Forums-->General game forum-->
1|2|3|4|5|6
| Author | Roulette Mathematics |
|---|
So if the most successful,pattern aware,player has 0.3% guess rate than the average Why the hell did he placed 12000 bets in vain if he has cracked roulette?
why not to wait his pattern to appear and what kind of pattern is it that appears once every 345 placed bets(BETS not spins)?
Easy answer,, because and why there is no pattern.
Well, that's not how the pattern apparently supposed to work (according to the 'pro-pattern')
It's more of a number Y more likely to appear after X appears, so it normally appears on average less than 38 times, which is of a higher chance than average
Which makes the 12000 failed bets makes sense, since he still wins more than average | He has a total of 357 wins.321 of them are the normal expected value (E[X]=n*p=12200*1/38).So pure average luck is responsible for those 321 wins and his hypothetical pattern gave him 36 more.That is one in 339 bets(not 345 sorry).Only for one win in 339 bets the "pattern" is responsible.the rest of the wins are expected anyway.So he places his bet when he sees the start of his pattern and expects the next number to appear in less than 38 times(on average) because of the pattern. But the pattern works once in 339.Isnt his fallacy obvious?Do you thing the probability to win at least one more time than the average in 339 games is zero?if you do the maths you have something between 37% and 50%.
Some more mathematic facts
-------------------------------------------------------
X the number of successes in 12200 bets
the successful gambler( player http://www.heroeswm.ru/pl_info.php?id=2161082):
P(X = 357)=0.00291195381743252 (have exactly 357 wins in 12200 bets)
P(X < 357)=0.976109496093327 (have less than 357 wins in 12200 bets)
P(X <= 357)=0.979021449910759 (have less than or equal to 357 wins in 12200 bets)
P(X > 357)=0.020978550089241 (have more than 357 wins in 12200 bets)
P(X >= 357)=0.023890503906673 (have more than or equal to 357 wins in 12200 bets)
!!!!you have about 2,3% probability to be as successful or more than him.
the average gambler:
P(X = 321)=0.0225596716173747
P(X < 321)=0.491092762258466
P(X <= 321)=0.513652433875841
P(X > 321)=0.486347566124159
P(X >= 321)=0.508907237741534
Someone with zero balance:
P(X = 339)=0.0132440379265326
P(X < 339)=0.838216414011042
P(X <= 339)=0.851460451937575
P(X > 339)=0.148539548062425
P(X >= 339)=0.161783585988958
Do you think those numbers are unreal and our community needs patterns to achieve them?
you can read about binomial distribution and do the maths yourself | P(X > 339)=0.148539548062425
Does this mean that there's approximatively 14,8%, let's say 15% people who have a positive balance at the roulette when they play 12200 bets ?
It looks ridiculous | Ok then, but when I say "this guy has to succeed let's say 13 times without loosing" I mean he has to win 13 times IN A ROW !! Before telling me I'm an idiot, be careful about what I wrote.
Yeah and that's the problem. The chance to win 13 straights in a row is almost non-existent. Whereas the chance to have a profit of 13*1000*CL is far far far more if you have placed thousands and thousands of bets, it is definitely possible. Unlikely, but possible. | for Lady sofiouta:
Can you show how you calculated those values?
Does this mean that there's approximatively 14,8%, let's say 15% people who have a positive balance at the roulette when they play 12200 bets ?
It looks ridiculous
Well remember one of the conditions is that you always bet the exact same amount and bet on completely random numbers. I think 15% seems kind of normal, nothing too unusual.
Most people you see will be at a negative because they don't bet exactly the way we are calculating, no wonder, and remember 15% people are just barely over the loss barrier, so 85% are at a loss sounds like quite a big deal to me.
Also remember out of this 15% actually only 2.3% are at huge profits, and that 2.3% means very very very few people as not a lot of people have bet 12000 times or more. | Well that is a different problem.
I thinkk you want to say that 15 from 100 players have positive balance.Lets rephrase that.
Among CL 15 players that have placed 12200 bets,the probability that a player has positive balance is 14,8% lets say 15%.We select 100 of those players. What is the probability 15 of them have positive balance?
the expected value E[15]=0.15*100=15.
the probabilities are:
P(X = 15)=0.11109095935 (exactly 15 have possitive)
P(X < 15)=0.45722420578 (less than 15 have possitive)
P(X <= 15)=0.56831516513 etc
P(X > 15)=0.43168483487
P(X >= 15)=0.54277579422 | @randomr1
it is simple binomial and cumulative probabilities calculations.The mathematics type is a bit complex to do it manually so i use a calculator for that.You can find one online. | for Lady sofiouta:
I see, I thought you did them manually :P |
This topic is long since last update and considered obsolete for further discussions.
1|2|3|4|5|6Back to topics list
|
