Math Help Please

here is the scenario, and it is a true one that happened in my FFL , and I am thinking the odds against are staggering but I am lost on how to calculate it.....

we have a ten team league, head to head, after week 1 we had 5 teams 1-0 and of course 5 teams 0-1... week saw ( the schedule was made up randomly by the CPU before the season) we had all 5 undefeated teams play all the winless teams (crazy) but even more odd was that all the winless teams won in week 2 making all ten teams 1-1. what are the odds of this happening ( throwing out the strength of each team)??? if someone knows how to compute this I'd be grateful.

thanks.
bdc42

Quote: bdc42

...we have a ten team league, head to head, after week 1 we had 5 teams 1-0 and of course 5 teams 0-1... week saw ( the schedule was made up randomly by the CPU before the season) we had all 5 undefeated teams play all the winless teams (crazy) but even more odd was that all the winless teams won in week 2 making all ten teams 1-1. what are the odds of this happening ( throwing out the strength of each team)??? if someone knows how to compute this I'd be grateful...

The probability of a team's winning is 0.5 assuming all teams are of equal strength. Then the probability that five particular teams would all win is 0.5 x 0.5 x 0.5 x 0.5 x 0.5 = 0.03125 = 1 in 32.

any help would be appreciated ..

thanks

thanks.... not nearly as impossible as it seemed to me.. but does that take into account the randomness that each team played a winless team?

Quote: bdc42

thanks.... not nearly as impossible as it seemed to me.. but does that take into account the randomness that each team played a winless team?

You're right; I missed that important point!

Now, I think the answer might be this: Prob. = ( 5 / 9 ) x ( 4 / 8 ) x ( 3 / 7 ) x ( 2 / 6 ) x ( 1 / 5 ) x ( 1 / 32 ) = 1 in 4032

(Call the five winners A, B, C, D, and E. The probability than A is assigned a loser to play against is 5/9. Then if A is assigned a loser, the probability that B is assigned a loser is 4/8, etc. Then the probability that each winner is assigned a loser is this product: (5/9)x(4/8)x(3/7)x(2/6)x(1/5). Multiply this probability by the probability of A, B, C, D, and E all winning to get the answer.)

Edit: And this answer assumes that a pair of teams that played the first game could also play against each other in the second game, which, I guess, is not allowed.

true , two teams couldn't face each other two times in a row...

thanks.

My solution: have 10 balls, one for each team; the ones representing the 1-0 teams are blue, and the ones representing the 0-1 teams are red.
Draw a team for the first game; there are 9 balls remaining, 5 of which are of the other color, so the probability of "game 1" having a 1-0 team against an 0-1 team is 5/9.
There are now 4 of each color left. Draw one for the second game; there are 7 balls remaining, 4 of which are the other color, so the probability of "game 2" having a 1-0 team against an 0-1 team is 4/7.
There are now 3 of each color left. Draw one for the third game; there are 5 balls remaining, 3 of which are the other color, so the probability of "game 3" having a 1-0 team against an 0-1 team is 3/5.
There are now 2 of each color left. Draw one for the fourth game; there are 3 balls remaining, 2 of which are the other color, so the probability of "game 4" having a 1-0 team against an 0-1 team is 2/3.
The last two teams are 1-0 and 0-1, so the fifth game is always one of each.
The total probability of such a matchup is 5/9 x 4/7 x 3/5 x 2/3 = 8/63, which is slightly better than 1/8.

The probability of all five 0-1 teams winning cannot be calculated as there is no way to calculate an accurate probability of the result of any particular game in advance. If the first week's games were 1 vs 2, 3 vs 4, 5 vs 6, 7 vs 8, and 9 vs 10 (where the numbers are the relative strengths of the teams), and the second week's is 2 vs 3, 4 vs 5, 6 vs 7, 8 vs 9, and 1 vs 10, then 4 of the 5 0-1 teams are expected to win.

Quote: ChesterDog

Now, I think the answer might be this: Prob. = ( 5 / 9 ) x ( 4 / 8 ) x ( 3 / 7 ) x ( 2 / 6 ) x ( 1 / 5 ) x ( 1 / 32 ) = 1 in 4032

Because it's 10 games, I don't understand why it would be so much greater than 1 / 2^10

Like if you were given God powers and could choose who won and lost each game to allow for this situation, you would simply be declaring the winner for each of the 10 games, without having to mess around with the schedule in any way

@ThatDonGuy:
I think your process of calculating 8/63 disregards bdc42's comment just above that, "true , two teams couldn't face each other two times in a row...."

Also, your comment that, "there is no way to calculate an accurate probability of the result of any particular game in advance" disregards the earlier parameter of "throwing out the strength of each team", i.e., assuming a 50-50 chance of winning.

In spite of critiquing others' efforts, I'm not going to try to solve this one myself.

Quote: Doc

@ThatDonGuy:
I think your process of calculating 8/63 disregards bdc42's comment just above that, "true , two teams couldn't face each other two times in a row...."

Sounds like a brute force counting problem - especially when you have to take into account the fact that the Week 2 schedule cannot then result in not being able to complete the Weeks 3-10 schedule without having a rematch.

Quote: ThatDonGuy

Sounds like a brute force counting problem...

I just now finished counting the number of ways of assigning 0-1 teams to 1-0 teams and the total number of ways of assigning teams for game 2. I got 44 and 544, respectively. So, assuming the teams are of equal strength, the answer to the original question might be ( 44 / 544 ) x ( 1 / 32 ) or about 1 in 396.

(By the way, counting the total numbers of ways of assigning teams for game 2 for different numbers of teams in leagues--4, 6, 8, 10, I got 2, 8, 60, and 544, respectively. Looking up this sequence in The Online Encyclopedia of Integer Sequences finds that this sequence is the solution for the problem "2n objects are initially paired in some way and then are re-paired so that no object is with its original partner," which fits the criterion that two teams cannot face each other two times in a row for games 1 and 2.)