Lords of Death Forums Lords of Death Forums LoD General Discussion Poker stars rigged?

True probability doesn't work like that. We are talking about individual draws in a 52 card pool coming up with one of 4 aces or other cards. So the probablility in a 52 card stack that you come up with and single set card is 4/52. But in probability it never is that simple. Since in this case we are going to treat each event as dependant on the previous event I will attempt here to describe the probability one draws 4 of any one card (Aces or 8s).
Probability (A) = p(A) which is to say that p(A) = 4/52
so the probability one draws a single Ace from a deck of
52 cards is 1/13th or 7.7% roughly.
Here is where it gets a bit more complicated though. Since we now have to find the exact probabilty that a SERIES of events will occur.
Event A is that the first card is an ace. Since 4 of the 52 cards are aces, p(A) = 4/52 = 1/13. Given that the first card is an ace, what is the probability that the second card will be an ace as well? Of the 51 remaining cards, 3 are aces. Therefore, p(B|A) = 3/51 = 1/17 and the probability of A and B is: 1/13 x 1/17 = 1/221. What is the probability the 3rd card in a series is an ace? Given the first two cards drawn were aces at 1/13 x 1/17 and that we have 50 cards remaining in which only two are aces the p(A|B|C) = 2/50 or 1/25. Again, we apply it to the equation and end up with 1/13 x 1/17 x 1/25 = 1/5525. Now finally we get to the probability that 4 cards will be aces. Following along the previous lines, there are now 49 cards left of which only 1 is an ace. Thus, p(A|B|C|D) = 1/49. And the final probability is going to be 1/13 x 1/17 x 1/25 x 1/49.
So the final probability of p(A)p(B)p(C)p(D) given that this final outcome is dependant on the three previous draws, making the variable dependant, is 1/270725.
My question here is, how the hell did you come up with 20%? Mathematically it is far less than 1%.

Nice... Someone who understands probability. <img src="/~stretch/ubbthreads/images/graemlins/smile.gif" alt="" />
If only you understood how to login you might get somewhere...

There is no reason to even bring up the KK in the story, since the AA had you both beat.
The chances of somebody having a better pocket hand than QQ in a 10 person game is roughly 20% of the time.
So when you go all in and lose, it is nothing unusual. Doesn't mean a website is rigged either.

Whats the odds that somebody gets pocket Q's. Then whats the odds that somebody gets pocket K's or Pocket A's or AK suited or AK unsuited.
The chances that somebody will have AA, KK, AKs, AKos is a lot more than you think.

You win some, you lose some, what else can you say? I believe the online poker sites are running programs that have set values, I don't think they are doing anything other than randomly dealing and shuffling and following set rules like a Flush beats a straight and so on, evertying else is just a roll of the dice just like in an RPG. It does suck to get bad beat, but it happens to all of us, and especially me. I had A-9 down, 9-A-9 comes on the flop, right off the bat I have full house 9's over aces, I go all in, someone calls, the fucker who called had pockets aces, so full house aces over nines, I lose. Shit happens man. *shrugs* I play on ultimatebet though.

Pocket pairs get delt out together, just about anytime someone has a pocket pair, it seems like someone else has one too, they come in streaks.

the odds of a given player having aa or kk are about 110 to 1
multiply that by 9 (players other than you)
9 in 110 or ~8%
.08 * .08 = .0064
the odds of 2 people out of ten is a little better than .08 * .08 because the chance of aa aa or kk kk while remote, does exist.
also remember that you have 2 cards that arn't a or k, which lowers the total cards down a bit and helps the odds
so lets estimate that the odds are about 8 of 1000 or about 1 in 125 in the case of Lev's question
and to answer your other question Lev, NO, it isn't rigged. and you retards that say the random shuffling isn't like real life, go back to school before you ever open your mouth again.

sorry rain but i like QQ up againest AK any day. AK suited then maybe ill take it over QQ. Infact I like any pair againest AK. I dont consider AK> QQ which means the only hands i consider better then QQ are KK and AA
as for bad beats. i went all in in a tourney yesterday with 68 suited trying to steal the pot. Guy calls wtih JJ. Flop comes JD 3D and like AD i won wi\th hitting a flush on flop. it helped me but i felt bad. sence this shit is so common. yea whatever :P

Rain, I was just speaking on pure probability that ONE person would draw any combination of 1 to 4 Aces. You lessen the odds even more by adding 10 players to the mix. Now you have to account for the number of cards in the total deck minus the number drawn each round and figure the probability that way.
My point was there is no way in hell it is even close to 20%. I can show you the math on it if you like. In order to do it I will need to know your order in the draw. Wait, I have a better idea, let me give you the equation and you can do the math.
N= number of outcomes
P(r) = probabilities A-Z of each individual event given they are depandant.
The binomial probability for obtaining r successes in N trials is:

where P(r) is the probability of exactly r successes, N is the number of events, and p is the probability of success on any one trial.

i've made $682.38 more money off AK then off QQ. <img src="/~stretch/ubbthreads/images/graemlins/frown.gif" alt="" />
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