Had to google 0% probability events after this thread. Very interesting concept, that is dominated by scenarios that are either paradoxial (it will rain and not rain at the same time tomorrow at the same location) or events outside the possible range of results (chance I throw a 7 on a standard die).
In Response to Re: The Probability Thread : hi CM, unfortunately your life is finite. if there were an unending line of CMs one of your descendents would press the button. but please don't just yet. Posted by aussie09
Fairly easy maths, but as Coxy, I think, said, the answer is tough to believe if you don't want to.
But a good suggestion for this use.
But we should also invoke the spirit of Occam's razor
i.e.
of the competing hypotheses
i) the RNG is rigged
ii) the regs have a boom box
iii) i just don't understand the maths
we should apply the appropriate heuristic and end up choosing iii as the most appropriate response before we invoke the spirit of fraudulent sky poker programmers.
couple of poker paradoxes people may not grasp at first:
----
if
A />B
and
B />C
is
A />C?
in logic, yes.
in poker, no.
A = JTs
B = 22
C = AK
using pokerstove:
JTs />22 [54% v 46%]
22s />AK [54% v 46%]
JTs<AK [40% V 60%]
------
bluff catching a river:
you are facing an all in bet, do you call knowing you will be beat 60% of the time. ie you are wrong more often than you are right?
starting stacks 150bb, you both have 50bb in the middle, on the rive villain overbet shoves all in for his and our remaing 100bb. you know enough about his ranges to know that your hand beats only 40% of his range. more often than not you will lose an extra 100bb and your ENTIRE stack.
action?
CALL!
you will lose our stack 60% of the time
40% of the time you will rake a pot of 300bb
your excpected stack size after calling then will be:
[0.4 x 300] + [0.6 x 0]
120 + 0
120bb
if you fold your stack size will always be 100
since
120 /> 100
we call despite knowing we are "wrong" more often then we are 'right'.
psychologically it is the hardest thing to do. we are losing our stack more often than not, and our edge only shows up after a LOT of volume. short-term we may get HAMMERED.
Lets consider the case when we go all in with AA and are up against 2-7, assume we run this 25 times and ignore rake.
Let's also assume that we win 5 in 6 times with this hand (83.3%), in our sample lets assume we win 21 times out of 25. We can assume for a significantly large sample size (i.e. >25) that this is normally distributed (google this).
We can work out a 97.5% confidence interval for the expected value (i.e. 97.5% chance that this interval contains the epectation).
We can work out the (sample) variance using a neat little formula and hence the standard variation (the sq. root of the variance). In our case we have that the sample variance is 7.66.
So using the fact that a 97.5% confidence interval is approximately 2 standard deviations away from the mean (which is +18.35 in our case) we get a 97.5% confidence interval for the expectation as [+3.4, +32]
We conclude that we can in fact have a negative expectation in our interval (i.e. values between +3.4 and +10)
Heres another more simplistic example. Assume we take 5 all ins with AA vs opponents 7-2 in a tournament and also assume that we are out of the tournament if we lose. Yes each individual flip will result in ~88% (5/6 ish) chance of surviving, but if we were to consider the chances of being knocked out of the competition in 5 of such All ins we see that the probability of winning all 5 all ins is,
Let's now consider that a gambler puts £1 on red or black on a roulette wheel (ignore the 0), he'll double his stake if he loses and leave the table if he wins, he continues this forever. Now one might think this is a sure way of winning money? Absolutely not.
If he loses 9 on the trot his next bet will be ~£1,000 (gettin expensive right?) lose 19 on the trot your next bet will be just over £1,000,000.
So, unless you have an infinite (isn't possible) amount of money you'll come unstuck.
If you meet someone and they tell you “I have two children, and at least one of them is a boy”, what is the probability that the other child is a boy? (Hint: It’s not 1/2). It’s actually 1/3.
The second child paradox. If you meet someone and they tell you “I have two children, and at least one of them is a boy” , what is the probability that the other child is a boy? (Hint: It’s not 1/2). It’s actually 1/3 . Posted by Swog
Comments
----
if
A />B
and
B />C
is
A />C?
in logic, yes.
in poker, no.
A = JTs
B = 22
C = AK
using pokerstove:
JTs />22 [54% v 46%]
22s />AK [54% v 46%]
JTs<AK [40% V 60%]
------
bluff catching a river:
you are facing an all in bet, do you call knowing you will be beat 60% of the time. ie you are wrong more often than you are right?
starting stacks 150bb, you both have 50bb in the middle, on the rive villain overbet shoves all in for his and our remaing 100bb. you know enough about his ranges to know that your hand beats only 40% of his range. more often than not you will lose an extra 100bb and your ENTIRE stack.
action?
CALL!
you will lose our stack 60% of the time
40% of the time you will rake a pot of 300bb
your excpected stack size after calling then will be:
[0.4 x 300] + [0.6 x 0]
120 + 0
120bb
if you fold your stack size will always be 100
since
120 /> 100
we call despite knowing we are "wrong" more often then we are 'right'.
psychologically it is the hardest thing to do. we are losing our stack more often than not, and our edge only shows up after a LOT of volume. short-term we may get HAMMERED.
Some really interesting things on here!