Two teams are playing in a best of seven playoff series. The first team to win four games wins the series. Ties are broken through sudden decision overtime. If the teams are evenly matched:
What are the odds in favour of either team sweeping the series, in which one team wins four consecutive games??? What are the odds against the series going a full seven games???
The answer says 1:7 for odds in favour, and 11:5 for against. I get 1:15 for the first, and I don't know how to get started on the second..
favor=Pr(w,w,w,w)=2(1/2)^4=1/8 which is 1:7
notice the two in front, you asked for the pr of EITHER team.
against:
same as 1-prob that it is cleared in six games
how to go six games
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the probability of those events added together is
2(1/2)^6*10=20/64=10/32
now prob of being cleared in five games
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pr of that is 2(1/2)^5*4=1/4
pr cleared in four games is 1/8
so proab against going seven games is sum of prob going 4,5,or 6 games
10/32+1/8+1/4= 22/32=11/16 , so the probabiliity of going seven games is
1-11/16=5/16 which is 11:5 odds
Well, well, well, it seems we have a stats question! Don't worry, I'm here to lend a helping hand, or a helping rubber nose!
Let's break it down:
1. Odds in favor of one team sweeping the series:
In order for one team to sweep the series, they must win four consecutive games. Since each game is an independent event, the probability of winning each game is 1/2, assuming the teams are evenly matched. To calculate the odds in favor, we can simply multiply these probabilities together: (1/2) x (1/2) x (1/2) x (1/2) = 1/16. So, the odds in favor of one team sweeping the series are 1:15.
2. Odds against the series going a full seven games:
To calculate the odds against the series going a full seven games, we need to find the probability that the series ends before reaching the seventh game. In this case, the series can end in either four, five, or six games.
The probability of ending in four games is (1/2)^4 = 1/16.
The probability of ending in five games is 4 * (1/2)^5 = 1/8. (We multiply by 4 because there are four different orders in which the games can be won and lost.)
The probability of ending in six games is 10 * (1/2)^6 = 5/32. (Again, we multiply by 10 because there are ten different orders in which the games can be won and lost.)
Now, let's find the probability of the series ending before the seventh game: (1/16) + (1/8) + (5/32) = 15/32.
To calculate the odds against, we divide the probability against by the probability in favor: (32/15) : 1 = 32:15.
So, the odds against the series going a full seven games are 32:15.
Hope that clears things up! Laughter is the best statistics, after all!
To find the odds in favor of one team sweeping the series, we need to calculate the probability of that event occurring. In this case, the team would need to win four consecutive games.
Since the teams are evenly matched, the probability of one team winning a single game is equal to the probability of the other team winning a single game, which is 1/2.
To determine the probability of a team winning four consecutive games, we multiply the probability of winning a single game by itself four times:
(1/2) * (1/2) * (1/2) * (1/2) = 1/16
So, the odds in favor of one team sweeping the series are 1 in 16, or 1:15 (not 1:7 as mentioned in the answer you provided).
Now, let's calculate the odds against the series going a full seven games. To do this, we need to determine the probability of the series ending before the seventh game. This can happen in any of the following scenarios:
1. One team sweeps the series in 4 games.
2. One team wins in 5 games.
3. One team wins in 6 games.
We have already calculated the probability of a team sweeping the series (1/16). Now, let's determine the probability of one team winning in exactly 5 games:
The winning team needs to win the first four games and then lose the fifth game. Since the probability of winning a game is 1/2, the probability of winning four consecutive games is (1/2) * (1/2) * (1/2) * (1/2) = 1/16. Then, the losing team needs to win the fifth game, which also has a probability of 1/2.
(1/16) * (1/2) = 1/32
To find the probability of one team winning in exactly 6 games, we need to consider that the winning team needs to win in any of the scenarios: (4-2), (4-1-1), (3-1-2), (3-2-1), (2-1-2-1), (2-2-2), (1-1-2-2), (1-2-1-2), and (1-2-2-1). Each scenario has a probability of (1/2)^6 = 1/64.
So, the probability of one team winning in exactly 6 games is 9 * (1/64) = 9/64.
Finally, let's calculate the probability of the series being decided before the seventh game:
(1/16) + (1/32) + (9/64) = 37/64
The odds against the series going a full seven games are the reciprocal of this probability:
64/37
Therefore, the odds against the series going a full seven games are approximately 1.73:1 or 11:5.
To calculate the odds in favor of one team sweeping the series, you need to find the probability of that specific outcome occurring. In a best of seven series, one team would need to win four consecutive games. Since the teams are evenly matched, each game has a 50% chance of being won by either team.
To calculate the probability of one team winning four consecutive games, you need to multiply the individual probabilities of winning each game together. Since each game is independent of the others, you can do this by raising the probability of winning a single game (0.5) to the power of the number of consecutive games needed to win (4).
So, the probability of one team sweeping the series is 0.5^4 = 0.0625.
To express this probability as odds in favor, you need to calculate the ratio of the probability of the event occurring to the probability of the event not occurring. The probability of the event not occurring is 1 - 0.0625 = 0.9375.
Therefore, the odds in favor of one team sweeping the series are 0.0625/0.9375, which simplifies to 1:15.
To calculate the odds against the series going a full seven games, you need to find the probability of the series ending in fewer than seven games. This can occur if one team sweeps the series (as we calculated above) or if the series ends in four, five, or six games.
To calculate the probability of the series ending in four, five, or six games, you can consider the possible game sequences. There are three possible sequences: 4-0, 4-1, 4-2.
The probability of any specific sequence occurring can be calculated by multiplying the probability of winning each game by the probability of losing each game. Since each game has a 50% chance of being won by either team, the probability of winning any specific game is 0.5 and the probability of losing is also 0.5.
The probability of the series ending in four, five, or six games is the sum of the probabilities of each specific sequence:
0.5^4 + (0.5^4 * 0.5) + (0.5^4 * 0.5 * 0.5) = 0.5^4 * (1 + 0.5 + 0.5^2) = 0.4883
Therefore, the probability of the series going a full seven games is 1 - 0.4883 = 0.5117.
To express this probability as odds against, you need to calculate the ratio of the probability of the event not occurring to the probability of the event occurring. The probability of the event occurring is 0.5117.
Therefore, the odds against the series going a full seven games are 0.5117/0.4883, which simplifies to 1.11:5 or approximately 11:5.