Playoff Tie (1 Viewer)

[Ignoratio Elenchi]

Lots of things could be less fair. The fact is that some kind of tiebreaker needs to be applied at this point. You can bemoan the fact that tiebreakers weren't in place beforehand, but that's just crying over spilled milk. Decimal scoring is as close to a perfect tiebreaker as you're going to get in this instance, because it's rewarding the owners for exactly what they thought they'd be rewarded for - the total quantity of fantasy stats their starting lineup produced. It's just measuring them to greater precision, which no one should take issue with. They wouldn't have managed anything any differently had they known this tiebreaker was going to be used ahead of time. It's vastly superior to something completely arbitrary like a coin flip, or something like bench points (which owners might have prepared for differently if they'd known it would be used as a tiebreaker), or a three-headed championship game (which screws the third guy who already won his matchup), or any of the other tiebreakers you proposed. The commissioner isn't "choosing" the champion - he's awarding the win to the team whose starting lineup amassed the most fantasy stats, which is exactly what was supposed to happen.

say you are team A. you have a better head to head record against team B, you are the higher seed, better record, better division and conference record, higher scoring starter, higher scoring bench player, more total points scored

but you loose if they implement decimal scoring as a tie breaker.

are you okay with that?

Of course. My team's starting lineup didn't accumulate as many total fantasy stats as my opponent's starting lineup. I knew going into the game that was exactly what would determine who was going to win. It happens all the time that the "better" team (by head to head record, higher seed, better records, higher scoring starter, higher scoring bench, more total points) loses a playoff game to the "worse" team. Happened to me in one of my leagues last week. That's fantasy football, sometimes it happens. We all know that going in. What I would not be ok with is letting the fate of my fantasy season rest on something completely arbitrary and unrelated to the game, like a coin flip.

 

[Utter Chaos]

Suppose some random tie-breaker is used and Team B advances. Team Z then loses to Team B in the championship but would have beaten Team A. Is that really fair to Team Z?

yes that is completely fair to all 3 teams because it is within the spirit of the single game elimination format.team Z has earned a 50% chance at the championship and 50% at 2nd place.

as long as you you give him his head to head matchup (50% chance at 1st, 50% chance at 2nd, 0% chance at anything else) then that is fair. i am facing the lowest point scorer of the playoff teams in my championship. if he lost the last game, i would be playing the highest point scorer. either outcome would have been fair from everyones point of view

I still think the best way is to split up the prize money in half and have two championships. Team Z vs Team A and Team Z vs Team B. If Team Z beats them both he's the undisputed champion. If he splits or loses both games then you have co-champs.

Doing it this way does not penalize Teams A and B by some tie-breaker created after the fact when none was in place. It gives each of them a shot of 1/2 the first place money which is better then no shot at all. It guarantees Team Z at least 2nd place with a chance to win half or all of first place.

I think that's a fair compromise that doesn't grossly reward or penalize any of the teams more than the others.

this is fair to team A and B but unfair to team Z and here is why...lets say 1st is $300, 2nd is $100, 3rd is $0

team Z: has already earned $100 for 2nd place, worse case scenario for him in either a normal head to head matchup, or your scenario

$100+...

Z beats A and B: .5*.5 = 25% at $200 = $50 expected dollars for team Z

Z beats A, looses B: 25% at $100 = $25 expected dollars for team Z

Z looses A, wins B: 25% at $100 = $25 expected dollars for team Z

Z looses A, looses B: 25% at $0 = $0 expected dollars for team Z

$100+(50+25+25+0) = $200 expected dollars for team Z

in a normal head to head matchup

Z beats A: $100 + (50% at $200) = $200 expected dollars for team Z

Z looses A: $100 + (50% at $0) = $100 expected dollars for team Z

$200+100 = $300 expected dollars for team Z. anything less is penalizing team Z

team Zs expected earnings decrease by 33% in your scenario

You lost me is your scenario. I don't think it's that complicated.Z beats A, an B: $150 + $150 = $300 winnings

Z beats A and loses to B = $150 + $50 = $200 winnings

Z beats B and loses to A = $150 + $50 = $200 winnings

Z loses to A and B = $50 + $50 = $100 winnings

In your normal head to head matchup you have Team Z expecting to win $300. I don't think that's correct.

