You will have to account for the probability of each card being missing in your strategy. This is a complicated (almost unanswerable) question as there is no known implementation of optimal strategy in Poker.īy seeing the cards that you know are not eaten (your cards + cards opened on the table), you get a list of cards of which you know for sure the 6 cards will be in. Thanks Tim Couwelier for the remind, i completely forgot about straight flushes. Well, I don't know the probability for any of those cases, but at least it would make me confortable enough to go 'all-in' on that hand. Going by the same principle that you can't make a combination if you don't have that card on the deck, we don't even need to recalculate the probability for straight flushes, because they're going to be even harder to get, so less than 0.0014% of chance. On those 47, what's the chance on the removal of 6 that one will be an ACE? Pretty low chance, lower than 0.0240%. If you have a 52 cards deck, and remove 5( 'KKKK2' ), there is 47 cards left. We need to know what is the probability of one of those 6 cards being an ACE, which would make the 'AAAAX' combination impossible. With that in mind, we can go to the second point: In Texas Hold'em ( or 2 cards poker ), the chance of making 'Four of a kind' is '4,164 : 1' or 0.0240%, and for Straight Flushes there are 0.0014% or 72,192 : 1 odds, which is even worse. With that in mind, only 'AAAAX' and Straight Flushes < K can beat you. That said, if you have the four kings in hand, nobody can make the 'AKQJT' combination. Not many hands in poker win from 'KKKK2': I don't play 5 Cards Poker, but i guess, if it goes by the same winning principle that Holdem have, the only hands that could win against that are: 'AAAAX'(let X be a random card), 'AKQJT'(all of the same kind, or Royal Straight Flush ) and Straight Flushes. I don't know for real the math behind it, but take in consideration a few things: To select 5 cards from a deck of 47, the amount of combinations without a single ace being in them is 43*42*41*40*39 = 115.511.760 The amount of combinations that does NOT hold at least one of the 5 cards is 42 * 41 * 40 * 39 * 38 = 102.080.160 For each of the above options, the occurence rate should be multiplied with the chances that set is actually possible with the remaining cards, or: the odds none of the required cards have been taken out.įor the aces, we need the odds that none of the 4 aces were in the 5 cards taken out of the set of 47.įor the straight flushes, we need the odds that none of the 5 cards for a given straight flush were taken out of the set of 47. This however assumes that all the required cards are left remaining. Odds are now 3.600 / 184.072.680 you lose to a straight flush. We also need to scratch A-5 of hearts and 2-6 of hearts, so 30 options left. There's 8 possible straight ranges (A-5 up to 8-Q) in 4 suits, so 32 possible straight flushes.
CARD SHARK QUESTIONS FULL
Hands to which you lose given a full deck:
What follows below assumes a variation of poker is played with no cards visible on the table
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