Just thought i would give you guys a problem that you try on your reak from poker. I guess its not very hard but thats because i study maths.
Probability of receiving pocket aces = 1/221
Which is about 0.45 %
OK so if we play 221 hands what is that probability you will have pocket aces at least one time?
EDIT: Maybe that was easy. Well differentiate
f(x) = x^x
i say:
63.2954%
what i did: 100%-chance of never getting pocket aces 221 times. hope that's right 
i fear that this is not the case. the exponent has to be independent of x in order to apply the formula (?) you used. and that's just not the case.
Originally posted by Praetor
norather:
x^x * (1+Log[x])
Well, almost correct: it's (1+ln(x))x^x
Not just a log(x) which is the 10th logarithm
I just stated it as Mathematica returns it
And there Log stands for the natural logarithm...
Originally posted by Yoghi
Log = 10log and Ln is natural logarhytm
Definate it however you want.
At the universit we had both for both. :/
According to Germany we belong to them so it shouldn't be different?
Yes it's a joke
Did anybody do it by hand?
EDIT.
OK here is another problem that bought a lot of discussion. I'm not sure how to format it so ill try this. Solve for x
x^(x^(x^(x^(x^(x ...infinite..))))) = 2
So x to the power of x to the power of x an infinite amount of times = 2
Or maybe you can see it better this way
...............
.............
...........
........x
......x
....x
..x
x = 2 => x = ??
hehe i know it, wrote a paper about power towers once, but i won't give the answer this time.
EDIT. concerning the question of doing it by hand, no i didn't
but i know how to do it ofc... (put it to form e^(...) then use rules for exp and products)
hmm... probability you have pocket aces at least one time is 1-0.9955^221
and can i say answer tends towards 1?
Originally posted by Yoghi
According to Germany we belong to them so it shouldn't be different?Yes it's a joke
Lol.
@Topic: I know why I stopped studying it.