One final thing about the coin flip. Suppose Team A lives out of state and Team B is best buds with the commish. The commish and Team B do the coin flip over the phone while Team A is on the line. Would you be OK with that if you're Team A?

Also, both teams probably are aware on which tie-breaking scenario they hold the advantage on and which ones they don't. I highly doubt that they would "agree" to a tie breaker to be used to settle it. They'd both would be arguing for the one that's more advantageous to them.

Now that I re-read it I think I know the problem. In the head to head scenario you're adding the $100 minimum two times. It should be.

in a normal head to head matchup

Z beats A: (50% at $200) = $100 expected dollars for team Z

Z looses A: (50% at $0) = $0 expected dollars for team Z

$100+100 = $200 expected dollars for team Z.

Also, fixed a problem with my numbers where I was giving 2nd place $200 instead of $100

 

[Ignoratio Elenchi]

Suppose some random tie-breaker is used and Team B advances. Team Z then loses to Team B in the championship but would have beaten Team A. Is that really fair to Team Z?

yes that is completely fair to all 3 teams because it is within the spirit of the single game elimination format.team Z has earned a 50% chance at the championship and 50% at 2nd place.

as long as you you give him his head to head matchup (50% chance at 1st, 50% chance at 2nd, 0% chance at anything else) then that is fair. i am facing the lowest point scorer of the playoff teams in my championship. if he lost the last game, i would be playing the highest point scorer. either outcome would have been fair from everyones point of view

I still think the best way is to split up the prize money in half and have two championships. Team Z vs Team A and Team Z vs Team B. If Team Z beats them both he's the undisputed champion. If he splits or loses both games then you have co-champs.

Doing it this way does not penalize Teams A and B by some tie-breaker created after the fact when none was in place. It gives each of them a shot of 1/2 the first place money which is better then no shot at all. It guarantees Team Z at least 2nd place with a chance to win half or all of first place.

I think that's a fair compromise that doesn't grossly reward or penalize any of the teams more than the others.

this is fair to team A and B but unfair to team Z and here is why...lets say 1st is $300, 2nd is $100, 3rd is $0

team Z: has already earned $100 for 2nd place, worse case scenario for him in either a normal head to head matchup, or your scenario

$100+...

Z beats A and B: .5*.5 = 25% at $200 = $50 expected dollars for team Z

Z beats A, looses B: 25% at $100 = $25 expected dollars for team Z

Z looses A, wins B: 25% at $100 = $25 expected dollars for team Z

Z looses A, looses B: 25% at $0 = $0 expected dollars for team Z

$100+(50+25+25+0) = $200 expected dollars for team Z

in a normal head to head matchup

Z beats A: $100 + (50% at $200) = $200 expected dollars for team Z

Z looses A: $100 + (50% at $0) = $100 expected dollars for team Z

$200+100 = $300 expected dollars for team Z. anything less is penalizing team Z

team Zs expected earnings decrease by 33% in your scenario

**there is one way this scenario is fair though.

If and only if 2nd place is exactly half of what 1st place, or zero is AND 3rd place is half of what 2nd place is or zero

ex:

400, 200, 0

400, 200, 100

400, 0, 0

Actually the expected winnings for Team Z are the same either way. Your probabilities are wrong which is probably why you're coming up with a different answer.Let F = the first place prize total, and S = the second place prize total. In a normal head-to-head matchup, Team Z's expected winnings would be (F+S)/2 (assuming any one team is equally likely to outscore any other team on any given week).

Under Utter Chaos's system:

There is a 1/3 probability that Team Z will beat both Teams A and B, for an expected value of F/3.

There is a 1/3 probability that Team Z beats one of them, and loses to the other, for an expected value of (F+S)/6.

There is a 1/3 probability that Team Z will lose to both of them, for an expected value of S/3.

Summing those gives F/3 + (F+S)/6 + S/3 = (F+S)/2. Exactly the same as the head-to-head matchup.

In the head-to-head scenario, of course, Team Z either wins ALL of the first place prize, or ALL of the second place prize, whereas in UC's system it's possible that Team Z will end up winning the average of the two. Team Z might object to such a compromise, and I think you'd have to honor his objection. But his expected payoff is the same either way, no matter what the prize amounts are.

