your playing a game of holdem you pick up ten ten ,

your opponent goes all in pre flop

you decide to call his all in

what are the chances of hitting another ten with five cards to come ?

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- August 12th 2009, 10:37 AMjickjokerhold em math
your playing a game of holdem you pick up ten ten ,

your opponent goes all in pre flop

you decide to call his all in

what are the chances of hitting another ten with five cards to come ? - August 12th 2009, 05:46 PMgarymarkhov
In my view, you shouldn't care much about the answer to this question. The only time you should care a lot about the probabilities involved is when you are choosing a betting strategy - in this instance, you should be interested in the probability that your opponent has a higher pair before calling his all in.

But maybe you're just curious and want to know the probability before seeing all the cards on the table. In that case, the answer seems to me to depend on how many players there are (how many people were dealt two cards before it was just you vs. your opponent), and whether you are "burning" cards (laying one card aside before the flop, another before the turn, and one more before the river). I'll think about that. And maybe someone else will chime in. - August 12th 2009, 05:46 PMbruxism
the chances of hitting your 3rd ten are 1 - (the chance of not hitting your 3rd ten).

which is

= 1 - (50/52)*(49/52)*(48/52)*(47/52)*46/52)

= 33.13%

(check the maths, i had to use a dodgy calculator)

So you should hit that 3rd ten about 33% of the time. Bare in mind that person with the Pocket aces will also hit their 3rd ace about 33% of the time.

with all this taken into account, the odds of your pocket 10's beating pocket aces (taking into account four of a kinds and straights) is about 20%.

Hope that helps. - August 12th 2009, 05:59 PMbruxism
Gary, From this post and the roulette post I'm going to suggest that you lay off the gambling for a little bit(Wink)

The number of players playing and whether or not you are burning cards has no effect at all on the probability of making a set.

Perhaps this somewhat strange example will give you an idea of why your logic is flawed.

Bruxism and Gary are playing heads up texas hold-em. Bruxism holds pocket tens, while Gary holds pocket aces and they both go all in. Just as they put all their money in the pot, a goat walks into the room and starts to eat the pack of cards! Thinking quickly, Bruxism manages to wrestle the goat to the ground and save the last 5 cards before he eats them. Just enough cards for a flop, turn and river.

Have the odds of hitting my ten been affected by the appearance of the goat?

You can think of all cards that you haven't seen as being eaten by the goat. Players who folded preflop, burnt cards and even people who are still in the hand whos cards you haven't seen yet.

I hope this makes things a little clearer. - August 12th 2009, 07:15 PMgarymarkhov
Haha. You know I wasn't suggesting the Martingale strategy was a good one, right? I thought it was clear that a near 1% chance ("the improbable") of losing your bankroll on any given fresh betting round is nothing to be cheery about if you play long enough.

That's a wonderful analogy! The intuition it gives is fascinating and, I think, correct. From my perspective, any card I have yet to see is akin to a card that the goat eats. Similarly, if you were to deal from the middle of the deck, two thirds into the deck, etc., the chances of pulling up a ten are the same as if you were to deal from the top of the deck. So, I agree, burned cards and cards other players hold or held are not relevant to our calculation.

However, let me humbly suggest that 1 - (50/52)*(49/52)*(48/52)*(47/52)*(46/52) may not be the right calculation. Because of the two tens I'm holding in my hand, I know that there are only 50 cards (the cards I've yet to see) that could possibly contain the other two tens. The calculation should continue to change as new cards appear face up on the table. There's a reasonably good chance that you will once again show me to be wrong, but I'll submit that the correct probability is 1-(48/50)*(47/49)*(46/48)*(45/47)*(44/46) = 19.2%.

Curious to know what you think. - August 12th 2009, 07:28 PMbruxism