Guys, a simple math question not related to poker

[revlis87]

Wrong forum but if I could just get a quick answer that'd be great. I contribute a lot of content so...

Let's say I've got a test with 13 possible essay topics. The prof will assign 5 of them, of which I can choose one. How many do I have to prepare to make it statistically like 90% that I get one that I can do?

First one who tells me gets $1 shipped to acct of their choosing.

Love,

Borntarun

[mosta]

number you have prepared versus percent chance he picks five from 13 without picking one you have prepared:

one 62%
two 36%
three 20%
four 10%
five 4%
six 2%


edit. I think I fixed it. don't count on me to have it right--but to inspire someone to correct me. I often screw it up.

[highhustla]

so there are (13 choose 5) combos the prof can choose = 1287

and you're screwed only if he chooses 5 that you don't know. and you wanna be screwed < 10% of the time.

so we want (x choose 5) / (13 choose 5) < 0.10

where x is how many you don't know, and x choose 5 is how many ways you can get screwed.

(9 choose 5) / (13 choose 5) = 0.0979020979

13-9 = 4 is my answer

[slik]

Should do 4 to guarantee 90%.
1-[(9/13)(8/12)(7/11)(6/10)(5/9)]=0.903

[Apathy]

As always theres more then math to the question... you should be able to predict which topics the professor is more likely to put on and also which essays have the most overlap and are therefore less likely to BOTH be on the exam.

[revlis87]

come on I'm not that big of a fish as to not know that... but this particular professor is a real oddball - i predicted last time and I literally got the 6 that I thought werne't gonna be on it... so i gotta just do math with this guy. thanks for all the help guys.

[Ratamahatta]

I like the way OP thinks!

[greg nice]

haha nice

[revlis87]

ok, i realized it was 15 questions... i 'm really math retarded (hence why i suck at poker). how does the math change?

[TheStandman]

You should do 9 to be 100%....