Is there an easy way to work out the odds of a particular card turning up in the last 2 “communal” cards in a game of “Texas Hold 'em” - i.e. not “the flop” - the next two.
I always maintain that since I can only see: my TWO plus the flop’s THREE, there are (52 minus 5) 47 cards left and as far as I know, any of them could come up.
The reason I ask is because a poker buddy of mine always maintains that if there is already three-of-a-kind showing, e.g three Ks, then the chances of the last “king” turning up as one of the last two cards in the centre, are pretty small - you would have to have dealt the four Ks into the flop to start - a pretty unlikely event.
I always respond “Bullshit” (or words to that effect), *as far as anyone knows * each card has the same chance of turning up in the last two places - since the cards are dealt at the start when their is a full, shuffled deck.
You are both correct. There is a 1/47 chance of the fourth king coming on the turn and 1/46 on the river. This is the same probability as any other particular card coming up. Your friend is right that is it not very likely. It would be better to draw to a hand that has more “outs” – cards that will improve it – left in the deck. For instance, if you had two suited cards in your hand and two came on the flop then you would have a 9/47 and 9/46 chance of making a flush.
Still, don’t throw that set of kings away. It’s probably good enough to win already, and the board might pair making you a full house.
You are correct. Absent any information other than the two cards in your hand plus the three in the flop, each remaining card has a 1/47 chance of turning up next, regardless of what those five visible cards are. Of course, your ability to read other players’ hands based on how they bet could influence those probabilities, but that doesn’t seem to pertain to the question you asked.
Well I think it depends on how you’re looking at the question. Sure, any particular card on a fair shuffle has an equal chance of coming up, but I think your friend is looking at the odds of a king versus any crappy old card, in which case he’s pretty correct. 1 in 47 versus 46 in 47. Is that how he’s looking at it?
This would make sense, but I don’t think he was thinking this way - he kept talking about the 3 Ks on display, and insisting that these cards had some bearing on the next cards - which is not true.
However, your point is well taken - that’s why it’s better to play to an open-ended straight / flush et cetera, than a close-ended one - the odds being 8 / 44 (1 in 5.5) as opposed to 4 / 48 (1 in 12).
As a major aside, does anyone know of a poker forum, for discussing these things in more detail ?
The reason I ask is because a poker buddy of mine always maintains that if there is already three-of-a-kind showing, e.g three Ks, then the chances of the last “king” turning up as one of the last two cards in the centre, are pretty small - you would have to have dealt the four Ks into the flop to start - a pretty unlikely event.
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Your friend’s error is that the probabililty of four kings given that 3 kings have already come out is quite different than the odds of four kings coming up in the first place.
To put it in more familiar terms. I flip a fair coin 9 times, I get heads each time. What is the probabilty that the 10th flip will be heads? It is 1/2. Wait, your friend says, the odds of getting 10 heads in a row is 1/1028. He is correct, but that is not the question, the question is what is the odds of the next flip being heads – which is equivalent to asking the odds of getting 10 heads in a row, given that 9 have already come up heads. 1/2 either way.
Sure, let him keep believing. Playing with stupid players is the easiest way to get an edge in poker.
If you get into an argument about it though, you might point out that his logic applies to ANY set of 4 cards, kings or not. If the flop is 6C, 5D and 10H, the odds of the last card being 9D is what? By his reasoning, it’s the same probability as getting dealt 4 kings.
I would more highly recommend www.twoplustwo.com - the signal to noise ratio there is much higher.
To specifically answer the question, here’s how you do it:
If the two events were completley independent, you could just multiply the odds together for your odds of getting your card with two draws. But this isn’t the case here, because if you draw a king on your first card, you can’t draw one on the second.
So the easiest way to calculate this is to calculate the odds of NOT getting that 4th king, which are 46/47 * 45/46, which equals 95.7%. So you will only draw your 4th king 4.3% of the time. Converting that to odds gives you about 22 to 1. Long odds indeed.
The same works for flush draws and straight draws. For example, the odds of making a flush with two cards to draw and a four flush in hand:
You have 9 ‘outs’ to make the flush. The odds of NOT making the flush then are 38/47 X 37/46, or 65%. So you’ll make the flush about 35% of the time, or odds of about 1.86 to 1.
If you’re trying to make an inside straight, you only have four outs. Doing the math again, you’ll find that your odds of filling your straight with two cards to come are a little worse than 5 to 1.
