# Tag Info

13

(oh well instead of commenting I may as well post this as an answer) You're asking two different questions and they have two different answers. Are burnt cards significant in odds calculation? No. They're not different than any other card still in the deck. there is also the possibility that some of our out cards have been folded by our ...

13

What are Pot Odds? Pot odds are the value of the pot (how much you stand to win) in comparison to the cost of you calling, and are most often used to evaluate the value of making the considered call. Calculating Pot Odds Pot odds are a ratio, Pot Value : Call Cost To convert this ratio to an equivalent percentage, divide the Call Cost by the sum of the ...

12

After the flop you've seen 4 cards of your suit, and 1 of another suit. This leaves 9 cards of your suit, and 38 of a different suit; your odds of completing your flush on the turn are thus 9/47, or 19.14%. If the flop hasn't completed your flush, your odds of completing it on the river are 9/46, or 19.5%. This means that the total odds for completing a ...

11

The simple case, for making your hand on the next turn, is to add up the number of cards that can give you the winning hand on the next turn, and divide by the total number of unseen cards left in the deck. This means that, in games like Stud, you need to remove any potential outs that have been folded by others from the cards that can give you a winning ...

10

Here is a nice tool called pokertablestats.com and propokertools.com to help you calculate your chances! A good rule of thumb I always use for calculating odds is multiply number of outs(13 with 4 on the flop and 2 on the turn. This way it will be much easier to remember and you are never that far away from the correct percentage. If you do not understand ...

10

No, those cards have no significance to odds calculation. If I shuffle a fresh deck, what is the chance the the top card is the Ace of Spades? 1 in 52. If I deal off the top 10 cards face down, what is the chance that the card on top now is the Ace of Spades? Still 1 in 52. The same probability will apply to the rest of the cards in the deck, including ...

8

Yes, you can and you should. The concept you're describing is called implied odds (the estimated profit you'll make if you make your hand). Notice is a much less concrete value as it is an estimation of whether your opponent will call when the draw comes and the amount he'll be willing to pay. There's also the concept of reverse implied odds which are the ...

7

I was thinking about how to explain my question a bit more and then realised I could work out the answer. I wrote a small python script to count all the badugis one can be dealt. 715 badugis in total 4 high: 1, 0.1% of tot. Cum 1, 0.1% of total 5 high: 4, 0.6% of tot. Cum 5, 0.7% of total 6 high: 10, 1.4% of tot. Cum 15, 2.1% of ...

7

To understand Implied Odds (IO) it's useful to clarify what It's counterpart is, Explicit Odds (EO).  EO describes how much you will win immediately in relation to what you have to risk. This is described in terms of a ratio, Total pot size : Amount we have to call. For example, current pot size is \$50. Your opponent bets \$50. Therefore, the current pot ...

6

The odds of getting a 4 of a kind given 7 cards (2 in your hand and 5 on the board) are (13 * (48 choose 3)) / (52 choose 7) or 0.00168067227. The probability of getting that specific 4 of a kind again are now (48 choose 3) / (52 choose 7) or 0.000129282482. The probability of both events occurring is 0.00168067227* 0.000129282482, which is 0....

6

"Power" of a hand is in practice an oversimplified notion. I will touch equity and your sub-question about what it says about a hand's goodness. If you're just trying to code equity, the Coding the Wheel article others have mentioned is mandatory reading for poker coders: http://www.codingthewheel.com/archives/poker-hand-evaluator-roundup As for the ...

6

Your question can be re-written like this: What are the odds of having 4 cards with the same suit, from the 5 available community cards? The answer can be found by using basic probabilities and computing probabilities of composed events. You can have 5 boards in which 4 cards are of the same suit: XXXXY XXXYX XXYXX XYXXX YXXXX Here, X means the suit ...

5

There are many elements of information that are vital to know here. Some of them are, Is the call (T\$700) the only chip amount you'll have to commit to call? Are any players All-In? Do you need to consider money from the blinds that is in the pot already? Is there a chance that you or your opponent will fold on later streets? and more... Technically, if ...

