The odds of being dealt a certain hand in poker are readily available. I’m more interested in the odds of making certain hands in five-card draw.
Some are easy to figure out, e.g. if you just need one card of a certain suit to make a flush, there’s a 1 in 4 chance of getting it (if you assume that the four suits are evenly represented in the undealt cards). Another way to look at needing one card to make a flush is that there are 9 cards left that could do it. If you compare this to needing 1 card to make an “outside” straight (you have four in a row), there are only 8 cards that could do it, so the odds are a bit worse.
But suppose your hand could is two cards away from either a flush or a straight - which should you try for? Let’s say the hand is 4C, 5H, 6D, 10D, QD. As far as the two cards to make a flush, the odds should be 1/4 X 1/4, or 1 in 16. But what about the two cards to make a straight? It could be filled by 2-3, 3-7, or 7-8. How would one calculate those odds, to compare to the odds of making the flush?
With one card to go with a flush draw, isn’t it like 19%? To say 1 out of 4 is like saying you have 12 outs to a flush draw, which is not possible when holding 4 suited cards, with only 9 outs to complete your flush.
The easiest way to approach this problem is to count the total number of ways to get the two cards you desire, and the total nuber of ways to get any two cards. The latter is easy to find. You have five cards, which means there are 47 cards unknown to you. There are then 47 choose 2 ways to pick two cards from them.
Finding the number of ways to get what you want is a little more difficult. First consider the flush. If you go for the flush you throw out the four and the five. There are 13 cards of each suit, and you have three of the diamonds, leaving 10. There are 10 choose 2 ways to pick two of them. This gives a probability of 10 choose 2 over 47 choose 2 of getting the flush. This equals 45/1081.
For the case of the straight there are three possibilities. You can get a two and three, three and seven, or seven and eight. In any case, there are four cards of each value remaining, and so there are 44=16 ways to get any of the three winning pairs. So there are 163 ways to win in this case, and still 47 choose 2 ways to pick two cards. This gives a probability of 48/1081, so you are more likely to get a straight than a flush, but only slightly so.
I’m resurrecting this thread I started nine years ago because I wasn’t really satisfied with the answers. Maybe I can clarify my question a bit to get around that.
I’m envisioning a game situation, with maybe five players. This is different from one person just drawing cards from a deck. Specifically, the comments about how many diamonds I hold in the example hand are off the mark, because given the other players it’s reasonable to assume an even distribution of all four suits among the five of us. I maintain that the chance is one in four to get a card of any given suit in a single draw, and thus the chance of getting two cards of a given suit, when drawing two cards, is one in sixteen.
The chance of getting one card of a given rank is one in thirteen. It’s the chance(s) of getting two cards of the various possible ranks to complete a straight that are beyond my ken. I think quelquechose was closing in on it, but I believe he approached it from the wrong angle I described, focusing on a deck with 47 cards remaining rather than a game with 25 cards dealt out and 27 left in the deck.
The only cards you know anything about are in your hand. It doesn’t matter if the cards you don’t know are organized into one deck, split into four other hands and the stub, or even hidden in forty-seven different locations around the room.
Yes. You’re wrong to assume that your situation is different than dealing to yourself. It doesn’t matter at all that the unknown cards are sitting in front of a person instead of sitting in the deck.
This makes no difference. You have 5 cards in your hand and there are 47 unknown cards out there. It doesn’t matter whether they are in other players hands or in the draw deck since you have no additional information, a given card could be in any one of the 47 places so the probability that it is on the top of the draw pile is 1/47.
Or if you want to make it more complicated, with 27 cards in the draw pile and 20 cards in other players hands, the probability that you draw the Ace of Diamonds (assuming it isn’t in your hand) is the probability that that card is in the draw pile (27/47) times the probability that it is at the top of the draw pile (1/27).
