PokerStars held its monthly $500 buy-in (with $30 entry fee) version of its Sunday Million tournament last night, and 2,617 entrants tried their luck, generating a total prize pool of $1,308,500. The first-place finisher of the event stood to pocket almost a quarter of a million dollars. The $500 buy-in version of the event had 2,644 entrants the last time it was held, on Aug. 26, 2007 (a month was skipped for September’s World Championship of Online Poker), meaning the tournament’s attendance has been holding fairly steady.

The final nine players had outlasted 2,608 other entrants, and had earned at least $10,000 for their troubles. It was the top two six-figure prizes that were the real targets here, though, and every player at the table was achingly close to seeing those figures in their **bitcoin casinos for us players** accounts. In the end, it was only blanconegro and Chris “SlippyJacks” Vaughn left in contention, and they’d each earned their six-digit paydays. SlippyJacks managed to outlast his final opponent to take the biggest chunk of change, the $240,633 top prize.

SlippyJacks’ win is particularly impressive considering he took down the $1 million-guaranteed event on Full Tilt just last Sunday, making this his second win in a Sunday major in as many weeks. His combined winnings for the tournaments come to almost $439,000. He also hopped within arm’s reach of the top-20 spots in the Card Player Online Player of the Year race. (He currently sits in 22nd place with 3,240 points.)

The final results were:

Chris “SlippyJacks” Vaughn — $240,633.15

blanconegro — $122,344.75

forcewithme9 — $80,996.15

PerPer — $66,733.50

yeamanyeaman — $53,648.50

LuckyLady519 — $40,563.50

domingo32 — $28,525.30

MrKorvOla — $18,057.30

Hasn82 — $10,468

Poker probability (Texas hold ‘em)In poker, the probability of many events can be determined by direct calculation. This article discusses how to compute the probabilities for many commonly occurring events in the game of Texas hold ‘em and provides some probabilities and odds for specific situations. In most cases, the probabilities and odds are approximations due to rounding.

When calculating probabilities for a card game such as Texas Hold ‘em, there are two basic approaches.

- Determine the number of outcomes that satisfy the condition being evaluated and divide this by the total number of possible outcomes. For example, there are six outcomes (ignoring order) for being dealt a pair of aces in Hold’ em: {A♣, A♥}, {A♠, A♦}, {A♠, A♣}, {A♥, A♦}, {A♥, A♠}, and {A♦, A♣}. There are 52 ways to pick the first card and 51 ways to pick the second card and two ways to order the two cards yielding 52 × 51 ÷ 2 = 1,326 possible outcomes of being dealt two cards (also ignoring order). This gives a probability of being dealt two aces of \begin{matrix} \frac{6}{1326} = \frac{1}{221} \end{matrix}.
- Use conditional probabilities, or in more complex situations, a decision tree. There are 4 ways to be dealt an ace out of 52 choices for the first card resulting in a probability of \begin{matrix} \frac{4}{52} = \frac{1}{13} \end{matrix}. There are 3 ways of getting dealt an ace out of 51 choices on the second card after being dealt an ace on the first card for a probability of \begin{matrix} \frac{3}{51} = \frac{1}{17} \end{matrix}. The conditional probability of being dealt two aces is the product of the two probabilities: \begin{matrix} \frac{1}{13} \times \frac{1}{17} = \frac{1}{221} \end{matrix}. (Note that in this case the total is not divided by 2 ways of ordering the cards because both cards must be an ace. Reordering would still require the first and second cards to be an ace, so there is only one way to order the two cards.)

Often, the key to determining probability is selecting the best approach for a given problem. This article uses both of these approaches.…