March 30, 2009

Poker Odds Tutorial: Lesson 1

If you hold two suited cards, what are the poker odds of flopping 2 of that same suit? How did you calculate this?

All unknown cards including burns are considered to be part of the deck. You calculate by using fractions, putting your outs on top (numerator) and the total cards remaining on bottom (denominator). For the 3 flop cards, the denominators are 50, 49, and 48.

Calculating something like flopping a set is simple addition … { 2/50 + 2/49 + 2/48 }. But with two of a suit, the numerator of the fraction changes when the suit hits. There are 11 of the suit left and you need to hit 2 of them. So, it's more complex than one single addition or multiplication problem.

Rather than write some massive nested expression, I'll try and explain this in a fun and interesting way. Let's say you're holding 2 clubs.

The first card of the flop is a club 11/50 times … 22% of the time, exactly. We go on from there.

22% of the time, you hit that first club. You will need only one of those other 2 cards to be a club … the odds of that are fairly high, { 10/49 + 10/48 }, about 42%.

The other 78% of the time, you'd need both the remaining cards to be clubs … the odds of that are fairly low, { 11/49 * 10/48 }, about 4.5%.

We multiply 22% by the first bracketed expression, which is how often it will happen.

We multiply 78% by the second bracketed expression, which is how often that will happen.

We add these two figures together and arrive at 12.7%, almost exactly 1 in 8 times.

If you also want to include situations where you flop all 3 of that suit, we add the odds of that, { 11/50 * 10/49 * 9/48 } which is very, very low, 0.8%, to the final mix.

Therefore we can say that the odds of flopping at least 2 of suit X, when you already hold 2 of suit X, is ever so slightly better than 1 in 7.5, almost identical to the odds of flopping a set.

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1 comment:

signals3_t5 said...

Let's make a stand for internet poker
so interesting blog

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