Never lost 2 games in a row

Both Golden State and San Antonio (NBA teams) never lost 2 games in a row (when the games really counted)

say Golden State finishes with 73 wins and just 9 losses

what is the probability they did NOT lose 2 games in a row
knowing they would end up with a 73 win season?

San Antonio had 65 wins and only 12 losses
(not counting the last 5 meaningless games of the regular season)
what is the probability they did NOT lose 2 games in a row?

some may say they tried to win against GS
make 65 wins and 13 losses then

show your work!

extra credit:
what is the average number of losing streaks of at least length 2
for both teams?

extra extra credit:
Oh, those Bulls
that went 72 wins and lost 10 times with MJ

what is the probability they did NOT lose 2 games in a row?

I assume you want to include the assumption that the probability of winning each game was constant in this problem. (Is that a double assumption?) In real life, the chances that this year's Warriors beat this year's 76ers are much, much higher than the chances that they beat this year's Spurs, but I don't think you intend to take each team's calendar into account to solve the problem...

That would be interesting to do any math that way.
But, keeping it simple the loss rate can be constant.

Looks like the 76ers may NOT win 2 in a row this year.

Sally

Quote: ThatDonGuy

The probability that a team with W wins and L losses did not lose 2 or more in a row can be determined as follows:

still trying to follow the concept. your answer looks too low

from here
http://math.stackexchange.com/questions/549309/probability-of-not-dropping-two-games-in-row

I get what is going on as a binomial probability question
lots of math without a spreadsheet
sum k(0:41) COMBIN(83-k,k) * p^k * q^(82-k)
p=9/82

in Excel I get the same result as using a calculator (streak)
https://www.wolframalpha.com/input/?i=streak+of+2+successes+in+82+trials+p%3D9%2F82

or just recursion
(the probability of a 2 loss streak so we subtract that from 1 for no streak)
like here in Excel in Google Sheets
(change the value of p in cell C1 (to 9/82 for example) and that changes the results in Column E)
https://goo.gl/JStr4p

more skittles
more solutions
Sally

Quote: mustangsally

I get what is going on as a binomial probability question
lots of math without a spreadsheet
sum k(0:41) COMBIN(83-k,k) * p^k * q^(82-k)
p=9/82

That's for calculating the probability of not having 2 consecutive losses in an 82 game season assuming that the probability of losing a game is 9/82. This is not the problem you stated; you stated that we already know they have 73 wins and 9 losses.

I counted the number of ways you can lose 9 games out of 82 without consecutive losses the same way as the StackExchange answer shows - (74)C(9), or, in your notation, COMBIN(74, 9).
I then assumed that each possible way of losing 9 games out of 82 games was equally likely; there are COMBIN(82, 9) of those.
Divide COMBIN(74,9) by COMBIN(82,9).

I agree with Don's answer, assuming that no given set of losses is more likely than any other given set of losses, as he noted. We don't actually need to calculate or use the 9/82 number at any point.

Don wrote the general case first. Here is the specific one.

For the denominator:
To find the total number of possible 9-loss seasons, we are picking 9 games to lose out of 82, so it's just 82C9.

For the numerator:
We know they lost 9 times. For each loss, we can say: "The team lost a game after win number X, before win number X+1." Since they lost 9 games, but never lost consecutive games, there will be 9 distinct values of X during the season.

X can range from 0 (meaning they lost their very first game of the season) to 73 (meaning they lost their very last game of the season). So there are 74 possible values for X, and we need to pick 9 of them.

Therefore, the numerator is 74C9.

Quote: ThatDonGuy

you stated that we already know they have 73 wins and 9 losses.

yes
i got it!
Yahoo