Pair vs. 4 to a flush
I guess my intuitive answer would be, only draw to the flush if you think your aces are completely smoked, not just losing.
Now if we're talking a lower pair vs. a flush draw, I'd be more inclined to draw to the flush.
If only one other player, keep the aces.
if a full table, go for the flush.
ZCore13
Quote: DRichIt all depends on the number of players you are playing against.
If only one other player, keep the aces.
if a full table, go for the flush.
You play in games where 5+ people see a draw? Sounds yummy.
Even three-way I'm drawing to the pair, four way is probably close.
However, if you really want to, I suppose you could factor in the probability of getting another Ace back to thereby result in the same hand, or better, so that's 2/47 = .0425532, giving a total of .1914894 + .0425532 = .2340426.
(As a practical matter, however, if you'd have gotten the Ace for a pair of Aces drawing one card, then you would have gotten it for---at least---Trips drawing three cards)
Holding the Aces, let's say the hand is (Ac Ad 10d 7d 4d)
The probability of the third card not improving the hand AND not being a 10, 7 or 4 is 36/47
The probability of the fourth card not improving the hand AND not being a 10, 7 or 4 is 32/46
The probability of the fifth card not improving the hand is 37/45, it doesn't matter what rank it is, at this point.
(36/47*32/46*37/45) = 0.43811285846
The probability of the third card not improving the hand and being a 10, 7, or 4 is 9/47
The probability of the fourth card not improving the hand and NOT being one of the other two ranks is 36/46
The probability of the fifth card not improving the hand is 38/45, rank doesn't matter.
(9/47*36/46*38/45) = 0.12654949121
The probability of the third card not improving the hand and NOT being a 10, 7 or 4 is 36/47
The probability of the fourth card not improving the hand and BEING a 10, 7 or 4 is 9/46
The probability of the fifth card not improving the hand is 38/45, rank doesn't matter.
(36/47*9/46*38/45) = 0.12654949121
The probability of the third card not improving the hand and BEING a 10, 7 or 4 is 9/47
The probability of the fourth card not improving the hand and BEING one of the two others is 6/47
The probability of the fifth card not improving the hand is 39/45, rank doesn't matter.
(9/47*6/47*39/45) = 0.02118605703
1-(0.02118605703+0.12654949121+0.12654949121+0.43811285846) = 0.28760210209
Thus, the overall probability of improving the hand is 28.760210209%.
Therefore, you are 0.28760210209-0.1914894 = 0.09611270209 or 9.611270209% more likely to have a better hand than a pair of Aces holding the Aces than going for the flush.
However, your argument concerns beating all of the other players in a game of five-card draw with four (or more) players. In a game of five-card draw, I'd probably hold the Aces and throw out a moderate sized bet to get slop to fold because slop is likely to either outdraw (and beat) me, or remain slop and fold during the post-draw betting, anyway, so I absolutely don't want any slop in the hand. However, if I put a bet out there and everyone calls, then I probably want to discard one of the Aces and go for the Flush because multiple people are probably drawing to Trips, 2P or another four-flush, anyway, so that's my best chance at the pot odds to draw to a hand that would beat that.
Others will notice you drew only 1 card this could affect the value of future betting.Quote: Mission146The first erroneous assumption is the statement that the four-flush will improve 1/4 times, specifically, you'll be left with nine Flush cards and the probability of improvement is 9/47 = 0.1914894, which is slightly under one in five times.
However, if you really want to, I suppose you could factor in the probability of getting another Ace back to thereby result in the same hand, or better, so that's 2/47 = .0425532, giving a total of .1914894 + .0425532 = .2340426.
(As a practical matter, however, if you'd have gotten the Ace for a pair of Aces drawing one card, then you would have gotten it for---at least---Trips drawing three cards)
Holding the Aces, let's say the hand is (Ac Ad 10d 7d 4d)
The probability of the third card not improving the hand AND not being a 10, 7 or 4 is 36/47
The probability of the fourth card not improving the hand AND not being a 10, 7 or 4 is 32/46
The probability of the fifth card not improving the hand is 37/45, it doesn't matter what rank it is, at this point.