5

There are thumb rules for the preflop equity (against a single random opponent) of pocket pairs and suited-connected combos. For the equity of a pocket pair, you calculate how many cards away from 2 your cards are (for example, Queens are 10 cards away from 2), multiply by 3 and add 50%. So QQ's preflop equity is approximately: (10 * 3) + 50 = 80% For ...

5

For an open-ended straight draw, there are exactly 8 cards that will give you your straight: 4 of each suit at the low end, and 4 of each suit at the high end. However, to make a flush, you have 9 cards available that will make your flush, as there are 13 cards to a suit, and you have 4 of them in your hand. At 3 cards to the set (again assuming an outside ...

5

As you can imagine, your equity in a heads up hand with no rake, where you bet preflop and deal out all community cards without betting, will be 50%. Other variations of this, such as the dealer winning ties or the introduction of a rake, will lower your equity (and since this is a casino game, I'm willing to bet that they have something in their favor). ...

5

Assuming you know nothing about the cards dealt, they don't matter, so the 66% holds up. In most calculations we would just ignore the folded cards since we don't have any definite information about them. If you want to factor them in, you can no longer calculate your exact pot equity, since you don't know how often your opponents are folding hands like Ax, ...

5

Before anyone speaks, no matter how many players there are the distribution is still totally random. Each card has, for example, exactly the same probability to be in anyone's hand. However as soon as someone speaks then things change... What does this do to the overall win/lose odds of a given hand? Simply put: as soon as a person folds the ...

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

Pot odds are a way of determining whether it's worthwhile to continue with a hand. Suppose that the pot contains \$10 at the river and your opponent bets \$2. You think that there's a 20% chance that you have the best hand, and that if you raise your opponent will fold a worse hand or reraise a better one, so your options are to call or fold. What should you ...

4

In tournaments, there are often factors that trump pot odds or implied odds when making decisions. Your stack size (and the comparative positions you'll be left in if you call-and-lose vs call-and-win vs fold) is often first and foremost in that list. In no particular order, other factors include table dynamics (e.g. are you going to have opportunities to ...

4

If your opponent has 10 000 in chips and the blinds are 200/400 your normal raise would be something around 1000 - 1350 which means that it's with the blinds something around 20% of your opponent's stack. He can go All in now and win your bet + blinds which is not much but can lose a lot (everything) if he gets called by you (or even reraised all in). If he ...

4

It all depends on pot odds. If you have better than 50% pot odds and have 50% equity versus your opponent's range, and you have the bankroll to handle the variance, then you should be looking to play for stacks. This will always produce a long term winning strategy, because you're getting >50% return on a 50% bet. Do you see why? The only situation where ...

4

The article is correct in the way it uses 4:1 and 5:1. Under their assumptions (actual value given their example is more like 4.2:1), you are "4 to 1" to make it while you are getting "5 to 1" on your money. I'd say that this is precisely because both are written / pronounced / thought of this way that it's convenient. If you check the Wikipedia article on ...

4

You need to calculate the odds of getting the exact flop that you need. Since the order doesn't matter, the first card dealt would have three possibilities, and then if you got one of those you would have two possibilities on the second card, etc. It would look like this: 3/50 * 2/49 * 1/48 = 1/19,600 = 0.005% EDIT Updated based on your comment/...

4

Based on your calculations... If you hold 4 cards (A to 4), you will ALWAYS make a pair? Estimate with 3/50 * 5 instead of the annoyingly similar fractions: (3/50 * 5) * 4 = 60/50 !? ===---=== The reason you won't get the right answer this way even though I can see the logic in your math is because of "double counting". You think the chance of hitting the ...

4

Note 1 in the article on Hold'em Odds elaborates on this a bit further: [Note 1] By removing reflection and applying aggressive search tree pruning, it is possible to reduce the number of unique head-to-head hand combinations from 207,025 to 47,008. Reflection eliminates redundant calculations by observing that given hands h_1 and h_2, if w_1 is the ...

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 TED....

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