 

[Utter Chaos]

In the head-to-head scenario, of course, Team Z either wins ALL of the first place prize, or ALL of the second place prize, whereas in UC's system it's possible that Team Z will end up winning the average of the two. Team Z might object to such a compromise, and I think you'd have to honor his objection. But his expected payoff is the same either way, no matter what the prize amounts are.

That's why I think this is the way to go. Team Z has the same expected payout. Under any tie breaker situation Team's A and B expected payout would not be the same. There's been many examples in this thread on why doing something now woul not be right. Decimal scoring probably seams like the fairest way but a) that could still end in a tie and b) if one team holds every other possible tie-breaker except decimal then good luck trying to get them to agree to that.

 

[Ignoratio Elenchi]

That's why I think this is the way to go. Team Z has the same expected payout.

I agree that this isn't a terrible solution, and if I was Team Z I'd personally be ok with it. But it's possible that someone else wouldn't be ok with it if they were Team Z (people are irrational creatures, after all) and I think you'd probably have to respect that.

Under any tie breaker situation Team's A and B expected payout would not be the same.

I don't understand this, can you explain? Their expected payout going into the game (and the tiebreaker) was the same. Their actual payout isn't supposed to be the same - one of them is supposed to advance to the championship, and the other isn't. This isn't like we're breaking the tie by flipping a loaded coin that favors one team over the other. Both teams had an equal and fair opportunity to amass as many fantasy stats with their starting lineups as possible. We're just awarding the win to the team that did a better job of that, just like we were going to all along.

There's been many examples in this thread on why doing something now woul not be right. Decimal scoring probably seams like the fairest way but a) that could still end in a tie and b) if one team holds every other possible tie-breaker except decimal then good luck trying to get them to agree to that.

Lots of other tiebreakers aren't fair to apply after the fact, because had the owners known that tiebreaker was going to be in play, they might have managed their team differently. Most bench points, for example, is not a good tiebreaker to apply after the fact, because someone might have traded away their depth to make a playoff run, and maybe wouldn't have done that if they knew bench points might be a factor. Maybe they left an inactive player on their bench this week, thinking it couldn't possibly make any difference, when they could have picked up a replacement to increase their bench scoring. But for decimal scoring, you're not changing anything. You're not applying anything retroactively that wasn't already essentially in place. You're just scoring exactly what you said you'd be scoring, just with greater precision. Re: (a), I think OP mentioned that decimal scoring would indeed break the tie, so in this case it's a nonissue. If that wasn't the case, then we'd consider other possibilities. Re: (b), I feel the same way about this as I do about gianmarco's thread - I can't imagine playing with owners that wouldn't agree with decimal scoring as a tiebreaker in this situation. It's the perfect solution to the problem. If you have people in your league that would refuse a fair solution just because it resulted in them losing, then you should probably reevaluate who you hang out with I guess.

 

[flc735]

Also, fixed a problem with my numbers where I was giving 2nd place $200 instead of $100

your changing the prize amount and that makes a difference herethis works ONLY if 1st place is exactly double what 2nd place is

AND 2nd place is either 0 or exactly double what 3rd place is

AND 3rd place is 0 or exactly half of what 2nd place is

so these payouts all work

300, 150, 0

300, 150, 75

999, 0, 0

550, 275, 0

4, 2, 1

800, 400, 200, 100

payouts do not work if payouts increase by more than 50% at each step up

300,100,0

400, 1, 0

350, 125, 75

400, 300, 200 (this gives a better advantage to team Z than he should get because 2nd place is more than 50% of first place)

 

[Ignoratio Elenchi]

Also, fixed a problem with my numbers where I was giving 2nd place $200 instead of $100

your changing the prize amount and that makes a difference herethis works ONLY if 1st place is exactly double what 2nd place is

AND 2nd place is either 0 or exactly double what 3rd place is

AND 3rd place is 0 or exactly half of what 2nd place is

so these payouts all work

300, 150, 0

300, 150, 75

999, 0, 0

550, 275, 0

4, 2, 1

800, 400, 200, 100

payouts do not work if payouts increase by more than 50% at each step up

300,100,0

400, 1, 0

350, 125, 75

400, 300, 200 (this gives a better advantage to team Z than he should get because 2nd place is more than 50% of first place)

It works no matter what the prize amounts are. I proved this a few posts up. Team Z's expected payout in UC's system is the same as if he faced a single opponent.

