the flames and avalanche are in the finals. each team hopes to win the best-of-seven games. the probability of the flames' win in any game is 80%. what is the probability that the flames will win the series in exactly 6 games???
2006-11-30 15:11:01 · 4 answers · asked by shishi 1 in Science & Mathematics ➔ Mathematics
OK... so the question isn't completely clear, but here's my best interpretation... The winner needs best of seven games, so the first team to rack up four wins will win the whole thing. If the Flames win in exactly 6 games, then that means that game 6 was their fourth win. In the previous 5 games, they must have won three and lost two. There are 10 different ways they could have gotten these three wins and two losses. It might have been Win, Win, Win, Lose, Lose. Or it might have been Win, Win, Lose, Win, Lose. And so on and so forth. The probabilty of the first pattern is 0.8*0.8*0.8*0.2*0.2. (20% chance of a loss). The probability of the second pattern is the same: 0.8*0.8*0.2*0.8*0.2 (although the order of the numbers is different the product is the same). There are ten patterns. So, the chance of winning 3 out of five is 10*0.8^3*0.2^2. Then, the chance of winning game 6 is 0.8. So the odds are 0.16384 or about 16%.
2006-11-30 15:31:53 · answer #1 · answered by pamgissa 3 · 0⤊ 0⤋
I agree with the first responder. With each succeeding game, the chances that the flames will win go down. You just keep multiplying to find out the probability in the 6th game.
1st game = 80%
2nd game = 80% x 80% = 64%
3rd game = 80% x 80% x 80% = 51%
4th game = 80% x 80% x 80% x 80% = 41%
5th game = 80% x 80% x 80% x 80% x 80% = 33%
6th game = 80% x 80% x 80% x 80% x 80% x 80% = 26.2%
This math applies in real life not just to betting odds, but to manufacturing as well. Suppose you were building an electronic device with 6 circuit boards. There's an 80% chance that any one board will work correctly. What's the chance that the completed device will work - 26.2%! Math is important in determining quality as well as in determining betting odds.
2006-11-30 23:41:03 · answer #2 · answered by keepthemhonest 2 · 0⤊ 0⤋
You're getting some incorrect responses. If they win in exactly six games it means they win exactly 3 of the first five then win the sixth.
If they win 3 then lose 2 then win 1 the probability looks like this:
0.8 * 0.8 * 0.8 * 0.2 * 0.2 * 0.8.
But those first five results can happen in any order. There are 5C2 = 10 ways this can happen.
The probability is 10 * 0.8^4 * 0.2^2 = 0.016384.
2006-11-30 23:55:52 · answer #3 · answered by MathGuy 3 · 0⤊ 0⤋
Probability of winning one game (0.8) * probability of winning another game (0.8) * probability.....
Ans: (0.8)^6 = 0.262 or about 26.2%
As you can see, the "margin for loss" stacks up as consecutive games are won. If only reality were that simple...
2006-11-30 23:20:57 · answer #4 · answered by John H 4 · 0⤊ 0⤋