Straight: 10

Flush: 10

Full House: 3

Quads: 0

Straight Flush: 0

Royal: 0

That's why I said straights probably aren't good in this game. I don't know if maybe we, 'Ran well,' on Straight+, but this was 23/100 hands just playing blindly.

More relevantly, I wanted to take an example 100 hands of something that a very aggressive player might play, which is a pair of Kings. We're going to make this player aggressive, but not stupid, so I also gave him some two card straight and flush outs:

Kc Kd Qc 10d 5c

The way I did this was that I still drew hands of eight cards, but duplicate cards would be ignored, so we would always get three cards not matching the above. Here are the results:

Kings: 18

Two Pair: 47

Trips: 8

Straight: 10

Flush: 8

Full House: 9

Quads: 0

Straight Flush: 0

Royal Flush: 0

Sample sizing could come into play, but what we see is the only thing keeping a pair helped this player do was get more Full Houses. Flushes actually decreased somewhat.

My next idea was to take a more conservative player, but not too much so, and give him a pair of Kings with four-to-a-flush as well as some two-card straight outs, not that you should be betting on straights in the final round, because you shouldn't. We're also giving him a Royal chance:

Kc Kd Qc 10c 5c

Kings: 6

Two Pair: 29

Trips: 4

Straight: 1

Flush: 47

Full House: 10

Quads: 2

Straight Flush: 0

Royal Flush: 1

The two quads and Royal happened, so they're in there. Actually, one of the quads was we hit the other three fives, so that was kind of neat.

As you can see here, many of our lower paying hands become flushes when holding four to a flush, so flushes aren't to be considered an excellent hand in this game. If you think about it, here are the cards we don't know:

Cards Known: 5

Cards Unknown: 47

Flush Outs: 9

With that, you figure that we can catch any one of nine cards, with three cards to do it and our hand is going to be at least a flush. That doesn't even foreclose us from quads or boats. Let's look at the probability of missing any kind of flush whatsoever, even though we could still make other hands:

nCr(9,0)*nCr(38,3)/nCr(47,3) = 0.520259019426457

With that, we are about 52% to not get a Flush of any variety, which includes Straight Flushes or Royals. Of course, there is still lots of potential to get a Full House, as you can see, because we got ten in the sample. Certainly some of them might have also been flushes, but the FH supersedes (I should have maybe tracked that), but most of them probably weren't.

Everyone get dealt 5 cards, you may use 0-5 of these hole cards in making your best 5-card poker hand.

Bet.

Flop with three cards. bet

Turn card. Bet

River card. Bet

Best 5-card poker hand wins.

In general, my experience was that it usually takes a full house to win.

Also, Omaha-5 which is very similar to the above game, but your final five card poker hand must include 2 of your 5 hole cards and three of the board cards.