If it’s just me with five cards and 47 sitting in the deck, then it’s KNOWN that (for the hand described) there’s a better chance of drawing a spade than of drawing a heart or a club, and a better chance of drawing a heart or club than of drawing a diamond. But in a real world situation, with several hands dealt out and the rest in the deck, in the big picture/in the long run/on average/however you want to say it, there will be an even distribution of the suits. If I’ve got three diamonds, someone else probably has three clubs, etc.
I can see a slight statistical difference from my having three of one suit, but it’s not a situation where there are 10 diamonds available and 13 of each of the other three suits available. Twenty cards are in other hands and are not available. The distribution of suits in the draw deck is generally going to be even, not lopsided in a 10:13 ratio.
Yes, because the ‘even distribution’ isn’t in the deck alone, it’s in the deck + other hands. The other hands dealt will not serve to even out the deck you’re drawing from. Statistically, they will have the same average composition.
The only way this is true is if you’re talking about a multi-draw game versus 5-card or another single draw game. In the former case, hands with flush possibilities will stay in longer than hands without. But even then, you have no idea whether any one of those hands has a flush possibility (and/or what was in the hands that were dropped) so you have to treat the deck as if it contains every card except the ones you already hold.
You might be getting screwed up by watching the WSOP on TV. When they show players’ outs, they show only the live cards remaining in the deck. They remove the players’ holecards from the calculation because they know they are not in the deck. That is not the same probability that you would calculate sitting at the table because you don’t know what the other players’ holecards are.
Since you correctly multiply by three, I assume by “both ways” you meant “all three ways.” Eight-Nine-Ten becomes a straight with Six/Seven or Seven/Jack or Jack/Queen. Change the Eight to a Seven and have a 2(8/47)(4/46) chance, or to a Six for a 1(8/47)(4/46) chance.
Those arguing against you are correct. Perhaps one way to see this is, when playing the solitaire version, to deal 15 cards face down before drawing. Do you think this affects the odds? Those 15 cards can represent hands of the other players. Your arguments about what opponents’ hands “probably have” apply equally to the 15 discarded cards in my scenario … or to the 15 cards left at the bottom of the stock in any scenario.
Don’t feel bad. It shows much mental acuity to remember a thread you started nine years ago. I can’t remember what I had for breakfast yesterday.
Here’s a trick to make a quick and dirty determination of the odds.
However many cards are out there that would help you is multiplied by 2 for each card you draw and that is the rough percent chance you have to make your hand. It doesn’t matter how many others are dealt in, unless you can see their cards the odds are always the same.
If you have 3 to a flush, there are 10 cards out there that could help you. The first card of the two you draw has a (2x10) 20% chance of being the right suit. If it is, the next card you draw has a (2x9) 18% chance of being that suit.
It’s not precise but close enough to be a useful shortcut because it’s so easy to do and it works for calculating any draw (turning 3 of a kind into quads or a full house,etc.)
I meant cards on both sides of your hand, but you’re correct in calling it three ways since if you’re holding J, Q, K you can make a straight on both sides, but not fully on the over so you’d multiply by 2 instead of 3.
Now for Part Two: Unless you are in the blind and the pot has not been raised you should almost always fold that hand. If it is a blind game and you are on the button and everybody else has folded you might try stealing. – That’s just about the only time you should voluntarily play this hand.
If the game is Jacks or Better to open it is folly to call a bet with such a slim chance to improve when you know someone already certainly has you beat with at least a big pair … and perhaps an even better hand than that which you hope to draw.
If any hand can open you should actually still stick to pretty much the same requirement to play a hand as if the game were Jacks or Better. Even with multiple callers and no raise the pot won’t be offering good enough odds to be making two card draws to mediocre straights and flushes; keep in mind that with multiple players, some of them may be drawing to better hands than you are and unless they are very poor players they probably have one card draws or big pairs or trips that will beat you every time you don’t improve.
If you are in the blind and somebody foolishly calls but doesn’t raise you should draw five if you can, otherwise hold the Q and draw four – or play it tricky and raise, then stand pat and bet out after the draw if your opponent calls your raise … unless your opponent is the tricky type who might be raising with nothing to steal your blind and might reraise to resteal.
