# Poker Probability Project

**Topics:**Playing card, Card game, Poker

**Pages:**2 (520 words)

**Published:**May 12, 2013

Probability Project

Strayer University

Math 104

Professor Stephen Vest

December 16, 2012

In this writing assignment I will explain how I determined the odds of winning a Texas Hold’em game or not, where I have a six and seven of diamonds. My opponent has a ten of clubs and a ten of spades. At the turn or fourth street, the cards on the table are a three of diamonds, four of clubs, nine of spades, and a ten of diamonds. These cards show that my opponent already has a three of a kind.

I have already bet $24,000 into the pot as well as my opponent. With the pot at $48,000 at the turn my opponent bets $10,000 more. The question is if I should call or fold? If I fold I lose $24,000. If I call the pot grows to $68,000 putting me all in. Before I call I need to calculate the odds of me winning vs losing.

There are 2,598,960 different possible poker hands. With my cards I can either possibly get a straight, one pair, flush, or nothing for that matter. I will take the one pair option out of the calculation since the one pair would not beat the three of a kind. The odds of getting a straight are 10,200 and 5,108 to get a flush. Add the two numbers and divide by 2,598,960 then my possibilities of winning are .00589. With the three of a kind and the possibility of getting another ten to make it a four of a kind or getting either a three, four, or nine to get a full house the odds of my opponent winning are: 54,912 (three of a kind), 3,744 (full house), 624 (Four of a kind) equals 59,280 and divide by 2,598,960 it will equal .02281. This shows that the odds of me winning are less.

My other strategy is to take in consideration the amount of cards that would give me a winning hand over my opponent. With the four cards on the table, my two cards and the two cards my opponent has there are 44 cards left in the deck. To beat the three of a kind I would need either a five (four cards), an...

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