 

[Utter Chaos]

Also, fixed a problem with my numbers where I was giving 2nd place $200 instead of $100

your changing the prize amount and that makes a difference herethis works ONLY if 1st place is exactly double what 2nd place is

AND 2nd place is either 0 or exactly double what 3rd place is

AND 3rd place is 0 or exactly half of what 2nd place is

so these payouts all work

300, 150, 0

300, 150, 75

999, 0, 0

550, 275, 0

4, 2, 1

800, 400, 200, 100

payouts do not work if payouts increase by more than 50% at each step up

300,100,0

400, 1, 0

350, 125, 75

400, 300, 200 (this gives a better advantage to team Z than he should get because 2nd place is more than 50% of first place)

I didn't change any of the prize amounts. I was using the example you used "lets say 1st is $300, 2nd is $100, 3rd is $0" however when I posted my message I inadvertantly was using $200 for 2nd place instead of $100. My "edit" was just to point out that I corrected my error.

 

[flc735]

Summing those gives F/3 + (F+S)/6 + S/3 = (F+S)/2. Exactly the same as the head-to-head matchup.

In the head-to-head scenario, of course, Team Z either wins ALL of the first place prize, or ALL of the second place prize, whereas in UC's system it's possible that Team Z will end up winning the average of the two. Team Z might object to such a compromise, and I think you'd have to honor his objection. But his expected payoff is the same either way, no matter what the prize amounts are.

i disagree with your head-to-head valueyour giving team Z a 50% chance at first place (good). your giving team Z a 50% chance at 2nd place (bad)

(300, 100, 0 prize moneys)

when you get to a championship, the absolute worse case scenario is 2nd place. so there is a 100% chance he comes away with $100 (2nd place), and a 50% chance at $300 (1st place).

so the value of a normal hed to head matchup should be S+(F/2) and not (S+F)/2

 

[Ignoratio Elenchi]

Summing those gives F/3 + (F+S)/6 + S/3 = (F+S)/2. Exactly the same as the head-to-head matchup.

In the head-to-head scenario, of course, Team Z either wins ALL of the first place prize, or ALL of the second place prize, whereas in UC's system it's possible that Team Z will end up winning the average of the two. Team Z might object to such a compromise, and I think you'd have to honor his objection. But his expected payoff is the same either way, no matter what the prize amounts are.

i disagree with your head-to-head valueyour giving team Z a 50% chance at first place (good). your giving team Z a 50% chance at 2nd place (bad)

(300, 100, 0 prize moneys)

when you get to a championship, the absolute worse case scenario is 2nd place. so there is a 100% chance he comes away with $100 (2nd place), and a 50% chance at $300 (1st place).

so the value of a normal hed to head matchup should be S+(F/2) and not (S+F)/2

No. There's a 50% chance you win first place ($300) and a 50% chance you win second place ($100). Your expected payout is (.5)300 + (.5)100 = $200. Or, if you prefer to have it phrased the way you've been wording it, there is a 100% chance he comes away with a minimum of $100, and there is a 50% chance that he wins an additional $200. So, (1)100 + (.5)200 = $200. Just two different ways of saying the same thing.

A third way of saying it is (F+S)/2. (300 + 100)/2 = $200.

 

[Utter Chaos]

That's why I think this is the way to go. Team Z has the same expected payout.

I agree that this isn't a terrible solution, and if I was Team Z I'd personally be ok with it. But it's possible that someone else wouldn't be ok with it if they were Team Z (people are irrational creatures, after all) and I think you'd probably have to respect that.

Under any tie breaker situation Team's A and B expected payout would not be the same.

I don't understand this, can you explain? Their expected payout going into the game (and the tiebreaker) was the same. Their actual payout isn't supposed to be the same - one of them is supposed to advance to the championship, and the other isn't. This isn't like we're breaking the tie by flipping a loaded coin that favors one team over the other. Both teams had an equal and fair opportunity to amass as many fantasy stats with their starting lineups as possible. We're just awarding the win to the team that did a better job of that, just like we were going to all along.