(36/47*32/46*37/45) = 0.43811285846
The probability of the third card not improving the hand and being a 10, 7, or 4 is 9/47
The probability of the fourth card not improving the hand and NOT being one of the other two ranks is 36/46
The probability of the fifth card not improving the hand is 38/45, rank doesn't matter.
(9/47*36/46*38/45) = 0.12654949121
The probability of the third card not improving the hand and NOT being a 10, 7 or 4 is 36/47
The probability of the fourth card not improving the hand and BEING a 10, 7 or 4 is 9/46
The probability of the fifth card not improving the hand is 38/45, rank doesn't matter.
(36/47*9/46*38/45) = 0.12654949121
The probability of the third card not improving the hand and BEING a 10, 7 or 4 is 9/47
The probability of the fourth card not improving the hand and BEING one of the two others is 6/47
The probability of the fifth card not improving the hand is 39/45, rank doesn't matter.
(9/47*6/47*39/45) = 0.02118605703
1-(0.02118605703+0.12654949121+0.12654949121+0.43811285846) = 0.28760210209
Thus, the overall probability of improving the hand is 28.760210209%.
Therefore, you are 0.28760210209-0.1914894 = 0.09611270209 or 9.611270209% more likely to have a better hand than a pair of Aces holding the Aces than going for the flush.
However, your argument concerns beating all of the other players in a game of five-card draw with four (or more) players. In a game of five-card draw, I'd probably hold the Aces and throw out a moderate sized bet to get slop to fold because slop is likely to either outdraw (and beat) me, or remain slop and fold during the post-draw betting, anyway, so I absolutely don't want any slop in the hand. However, if I put a bet out there and everyone calls, then I probably want to discard one of the Aces and go for the Flush because multiple people are probably drawing to Trips, 2P or another four-flush, anyway, so that's my best chance at the pot odds to draw to a hand that would beat that.
Quote: AxelWolfOthers will notice you drew only 1 card this could affect the value of future betting.
But you will also draw 1 card with two pair, and you should be drawing 1 card with trips. You typically don't want to be drawing to flushes, but i think it's okay if you were a predraw raiser and somehow get stuck in a 4 or 5 way pot.
Quote: AxelWolfOthers will notice you drew only 1 card this could affect the value of future betting.
I agree with you, I'd typically check no matter what if I was first to act after the draw, which sets me up either to bluff or go over-the-top if anyone bets, depending on how strong I think the bettor is. If I act later, then it's just in accordance with what everyone else does, probably a small value bet if everyone checks around to me.
1-(46/47*45/46) = 0.04255319148 to hit Quads
(40/47*3/46) = 0.05550416281 (Full House without either of the two original cards)
(6/47*2/46) = 0.00555041628 (Full House with two of the discard ranks)
0.04255319148+0.05550416281+0.00555041628= 0.10360777057 or 10.361% overall probability of improvement.
(Holding Trips + Any)
(1/47) = 0.02127659574 (Quads)
(3/47) = 0.06382978723 (Full House)
0.06382978723+0.02127659574 = 0.08510638297 or 8.5106% overall probability of improvement.
Five-Card draw is much different than VP in that the extra potential for the 4OaK doesn't have much value, in this case. Whether you have 4OaK or FH, it is extremely likely that you have the best hand after the draw. If you discard two cards and your opponent(s) ends up with Two Pair, there is absolutely no way they are going to bet into you or call you, so the 1.8504% greater chance of improving the hand by only holding three cards simply isn't worth the element of unpredictability you maintain by only drawing one card.
Quote: RSWhy would you draw 1 card with trips?
Deception.
Hide the straight of your had for 1 reason, also a FH will win most of the time.Quote: RSWhy would you draw 1 card with trips?
Quote: RSWhy would you draw 1 card with trips?
Drawing two cards telegraphs your hand too much. You're not going to draw two cards to try and make a straight or a flush.
Quote: Mission146
However, your argument concerns beating all of the other players in a game of five-card draw with four (or more) players.
Hey thanks for all this it really helps. I'm not assuming that all players stay in the game, but I am assuming that the best one or two would. My observation is that with five players you need at least a pair of kings, six a pair of aces, and seven 2 pair to expect to stand a chance at showdown. The odd pair of jacks or nines will win occasionally but you certainly shouldn't expect to.
