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wayotogo
10-03-2005, 05:56 AM
I've read that the chance of hitting an A or a K by the river when you hold AK is 50% (http://www.learn-texas-holdem.com/texas-holdem-odds-probabilities.htm) but how do you arrive at this number?

What is wrong with my own simple approach:
6/50 + 6/49 + 6/48 + 6/47 + 6/46 = .6255 = 62.55%

I'm not concerned with the chances of hitting more than one A or K, so that does not need to be discounted.

Also is there a good book out there that goes through these kinds of basic scenarios and numbers for the not so math adept people such as myself?

BruceZ
10-03-2005, 06:14 AM
[ QUOTE ]
I've read that the chance of hitting an A or a K by the river when you hold AK is 50% (http://www.learn-texas-holdem.com/texas-holdem-odds-probabilities.htm) but how do you arrive at this number?

[/ QUOTE ]

1 minus the probability of not hitting an A or K:

1 - (44/50 * 43/49 * 42/48 * 41/47 * 40/46) =~ 48.7%.

www.learn-texas-holdem.com/texas-holdem-odds-probabilities.htm (http://www.learn-texas-holdem.com/texas-holdem-odds-probabilities.htm) is a little off, and it has a number of other errors (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Board=genpok&amp;Number=2947988&amp; Forum=f28,f1,f6,f30,f24,f7,f8,f10,f2,f4,f3,f21,f5, f25,f9,f22,f23,f11,f12,f13,f14,f35,f37,f34,f26,f15 ,f16,f17,f18,f19,f31,f36,f27,f33,f32,f20&amp;Words=web site&amp;Searchpage=0&amp;Limit=25&amp;Main=2946573&amp;Search=tru e&amp;where=bodysub&amp;Name=197&amp;daterange=1&amp;newerval=4&amp;ne wertype=m&amp;olderval=&amp;oldertype=&amp;bodyprev=#Post29479 88) as well.

[ QUOTE ]
What is wrong with my own simple approach:
6/50 + 6/49 + 6/48 + 6/47 + 6/46 = .6255 = 62.55%

[/ QUOTE ]

You would have to multiply each probability by the probabiilty that we missed on previous cards:

6/50 +
(44/50)*6/49 +
(44/50 * 43/49)*6/48 +
(44/50 * 43/49 * 42/48)*6/47 +
(44/50 * 43/49 *42/48 *41/47)*6/46

=~ 48.7% as before.

The probability of each board card being an A or a K before any of them are dealt is 6/50. If we add the 5 values of 6/50, or multiply 6/50 by 5 for the 5 cards, this will over count the times that we get more than one A or K. We cannot sum probabilities of events that are not mutually exclusive (more than one can happen).