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Probabilities you are beaten

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Pingviini

03-17-2005, 12:05 PM Post subject: Probabilities you are beaten

#1 (permalink)

Full House

Join Date: Feb 2005

Location: Bangkok

Posts: 1,090

This is something I have been arguing with my friends quite a bit.

Here is the scenario:You are holding KK. Is it more likely that one of your opponents has a pocket A(and has you beaten) when there is a)1 b)2 c)3 aces on the board. (and you do not hit the set)

I would calculate the scenario like this: lets say there is 6 players alltogether. After the flop there is one A showing, you can calculate that the probability that one of your 5 opponents (5 opponents having alltogether 10 cards, hence 10) has A is 3/47/10, if the turn is A as well the probability will be down to 2/46/10 and when the river is A as well the probability will be 1/45/10.

IMO the common sense tells you the same, you are more likely to win with you KK if there is AAA showing compared to just 1 A.

Am I wrong and if so, how do you calculate this?

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arkana

03-17-2005, 01:10 PM

#2 (permalink)

Full House

Join Date: Mar 2005

Posts: 1,109

If there is one A showing on the flop that means there are 3 remaining Aces among the unseen cards. There are 5 seen cards (the board and your 2).

That leaves 47 unseen cards... The probability (the value ranges from 1 to 0, 1 being 100% and 0 being 0%) that one of your opponents is holding an Ace is equal to 1 - the probability that none of them are holding an Ace.

I am calculating this in reverse to simplify it, the reason is that you cant simply calculate the odds of one opponent having an Ace and multiply that with 5 because the probability of event A occuring or event B occuring equals the probablity of A occuring added to the probability of B occuring minus the probability of both of them occuring: P(A or B) = P(A) + P(B) - P(A and B). Now with more than two events this gets even more complex. Its just easier to calculate the probability of none of them having an Ace and then subtract that from 1.

The 10 random cards your opponents are holding are basically a random sample from the 47 unseen cards. The probability of a sequence of events occuring is the probabilities of the events multiplied. So if we are picking 10 random cards out of the 47 card deck (which can only contain 3 Aces) the probability that none of them is an Ace would be:
(44/47) * (43/46) * (42/45) * (41/44) * (40/43) * (39/42) * (38/41) * (37/40) * (36/39) * (35/38) = 0.479185938945421
So the probability of one of them having an Ace would be
1 - 0.479185938945421 = 0.520814061054579 = 52% (1 Ace showing)

If there were 2 Aces on the board this would become:
(45/47) * (44/46) * (43/45) * (42/44) * (41/43) * (40/42) * (39/41) * (38/40) * (37/39) * (36/38) = 0.616096207215541
So the probability of one of them having an Ace would be
1 - 0.616096207215541 = 0.383903792784459 = 38% (2 Aces showing)

With 3 Aces on the board:
(46/47) * (45/46) * (44/45) * (43/44) * (42/43) * (41/42) * (40/41) * (39/40) * (38/39) * (37/38) = 0.787234042553191
So the probability of one of them having an Ace would be:
1 - 0.787234042553191 = 0.212765957446809 = 21% (3 Aces showing)

If the other 5 players were fish then I would say the probability is 100% since all 5 of them are seeing the flop and we know how much a fish loves an Ace.

That is why I personally believe that a poker bot just wont be able to beat a good human player. Human brains are just so much better at pattern recognition than machines are. Well at least currently...

Hope this helps, if you dont understand anything i will be happy to try and explain it to you.

Disclaimer: I cannot accept any responsibility for financial losses caused by using the above values since they were calculated from my knowledge obtained at university which was basically one big drunken haze.

Oh and if my maths is wrong, feel free to humiliate me as long as you explain to me why.

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