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[Closed] full tilt sit n gos all in

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nice conversation....lol, many players try to trap but it traped them self. no1 will call with 27 when u raise big enough (unless they crazy enough or a donk) donks win 10% against skilled players. don't be mad. still have 90%.


fembot26
Joined: 25.08.2010

Time for some mathematical nerdiness (precision, rigour, hair-splitting, what you will):

(Actually, I'm kind of interested myself to see what the probabilities are)

Using the binomial distribution, the probability of eight out of 16 dominated hands winning all-in showdowns is
C(16,8) x .27^8 x .73 ^ 8 ~ 2.93%,
and the probability of seeing eight or more (8, 9, 10, ..., or 16) such wins is about 4.20%.
But if the hands that are behind are on average 35% to win, which is a bit more realistic, then the probability of seeing eight or more wins out of 16 by underdog hands is about 15.94%: you could expect to see this result in about 3 out of twenty SNGs, on average. So it's not ridiculously unlikely.

Even if you'd seen ten or more wins out of 16 by underdog hands, the probability of this is about 2.29%, so you can expect this in about one in 44 SNGs.

Coincidences are less rare than most people assume. The chance of you reading this exact post in your lifetime is pretty damned low, but the chance of you one day being on a site where some schmuck feels the need to show off his elementary-level maths knowledge is very high. ;)


Originally posted by fembot26
Coincidences are less rare than most people assume. The chance of you reading this exact post in your lifetime is pretty damned low, but the chance of you one day being on a site where some schmuck feels the need to show off his elementary-level maths knowledge is very high. ;)


belthazorrrr
Joined: 12.02.2011

Originally posted by fembot26
Time for some mathematical nerdiness (precision, rigour, hair-splitting, what you will):

(Actually, I'm kind of interested myself to see what the probabilities are)

Using the binomial distribution, the probability of eight out of 16 dominated hands winning all-in showdowns is
C(16,8) x .27^8 x .73 ^ 8 ~ 2.93%,
and the probability of seeing eight or more (8, 9, 10, ..., or 16) such wins is about 4.20%.
But if the hands that are behind are on average 35% to win, which is a bit more realistic, then the probability of seeing eight or more wins out of 16 by underdog hands is about 15.94%: you could expect to see this result in about 3 out of twenty SNGs, on average. So it's not ridiculously unlikely.

Even if you'd seen ten or more wins out of 16 by underdog hands, the probability of this is about 2.29%, so you can expect this in about one in 44 SNGs.

thanks for that
Coincidences are less rare than most people assume. The chance of you reading this exact post in your lifetime is pretty damned low, but the chance of you one day being on a site where some schmuck feels the need to show off his elementary-level maths knowledge is very high. ;)


AussieIan
Joined: 10.12.2007

Another way to look at it is to use the 16 hands in total.

If the hands behind have an avg of 30% to win, then you would expect 5 of the 16 behind hands to win. The fact that 50% of the hands that were behind won isn't that remarkable when you consider it that way, and no doubt you'll find another sng where 10-12 of the favourites may have won.

Hope that makes sense in print rather than just in my head!


konkey
Joined: 29.12.2009

And for those about to think every better hand should win...
Probability of winning 16 out of 16 70% favourite hands is ~0.33%