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Suppose you want to calculate the equity for a hand 4 and 8 (the color does not matter). The cards on the flop are A,2,3. To calculate the probability to have a straight you first consider the outs for a 5. Here the outs are 4. With the usual methods you multiply by 2 to get the probablity to get a four at the turn and by 4 to get it at the the river. However what I am considering is that on a table of 9 players, there are 5 known cards on the flop (your hand and the 3 on the floop). There are 16 cards of the total 47 unknown cards that are on the players hand (8 players x 2). Therefore the probability that the card that you need, 5 in this case, only has a probablity of 31/47 to be on the "remaining cards yet to be played" at the table (since 16 of 47 cards are on the players' hands). Therefore when you you multiply the outs for a Gutshot straight by four and get 16% shouldnt you also multiply by 31/46 considering the possibility that this card could be on the players' hand ?

Consider an extreme case which there are 22 other players and only one card remains to be played your odds in this case to get that one card you need is obviously very small. How does multiplying by 2 and 4 considers the fact that as the number of players increase your odds decrease

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The number of players has zero effect on the probability of seeing your outs. After the flop, there are 47 cards you cannot see. It does not matter where they are, whether they are in other players' hands, the deck, or the burn pile. Whatever card you need is equally likely to be any of those 47 cards, and all that matters is whether is the next card on the turn.

There are 47 possible positions of a single out card which are all equally likely, so the probability that it is in any one particular location is 1/47. The out card has a 1/47 chance of being the left card in Opponent X's hand, it has a 1/47 chance of being the card that was burned pre-flop, and it has a 1/47 chance of being the next card to appear on the turn. Note that the probability does not depend at all on where the other 46 cards are, or how many players there are.

You can also recognize from this that the burn cards have no effect on probabilities, either. You could burn an arbitrary number of cards before flipping community cards, and also not affect the probability of seeing your out - all that matters is if your out happens to be in the 1 position you reveal among 47 hidden choices. Since the cards are randomized, there's no way to increase your odds by picking the second card to reveal, or the 10th, or a random card. The rules state you need to burn 1 and flip from the top of the deck (i.e. always reveal the second card), but the odds would be no different if you just picked a random card from anywhere in the deck (or a random card from the top half of the deck).

With more players, it does become more likely that your out is in someone's hand, but it does not affect the likelihood that your one out is the next card on the turn - it's always 1/47.

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Here is a perfect scenario where the player does not understand why multiplying by four is done.

The probability of hitting a five by the river is the complement of not hitting a five by the river; it is calculated this way since it is much less tedious.

You would miss the turn if a non-five showed up, and you would miss the river if a non-five showed up. Out of the 47 possible cards, 43 of them are non-fives on the turn, and supposing that is the case, 42 of them are non-fives on the river. Thus, the probability of hitting a five by the river is 1-(43/47)(42/46), which is roughly 16.47% -- somewhat close to what multiplying by four gives you.

Multiplying by four is a simple way of getting a somewhat close approximation of your chances. Suppose you have n number of outs on the flop and you want to see what the chances of hitting at least one of those outs are by the river. Similar to the math done above, we would have 1-((47-n)/47)((46-n)/46)=(93n-n^2)/2162, and we can see that multiply the number of outs by four is a really, really good approximation if you have ~8 or fewer outs.

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  • If anyone knows how to format the math a little bit more nicely like they do at math.SE, please let me know. – Andrew Chin May 28 at 13:42
  • But how does this takes into account the number of cards that are in the players hands ? Suppose there were acutally 23 other players there would be only one card left to be played and your odds are very small. How would multiplying by four give you a real approximation of the probability which for this case wouldnt not result accurate – qubitz May 28 at 17:17
  • Any unknown card has an equal probability of either being still in the deck or in someone's hand. It is effectively disregarded (unless you have reads that give you probabilistic knowledge of others' hands, in which case the math becomes slightly more complex). – Andrew Chin May 28 at 17:21
  • Last thing, if the card has an equal probability of being in the deck or in someone's hand there 50/50 should you mulitiply the odds by 0.5 ? How does one reconcile this with the fact that an increasng number of players decreasen your odds ? – qubitz May 28 at 17:28
  • That is the misconception; the probability of a card being in a particular position in the deck (whether it is or is not to be dealt) is independent of the number of players at the table. – Andrew Chin May 28 at 17:31

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