Quote: Mission146The first erroneous assumption is the statement that the four-flush will improve 1/4 times, specifically, you'll be left with nine Flush cards and the probability of improvement is 9/47 = 0.1914894, which is slightly under one in five times.
You are right I was assuming a 25% chance and not accounting for the four in my hand. So am I correct to summarize, the odds of improving a pair (to 2 pair or higher, not just a better kicker) should equal the odds of drawing a pair from 3 cards + the odds of drawing trips from three cards + the odds of getting the remaining two of the pair?
Quote: wicklundaHey thanks for all this it really helps. I'm not assuming that all players stay in the game, but I am assuming that the best one or two would. My observation is that with five players you need at least a pair of kings, six a pair of aces, and seven 2 pair to expect to stand a chance at showdown. The odd pair of jacks or nines will win occasionally but you certainly shouldn't expect to.
First of all, you're welcome.
Basically, what you're getting from us is that you should make your decision predicated upon how many players stay in the hand as well as your knowledge of who is drawing how many cards. Generally, with four or five players, you're going to want to go for that Flush draw because you'll see that it takes a straight or a flush to win a hand more often in games like that. Essentially, Aces that don't improve and two pairs aren't going to be any good.
Betting/Acting position is a factor, though. If you're last to act and nobody has bet, then you want to throw in a pot-sized bet, I would say, assuming it is an all-ante game, just to get complete slop out of the hand, perhaps even second to last to act. The last thing you want is to give slop a chance to draw to something, and that's also going to be more hands out there that can beat you.
If you're last to bet, you're also last to act, which means that you've had the chance to see how many cards other people have drawn. If everyone who stays in the hand (except one player who draws one) draws three cards, then you know they're probably playing a pair, but you have the best pair...the player who draws one might have 2P, but then he could be drawing to a straight or Flush himself, point is, if your Aces improve to 2P, or better, the probability is that you'll beat that one player who only took one card, either way....and he probably doesn't have 2P, or he'd have likely bet it or raised you.
Simply put, you toss one of the Aces and keep the four-flush if you have reason to believe that 2P or 3OaK is not going to be good enough at showdown, if you think that will be good enough at showdown, you keep the Aces.
If you keep the Aces, you have a 100% chance of having at least a Pair of Aces, which encompasses the 71.24% chance that they do not improve and the 28.76% chance that they do improve. If you toss an Ace, you have a 19.15% chance of hitting a Flush, 4.26% chance of ending up with a Pair of Aces, anyway, and a 76.6% chance of ending up with garbage. (Errors due to rounding)
Essentially, you only go for the Flush if you strongly believe that the Aces alone, or Two Pair, aren't going to be enough to get the job done because the probability of hitting a Flush is greater than your probability of hitting Trips, FH or 4OaK.
Oh, and you probably play over-aggressively if you think a Pair of Kings has a chance at showdown in a five player game.
Quote: wicklundaHey thanks for all this it really helps. I'm not assuming that all players stay in the game, but I am assuming that the best one or two would. My observation is that with five players you need at least a pair of kings, six a pair of aces, and seven 2 pair to expect to stand a chance at showdown. The odd pair of jacks or nines will win occasionally but you certainly shouldn't expect to.
Keep in mind a lot of our discussion is focused on how many players are actually calling the initial bets to see the draw. If you are only up against two other opponents drawing, I'd keep the aces nearly every time. Obviously if they both draw 3 in front of you.
Quote: tringlomaneKeep in mind a lot of our discussion is focused on how many players are actually calling the initial bets to see the draw. If you are only up against two other opponents drawing, I'd keep the aces nearly every time. Obviously if they both draw 3 in front of you.
I'd keep the Aces even if both opponents were taking one because the probability is that, drawing to any combination of Flushes, Straights or trying to improve Two Pair to a Full House, they miss and my Aces improving to any other hand rank will be good.
Quote: Mission146I'd keep the Aces even if both opponents were taking one because the probability is that, drawing to any combination of Flushes, Straights or trying to improve Two Pair to a Full House, they miss and my Aces improving to any other hand rank will be good.
I likely would as well. The only one I can think where you might want to draw 1 is if someone tries to be sneaky and calls predraw and "pats" (draws 0). And that's player dependent.