There's been many examples in this thread on why doing something now woul not be right. Decimal scoring probably seams like the fairest way but a) that could still end in a tie and b) if one team holds every other possible tie-breaker except decimal then good luck trying to get them to agree to that.

Lots of other tiebreakers aren't fair to apply after the fact, because had the owners known that tiebreaker was going to be in play, they might have managed their team differently. Most bench points, for example, is not a good tiebreaker to apply after the fact, because someone might have traded away their depth to make a playoff run, and maybe wouldn't have done that if they knew bench points might be a factor. Maybe they left an inactive player on their bench this week, thinking it couldn't possibly make any difference, when they could have picked up a replacement to increase their bench scoring. But for decimal scoring, you're not changing anything. You're not applying anything retroactively that wasn't already essentially in place. You're just scoring exactly what you said you'd be scoring, just with greater precision. Re: (a), I think OP mentioned that decimal scoring would indeed break the tie, so in this case it's a nonissue. If that wasn't the case, then we'd consider other possibilities. Re: (b), I feel the same way about this as I do about gianmarco's thread - I can't imagine playing with owners that wouldn't agree with decimal scoring as a tiebreaker in this situation. It's the perfect solution to the problem. If you have people in your league that would refuse a fair solution just because it resulted in them losing, then you should probably reevaluate who you hang out with I guess.

To answer the question in red what I was trying to say was neither A or B "lost" their previous game under the current rules. If a tie-breaker is used (decimal for example) then one of the teams will have an expected payout of (1st + 2nd) / 2 while the other would have an expected payout of (3rd + 4th) / 2. If some other tie breaker is used that puts the other team in the finals then their expected payouts are reversed. Under my suggestion they both have the same expected payout (0.5*((1st + 2nd) / 2)) + (0.5*((3rd + 4th) / 2)).I agree the decimal scoring is by far the best scenario to use if it breaks the tie and everybody agrees with it. I hate the idea of a coin flip or bench points to be used.

 

[flc735]

I didn't change any of the prize amounts. I was using the example you used "lets say 1st is $300, 2nd is $100, 3rd is $0" however when I posted my message I inadvertantly was using $200 for 2nd place instead of $100. My "edit" was just to point out that I corrected my error.Now that I re-read it I think I know the problem. In the head to head scenario you're adding the $100 minimum two times. It should be.in a normal head to head matchupZ beats A: (50% at $200) = $100 expected dollars for team ZZ looses A: (50% at $0) = $0 expected dollars for team Z$100+100 = $200 expected dollars for team Z.

ah yes, your right, im wrong. nice work!now i just don't get why i am agreeing with you and disagreeing with ignoratio

 

[flc735]

nevermind, your both right. im wrong

im going to go enroll in a math class again

 

[Ignoratio Elenchi]

I didn't change any of the prize amounts. I was using the example you used "lets say 1st is $300, 2nd is $100, 3rd is $0" however when I posted my message I inadvertantly was using $200 for 2nd place instead of $100. My "edit" was just to point out that I corrected my error.

Now that I re-read it I think I know the problem. In the head to head scenario you're adding the $100 minimum two times. It should be.

in a normal head to head matchup

Z beats A: (50% at $200) = $100 expected dollars for team Z

Z looses A: (50% at $0) = $0 expected dollars for team Z

$100+100 = $200 expected dollars for team Z.

ah yes, your right, im wrong. nice work!now i just don't get why i am agreeing with you and disagreeing with ignoratio

Because you're saying that there's a 100% chance of the second place money ($100), and a 50% chance of the first place money ($300). That's not true. If you win first place, you don't get the $100 and the $300. You just get the $300 (or, put differently, you get the $100 and an additional $200).

 

[Novice2]

Two brackets. Both facing off in the Championship game against the other team. If both teams win they both get the payout and you have to explain to your children there will be no Christmas this year because you weren't thorough enough with your fantasy football leagues bylaws.

Highest scoring team of the 3 gets ''league champ'' bragging rights.

 

[Sudoku_in_the_Bathtub]

The other 2 in Week 15 did not advance. Give other team a Week 16 bye and is the league champion by default.

No coin flips or any of that nonsense. Rules are rules. Happy Holidays.