Probability question..>?!?

Question

Question- Two evenly matched hockey teams in the NHL playoffs square off in a best-of-7 series. The winner is the first team to win four games.

1- What is the probability that one team loses the first two games and then comes back to take the series by winning four straight games?

2- What assumptions are made to calculate these probabilities?
(I dont understand what the question is asking plz help!thnx)

Answer 1

The probability of winning any game is 0.5.
The probability of loosing the first two is then .5x.5 =.25
The probability of loosing the first 2 and winning the next four is
.5x.5x.5x.5x.5x.5 = .0156 or 1.56%

The assumption is that the teams are evenly matched for every game.

Answer 2

I disagree with most answers.

+ add
I took another look and the ginalino and I agree. He's made a clearer statement of the sequences.

The underlying model is Bernoulli trials. Here there are 7 Bernoulli trials.

Also, is it necessary that all 7 games be played?

The sequence from the perspective of one team, assuming all 7 games must be played:

WWLLLLX, where X means it could be W or L

Now I realize that if the 7th game goesn't matter then it contributes 2/2 into the calculation and so we only need to consider the first 6 games:

This sequence has prob: (1/2)^6.

But the sequence
LLWWWWX also qualifys but it means that the other team wins the series. Adding the probability of these two sequences:

Thus

2*(1/2)^6 = (1/2)^5 = 1/32

is the correct probability.

[ To see this, consider flipping a coin twice. The probability of either coin getting two heads is 2(1/2)^2 = 1/2 ]

2. The assumptions are that each game is an independent bernoulli event with p=1/2, and that it doesn't matter what happens in the 7th game.

Answer 3

1- What is the probability that one team loses the first two games and then comes back to take the series by winning four straight games?

(1/2)^4
=1/16

2- What assumptions are made to calculate these probabilities?
(I dont understand what the question is asking plz help!thnx)

The probability of each team winning a game is always 1/2.

Answer 4

1)Each time they play, there is a 50-50 chance or half a chance that they will win. I think doing it in this particular pattern, drawing a probability tree shows that the chance is 1/2*1/2*1/2*1/2*1/2*1/2 (1/2 chance that will result in a win or a loss- be it the first 2 games a loss and then the last 4 wins)
This adds up to a 1/64 chance.

2)That they cannot draw (If you can do in hockey playoffs- or that there best player doesn't get injured!)

Me thinks that's the answer

Answer 5

This just says that it's a random result. So you might as well toss a coin. What is the probability that you get the same result for the first times and then the opposite result 4 times in a row. You have exactly 2 combinations AABBBB or BBAAAA out of 64, hence one chance in 32.