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Originally Posted by Wedge Rock If I have K-Q against J-J and the flop misses us both, there are 7 cards out.
The chances of me making another {K or Q} are (1-(39/45)) + (1-(38/44)) = 24.7%
If someone tells me they folded a K (or a Q), my chances are (1-(39/44)) + (1-(38/43)) = 20.6% |
The math you are using to calculate probabilities is fundamentally flawed, Wedge Rock.
Those probabilities are wrong, and not because they ignore the slim chance of a Jack
helping your opponent. You are adding probabilities that do not give you the correct
answer. You can't just add the chance of hitting an out on turn and hitting an out
on the river, because you need to WEIGHT the latter probability according to how often it happens.
To clarify -- consider what would happen if you use your reasoning to find out the
probability of getting heads if you flip a coin two times. Using your reasoning, you
would say that your chances of getting heads on the first or second toss (or both)
is 1/2 + 1/2 = 1 = 100%, but we know that's not true.
The correct calculation of hitting a K or Q on turn or river (or both) is the sum
of the probability of hitting a K or Q on the turn (6/45) plus the probability
of hitting a K or Q on the river GIVEN that the first card was not a K or Q:
Your method:
P(hit out knowing nothing) = (6/45) + (6/44) = 26.9%
P(knowing one out is gone) = (5/44) + (5/43) = 22.9%
Correct method:
P(hit out knowing nothing) = (6/45) + (39/45)(6/44) = 25.2%
P(knowing one out is gone) = (5/44) + (39/44)(5/43) = 21.7%
Your overall point that learning one out is gone only affects your chances by about
4% is still valid.
But your seat-of-the-pants method of adding probabilities is not
something you should use in play, since it is going to always
lead you towards OVERSESTIMATING your chances of hitting your
outs, and as a result you are probably UNDERESTIMATING the pot odds
you need to justify calls. In the long run, this might be causing you
to make the fundamental mistake of calling when you don't actually have
the correct pot odds.
Checking the Math
Another way of thinking about this, and a way of checking the math above, is to
find the chances that you are NOT helped on the turn and NOT helped on the river,
and subtract that from 1.
If you know nothing about folded outs (45 unseen cards):
P(hit on turn or river or both)
= 1 - P(nonout on turn, nonout on river)
= 1 - [(39/45) x (38/44)]
= 1 - 0.748
= 0.252
= 25.2%
If you learn that one of your outs was folded (44 unseen cards):
P(hit on turn or river or both)
= 1 - P(nonout on turn, nonout on river)
= 1 - [(39/44) x (38/43)]
= 1 - 0.783
= 0.217
= 21.7%
The same approach can be used with the coin toss scenario:
P(heads on first or second toss or both)
= 1 - P(tails, tails)
= 1 - [(1/2) x (1/2)]
= 1 - (1/4)
= 0.75
= 75%
This agrees with our understanding that 3 out of the 4 possible outcomes
(HH, HT, TH, TT) involve at least one head showing up. Only 1 out of the
4 possible outcomes is unfavorable.
Linear Mode
