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well it's counter intuitive for me at least [img]/images/graemlins/confused.gif[/img] and not so basic! [img]/images/graemlins/tongue.gif[/img]
so i'm trying to figure out the % chance of hitting an Ace or a King on the flop when you hold AK and are heads up and you're assuming your opponent doesn't hold an A or a K.
there are 3 remaining Aces and 3 remaining Kings available and i thought the equation was 6/48 + 6/47 + 6/46 = .383?
but that's wrong apparently?
could someone please give me the correct formula?
and if anyone wanted to explain why the 'intuitive' formula is wrong that would be appreciated too!
cheers
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To do it your way, we must do it like this:
P(flopping A or K) =
P(1st card is A or K) +
<font color="red">P(1st card is NOT A or K)*</font>P(2nd card is A or K when 1st is not) +
<font color="red">P(1st and 2nd cards are NOT A or K)*</font>P(3rd card is A or K when 1st and 2nd are not)
The probability of getting an A or a K on just the first 2 flop cards, which we will call P2, is:
P2 = 6/48 + <font color="red">(1 - 6/48)*</font>(6/47) =~ 23.67%
and then for all 3 flop cards it is:
P2 + <font color="red">(1 - P2)*</font>6/46 =~ 33.6%.
We must multiply the 6/47 by (1 - 6/48) which is the probability that we miss on the first card since the 6/47 only applies to the times that we miss on the first card. Otherwise we would be double counting the times that we hit on both cards since these are already accounted for by the 6/48. Similarly, the 6/46 only applies to the times that we miss on both the first and second cards, which has probability 1 - P2.
Note that instead of (1 - P2) above, we could have written 42/48 * 41/47 since this is also the probability of no A or K on the 1st and 2nd cards.
A much easier way to do this problem is to take 1 minus the probability of NOT getting an A or K on any of the 3 cards. This is:
1 - (42/48 * 41/47 * 40/46) =~ 33.6%.
This post answers the same type of question for completing a flush draw.