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Test your math knowledge

• Bronze
Joined: 25.11.2006
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
• 19 replies
• Bronze
Joined: 01.09.2007
i say:

63.2954%

what i did: 100%-chance of never getting pocket aces 221 times. hope that's right
• Black
Joined: 10.09.2007
It's just 1 - (220/221)^221?
• Bronze
Joined: 25.03.2005
You want to differentiate x^x?

x*x^x-1?
• Bronze
Joined: 01.09.2007
no rather:

x^x * (1+Log[x])
• Bronze
Joined: 25.03.2005
Well I guess my solution counts only for x € |N right?
• Bronze
Joined: 01.09.2007
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.
• Bronze
Joined: 25.03.2005
EDIT: Long time since I had this in school (4 years?).

But I think I got it.
• Bronze
Joined: 15.06.2007
Originally posted by Praetor
no rather:

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 :p
• Bronze
Joined: 01.09.2007
I just stated it as Mathematica returns it
And there Log stands for the natural logarithm...
• Black
Joined: 10.09.2007
Log = 10log and Ln is natural logarhytm
• Bronze
Joined: 25.03.2005
Originally posted by Yoghi
Log = 10log and Ln is natural logarhytm
Definate it however you want.

At the universit we had both for both. :/
• Black
Joined: 10.09.2007
Weird, I thought that was universal.
• Bronze
Joined: 15.06.2007
There's the universe and there's Germany.
• Black
Joined: 10.09.2007
According to Germany we belong to them so it shouldn't be different?

Yes it's a joke
• Bronze
Joined: 07.03.2007
log(x)=log.10.(x)
ln(x)=log.e.(x)
• Bronze
Joined: 25.11.2006
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 = ??
• Bronze
Joined: 01.09.2007
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)
• Bronze
Joined: 03.10.2007
hmm... probability you have pocket aces at least one time is 1-0.9955^221

and can i say answer tends towards 1?
• Bronze
Joined: 25.03.2005
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.