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5

My short answer is this: as long as your opponents have 100BBs, it doesn't matter how many BBs you have (as long as you have them covered). You will only play for their 100 BBs. Shoving every hand when you have 10000 BBs is as bad as shoving every hand when you have 100 BBs. The difference is that in former case you don't have to re-buy every time after you ...


4

When you are dealt a card, it has 13 possible ranks. When you are dealt a second card, for it not to be a pocket pair, that can be of any rank different to your first hole card - i.e. 12 different card ranks. For the division by 2, the calculations are for the number of combinations of different types of hands and a combination disregards order. If you take ...


3

First off, I believe it is easier to understand the probability of getting dealt a pocket pair this way: Your first card can be any of 52 cards, so it is not relevant in the calculation. The 2nd card must be one of the 3 cards that match the value of your other card. There are 51 cards left in the deck now, so the probability of being dealt a pocket pair is ...


3

There are 10 players at the table, any one of them has a 6% chance of having a pocket pair. So the chance that no-one is dealt a pocket pair is 0.94^10 ≈ 0.54, ie chance of someone having a pocket pair at a FR table ≈ 0.46. Excluding card removal for ease: You have a pocket pair, what is the chance that out of 9 players none of them has a pocket pair ? ...


3

The probability would be calculated by taking the odds of having 1 card pair and then the other two cards NOT pairing or forming a set. Calculate this for the three combinations (1st card pairs, 2nd card pairs, and third card pairs) and add them together: (6/50) * (44/49) * (43/48) = 0.096530612 (44/50) * (6/49) * (43/48) = 0.096530612 (44/50) * (43/49) * ...


3

This is actually an interesting question and not because it might relate to good strategy when considering hand equity match-ups or Expected Value, (let's face it, random All-In shoves every hand don't fair well in terms of overall strategies) but because it's a good example of bankroll management, and how people misapply it. The key concept here is how ...


2

This question is actually very similar to the "double your bet" strategy for baccarat, or any other close to even odds game. Casinos typically will always have a table max bet for these games to prevent some billionaire from coming in and keep doubling their bet until win back all of their loses. The number of times you can lose before you hit the limit is ...


1

TL;DR Given a flush on the board, five players will split 19.7% of the time, or about once every 5 hands. Given a fresh deal, the deal will come up with a flush showing and a five-way split 0.0389% of the time, or once every 2,569 hands. Some general probabilities Overall odds that none of the five players have a given suit (say, hearts): comb(39,10) / ...


1

I think the real question here is with a 5 card flush on the board what is the chance that out of 5 players none have one of the remaining 8 flush cards? The answer is 12.3% The odds of player one’s first card is not a flush suit is 39 of the remaining 47. We can see 5 cards, so 52-5 = 47 available cards. Of those 47 there are eight flush cards remaining so ...



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