Hey

If the odds in favor 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?

Can someone give me an answer with explanation Plz.

Thanx

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- Apr 19th 2008, 07:20 PMballin_sensationProbabilities Plz Help
Hey

If the odds in favor 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?

Can someone give me an answer with explanation Plz.

Thanx - Apr 19th 2008, 08:38 PMSoroban
Hello, ballin_sensation!

Quote:

If the odds in favor 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?

There are three ways in which she can win three games.

(1) She wins the first three games:

. . . $\displaystyle \left(\frac{4}{9}\right)^3 \:=\:\frac{64}{9^3}$

(2) She wins two of the first three games, then wins the fourth game.

. . . $\displaystyle {3\choose2}\left(\frac{4}{9}\right)^2\left(\frac{5 }{9}\right)\cdot\frac{4}{9} \;=\;\frac{960}{9^4}$

(3) She wins two of the first four games, then wins the fifth game.

. . . $\displaystyle {4\choose2}\left(\frac{4}{9}\right)^2\left(\frac{5 }{9}\right)^2\cdot\frac{4}{9} \;=\;\frac{9600}{9^5}$

Therefore: .$\displaystyle P(\text{Elena wins}) \;=\;\frac{64}{9^3} + \frac{960}{9^4} + \frac{9600}{9^5} \:=\:\frac{23,424}{59,049} \;=\;0.396687497 \;\approx\;40\%$

- Apr 19th 2008, 08:48 PMTheEmptySet
This seems like a binomial probabilility so

the prob of x successes in n trials is

$\displaystyle P(x)=\binom{n}{x}P^{x}(1-P)^{n-x}$

where P is the prob of success

since we want an upset victory she needs to win 3,4, or 5 games

so we can add P(3) +P(4)+P(5) or take 1 -P(0)-P(1)-P(2)

Now lets set up the formula:

The odds are 5 to 4 against her so $\displaystyle P=\frac{4}{9}$

and they are playing the best of 5 so

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

$\displaystyle P(3)=\binom{5}{3}\left( \frac{4}{9}\right)^{3} \left(\frac{5}{9} \right)^{5-3}=\frac{5!}{3!2!} \cdot\left( \frac{4}{9}\right)^{3} \left(\frac{5}{9} \right)^{5-3}= 10 \frac{(4^3)(5^2)}{9^5} \approx .2710$

$\displaystyle P(4)=\binom{5}{4}\left( \frac{4}{9}\right)^{4} \left(\frac{5}{9} \right)^{5-4}=\frac{5!}{4!1!} \cdot\left( \frac{4}{9}\right)^{4} \left(\frac{5}{9} \right)^{5-4}= 5 \frac{(4^4)(5^1)}{9^5} \approx .1083$

$\displaystyle P(5)=\binom{5}{3}\left( \frac{4}{9}\right)^{5} \left(\frac{5}{9} \right)^{5-5}=\frac{5!}{5!0!} \cdot\left( \frac{4}{9}\right)^{5} \left(\frac{5}{9} \right)^{5-5}= 1 \frac{(4^5)(5^0)}{9^5} \approx .0173$

so P(3)+P(4)+P(5) $\displaystyle \approx .3966$ - Apr 19th 2008, 08:49 PMTheEmptySetDang!
Too slow HAHA(Rofl)

- Apr 20th 2008, 07:23 AMballin_sensation
Thanx for the help guys, really appreciate it.