Hot answers tagged

5

I get about once every 325 hands. Not that uncommon.


5

When facing an "at least one of" problem, you can't just add probabilities. You have to calculate the probability of missing everything, then subtract from one. Assuming 47 unknown cards, 6 of which are outs, the exact probability of missing both the turn and river is 41/47 * 41/46, or 1681/2162, so the probability of hitting either or both the turn and ...


4

Sometimes during play you want a quicker method that isn't 100% but gives you very close percentages. The rule of 2 and 4. If you have 2 cards left to reveal (ie the turn & river) then you multiply your outs by 4. If you have 1 cards left to reveal (ie just the river) then you multiply your outs by 2. So in your example, 6 outs with 2 chances left = ...


4

It turns out that each shuffled deck is in the order that may have never before existed in the history of the universe! :) The odds of you getting two 52 card decks arranged in the exact same order are 52! ~= 8 x 10^67, which is waaay more than the number of atoms on Earth (~ 10^50). For a detailed explanation, please check out a great video answer on ...


3

This depends on number of things you have to consider, not only the direct odds. What I mean: The pure odds you calculate should be used if you expect your opponent to check the turn and you see free river. Always calculate implied odds! People usually bet on turn and river! If you expect your opponent to bet on the turn, you have to calculate this too - ...


3

That depends on what you're contemplating, and what you think future action will be. For example, if you're contemplating a bet that will put you or your opponent all in, then the odds of the next two cards are what matters, because you're buying the right to see both of them. But if you're contemplating calling for a draw, and you both have stacks, well ...


3

Frisbee's answer is correct. But it only answers for eight players (although the other numbers are hidden in that table). Since the OP asked for 8-10, I will add the answers for nine and ten players. The odds of an ace not occurring in the next x cards dealt (after you're dealt two aces) is ( 48!/(48-x)! )/( 50!/(50-x)! ). That's the probability that no one ...


3

This answer assumes you have no additional knowledge of the opponent's cards. You have five cards in your hand. Of those, three are spades and two are not spades, so you need two more spades to complete your flush. There are 47 cards that you haven't seen, and 10 of those 47 cards are spades. You discard the two non-spades. The odds of drawing two more cards ...


2

You need more training not the dealers. Actually its neither, it is the people who set policy on the amount of time a deck is used, the policy for replacing stored cards etc. that could use a little more training. When a shuffle is poor, the cards are being shuffled into fairly big clumps. Now since the dealer deals one card at a time, if the deck is clumpy ...


2

Tie is neither win nor loss. Therefore, you should have a counter for ties the same way you have a counter for wins. In any case a round with a tie should be removed from your simulation.


2

You are calculating pot odds a very unusual way. Your formula is mostly correct (but only works some of the time), and I'll get back to that in a moment, but typically you would just use two variables: costToCall and sizeOfPot. Pot odds don't depend on the number of players to have called the bet. One player putting 400 in is the same as four players each ...


2

These are the chances (assuming you have no ace): These are valid only preflop assuming that there are 50 cards left in the deck (you are holding 2) and you are one of the players (2 players = 1 opponent with 2 cards; 3 players = 2 opponents with 4 cards and so on)


2

Pick is it 8 or 10 At 8 I think it is 0.4114 Or 1.4306 : 1 This is for exactly one ace out - (not two) Using combination (2/1) * (48/13) / (50/14) (2 aces need 1) (48 non ace need 13) / (50 cards need 14) Both other aces out would be 0.0743 Add them for 1 or 2 aces out = 0.4857 You can get that same number with 1 - 48 / 50 * = ...


2

I have seen it on TV a few times Based on hand independent it is 0.000256 (quad) * 0.0017 (boat) = 0.00000044 From the boat there are 45 cards out and need to match the pair on the board So the odds of getting quadded are: (2/2) / (45/2) = 0.00101 = 0.1% Which also = 2/45 * 1/44 Highest or lower boat is immaterial. Quads beat a full house.


2

Not really, I'd have played it exactly the same, especially given this is a multi-way pot, you want to isolate here against baby aces and small pairs hitting a set on subsequent streets. A King is well within any of the other players' ranges too. It's the right play.


2

You can use the rule of 2/4 as described above, or by thinking in reverse, do some simple math to get the odds of completing your hand by the river. Instead of directly calculating the odds of building your hand, think of the odds of NOT getting your card -- it makes it much simpler. As an example, after the flop you have 4 to a flush. What are the odds of ...


1

I'm not familiar with .NET and stuff. I probably should mention that I'm not a professional programmer myself. But there surely are a lot of great libraries in pure C. Using C++ and i5 CPU I personally get about 4.4 million hands/s in Monte Carlo with 3 players and 2.3 million with 6 players. But that's just raw unoptimized Monte-Carlo. Brute-force ...


1

Features important to me: total and pct on win loss tie ratio for hand odds give a tie 1/2 win break down on all the hands percent straight flush down to pct high card need this for debug anyway further break down on all hands when it wins / ties did it do so as what hand (straight flush down to high card) (my app does not currently do that - going to ...


1

Those number are wrong 19.15% turn 19.57% river 34.97% turn or river (I think you are missing that 5 cards are out) At the flop if either are all in then you can count on no more bets on the turn. You have to base it on what you think your opponent will do on the turn If you don't hit If they bet in to you on the flop then highly likely they are going ...


1

Squeeze play is a specific move that works under the following circumstances: A very aggressive and loose player raises preflop (wide range) A loose player after him calls, since he knows the raiser will have a wide range You currently have a relatively tight table image What you do with a squeeze play is re-raising the pot big, such that the original ...


1

Everyone check to the button, no one is showing strength. Someone wants a free/cheap draw The button's job is to show strength, he might have anything, even after raising pre-flop. You know he has something, probably a suited-connected hand or a small pair, it's unlikely that he has a high pair in this action, although it's always possible. We later come ...


1

You got your money in with the best hand. And he was not getting pot odds to call. Even if you knew what was in his hand you make the right play. At the flop you were 68%. Even if he filled up the straight you still had 4 outs yourself for a full house. On that board you had a very good hand. I would not have put a player on 24. And I doubt they put ...


1

I do not know if the callers are big/small blind, so I am going to be pessimistic and say the pot is 4*6= 24$ after the flop. On the flop, before you shove the pot is 24+15+15=52$. You shove with 48$, hence he has to call with 33$ the pot odds are 33/100 or 33%. He has 8 outs, hence the probability of hitting is aproximatley 8*4= 32%, or more precise 30% ...


1

It's a reasonable improvement for the overcards when they are suited. In the example of AK versus pocket tens, you improve from about 42.5% chance of winning to about 45.9% chance. AK does a little better against 99 than against tens, because the 10 blocks the possible straight. Here are some examples from pokerstove. First, unsuited AK against pocket ...


1

Let C(n,m) = n!/(n - m)!. That is, the number of ways to choose m items from n. The calculation you did for the first value is correct: 1 - C(52-4, 30)/C(52, 30) = 97.3% You can arrive at that calculation by first calculating the probability that no eights are drawn. That is the number of ways to choose 30 cards from the deck excluding the 4 eights, ...


1

As far as the math for determining this, it gets very involved and complicated. The basic formula is the same though, the cards you can hit to win / the total remaining cards unseen. Now I'm sure the problem your having in the situation you posted is that you don't know the other players hole cards. That throws a wrench in the works. Since you don't ...



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