Probability Question - Page 2

like .000000000000000000000000000001% chance to go bust. Heres my logic...go google percentages of hands, then go check pt. You're suppose to get pocket aces about X% of times, so take the # of times you got aces, put it over the total amount of times, and it should be within a very small margin of the odds of getting it.... If you 51% of the time, the theory of averages amd all that say's you will always win more than the house does. Unfortunately, the house has better odds than you IRL, but it is the most profitable game to play at a casino (as in you have the best odds compared to slots/roulette/craps etc...)

In conclusion, check PT to confirm the actual probability of getting cards to prove that probabilty is basically always right. With that said, if you have 51% odds, you will win more than the house and will not go busto.

i figured it out

RAKE WINS!

umm ok i think that is has to be a function it cant be a single number beacuase u increase or decrease ur total money with every hand. the base number would be the probability of loosing 51-49 1,000 in a row times that would be (x), but tyhats just a theory cuz im REEEEEALLY bad at math especially anything higher than trigonometry. ask Daut he should know

my point is if the probability of going broke varies like losing 1000 time in a row is lets say 1% the number will increase to go broke once u lose 1 time making it lets say 1.001% but if u win the hand it would decrease making it 0.999999%

hmm so taking this is mind then teh answer would be probability of losing = [49% (cuz of teh last hand u will play when u have 10 bucks left) > probability of losing 49-51 1000 times in a row > infinite]

ship teh prize?

hmm..i think this is a markov chain question

so simplified (49% > a > infinite) > 0

where a = prob of losing 1k 49-51 flips in a row


edited cuz forgot to put 0

lol now im hyped i wanna know if im right some1 tell me the answer wiiiii

if there is any chance, no matter how small, over an infinite number of trials you will go busto.

well yes because u have to keep playing always, your purpose is not to win money is just to bet 10 dollars over an infinite if your purpose was to make it profitable you would take out the original invesment after u have profited a decent amount.

Its the same thing we do with the martin gale if we had an infinite BR we could keep doubling our bet and eventually we will win even if our flip is 1 - 99 in favor of the house, hence they house puts a cap or maximum bet

Applying the concept of percent to this game will yield counterintuitive results since there are infinite possible game outcomes. The problem is that there are infinitely many possible games in which you bust and infinitely many in which you never bust (it is trivial to construct a scenario in which you never bust - simply hit your 51% every time). To try to divide this infinite set into parts that can be meaningfully compared to one another is problematic.

If you wish to argue either 100% or 0% is the answer, then I submit that by the same argument you can conclude that either 0% of all integers or 100% of all integers contain the digit 3.

it can never be greater than 49% cuz thats the probability of losing the hand that would declare you busto it can never be 0% because theres always a chance to run extremly bad and go busto. but the point remains if the purpose is to play infinite u will indeed go busto if the purpose is to make a profit then u have probability of losing 1k flips in a row of going busto.... AHH my heard hurts to much thinking

That's not to say that an argument that arrives at 0% busto is wrong. I think 0% is actually the correct answer here. It's simply a consequence of trying to apply concepts used for regular sets on infinite ones. If you want to try to use percentages when talking about a set such as the integers, the result you obtain is actually that 100% of all integers contain the digit 3.

On September 10 2008 18:26 CrownRoyal wrote: so this means its theoretically possibly impossible to win at poker, investing, or anything really amirite

On September 10 2008 18:26 CrownRoyal wrote: so this means its theoretically impossible to win at poker, investing, or anything really amirite

if you don't win someone else does, the house in this case.

On September 10 2008 19:29 Jorge wrote: it can never be greater than 49% cuz thats the probability of losing the hand that would declare you busto it can never be 0% because theres always a chance to run extremly bad and go busto. but the point remains if the purpose is to play infinite u will indeed go busto if the purpose is to make a profit then u have probability of losing 1k flips in a row of going busto.... AHH my heard hurts to much thinking

No it's the other way around. The probability of going bust is AT LEAST and definitely greater than 49% since that's the chance we go busto on the first hand. Since we can also go bust on the 3rd hand etc we see that it is definitely represented by an increasing series with decreasing extra terms in the sequence of partial sums. The question is whether it has a limit. I've been working on it and I'm fucked if I know. It's a great question though.

edit: that's a limit <100%

wait what? u can never go bust on the first 999 hands cuz " Each game you must bet $10 "

edit: its the chance of loosing all ur moneys not the hand sir

oh whoops sorry I missed that, I was assuming starting from $10. The principle of the theory stays the same for $10,000 I imagine but it just gets a whole lot uglier.

yo it's a standard risk of ruin calculation. the winrate is easy, it's $0.10/hand, the bankroll is $10k and I'm too tired to work out how to work out the standard deviation and dig up the equation; but that's all the variables I believe...

u dont have to work out anything it's 100 percent.