# Odds and probability question!

• Mar 17th 2010, 03:30 PM
ibetan
Odds and probability question!
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?

I found the probability of Elena winning to be 4/9. Then for winning the first 3 games of the series to be (4/9)^3= 0.088. However the textbook's answer key says the correct answer is 7808/19683 or .397. Can anyone help me please
• Mar 17th 2010, 05:42 PM
Soroban
Hello, ibetan!

Quote:

f 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 tournament?

I found the probability of Elena winning to be 4/9.
Then for winning the first 3 games of the series to be (4/9)^3
This is true, but it is not the only way Elena can win the tournament.

A best-of-5 tournament means the first to win 3 games is the winner.

Elena can win in three games.
There is one way: .$\displaystyle EEE$
Hence: .$\displaystyle P(\text{Elena, 3 games}) \;=\;\left(\frac{4}{9}\right)^3 \;=\;\frac{64}{729}$

Elena can win in four games.
There are three ways: .$\displaystyle BEEE,\;EBEE,\;EEBE$

Hence: .$\displaystyle P(\text{Elena, 4 games}) \;=\;3\cdot\left(\frac{4}{9}\right)^3\left(\frac{5 }{9}\right) \;=\;\frac{960}{6,\!561}$

Elena can win in five games.
There are 6 ways: .$\displaystyle BBEEE,\;BEBEE,\;BEEBE,\; EBBEE,\;EBEBE,\;EEBBE$

Hence: .$\displaystyle P(\text{Elena, 5 games}) \;=\;6\cdot\left(\frac{4}{9}\right)^3\left(\frac{5 }{9}\right)^2 \;=\;\frac{9,\!600}{59,\!049}$

Therefore: .$\displaystyle P(\text{Elena wins tourna{m}ent}) \;=\;\frac{64}{729} + \frac{960}{6,\!561} + \frac{9,\!600}{59,\!049} \;=\;\frac{23,\!424}{59,\!049} \;=\;\frac{7,\!808}{19,\!683}$