• May 7th 2008, 05:54 PM
ives
If the odds in favour of Boris beating Elena in a chess game are 5 to 4, what is the probability that Elena will win an upset victory in a best-of-five chess tournament?
----the answer is 7808/19683, i don't understand the chess thing basicly, i only can guess out what is the 19683, 3^9, but i don't know how to explain it. Can somebody explain me the chesses, and how to get the answer.
thanks a lot
• May 7th 2008, 06:00 PM
Jhevon
Quote:

Originally Posted by ives
If the odds in favour of Boris beating Elena in a chess game are 5 to 4, what is the probability that Elena will win an upset victory in a best-of-five chess tournament?
----the answer is 7808/19683, i don't understand the chess thing basicly, i only can guess out what is the 19683, 3^9, but i don't know how to explain it. Can somebody explain me the chesses, and how to get the answer.
thanks a lot

what's an "upset victory"?
• May 7th 2008, 06:04 PM
ives
i don't know either..........
• May 7th 2008, 06:05 PM
ives
that's the key to the answer. i think
• May 7th 2008, 06:21 PM
Jhevon
Quote:

Originally Posted by ives
that's the key to the answer. i think

not really

i think an upset victory just means the person who is least likely to win, wins.

thus you need to find the probability of Elena winning the best of 5 games, which means she wins three

so you want the probability of her winning 3 games

that should give you the answer, but don't quote me on that. as i said, i am a noob with probability
• May 7th 2008, 07:30 PM
TheEmptySet
This is a binomial probability(the formula is set up below)

$\displaystyle p(x)=\binom{5}{x} \left( \frac{4}{9}\right)^x \left( \frac{5}{9}\right)^{5-x}$

We want the prob of her winning at least 3 games so we want
P(3)+P(4)+P(5)

$\displaystyle p(3)=\binom{5}{3} \left( \frac{4}{9}\right)^3 \left( \frac{5}{9}\right)^{5-3}=\frac{16000}{59409}$

$\displaystyle p(4)=\binom{5}{4} \left( \frac{4}{9}\right)^4 \left( \frac{5}{9}\right)^{5-4}=\frac{6400}{59409}$

$\displaystyle p(5)=\binom{5}{5} \left( \frac{4}{9}\right)^5 \left( \frac{5}{9}\right)^{5-5}=\frac{1024}{59409}$

so we add all these up to get

$\displaystyle \frac{1024}{59409} +\frac{6400}{59409}+\frac{16000}{59409}=\frac{2342 4}{59409}=\frac{7808}{19803}$
• May 8th 2008, 02:26 AM
ives
thanks...
i haven't leaned the binomial things