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[timchen]

Okay. Think I'm counting right again now? 3p pod means 216 order-invariant possible result lines, with 8 of these giving perfect scores, for a ~3.7% chance. 4p pod means 13824 lines with 216 of them giving perfect scores, for a rate of only 1.5625%. As the number of possibilities grows, though, it becomes increasingly difficult to count them, because there's LOTS of different cases, especially for the middle ones like 18 points.

So first you are trying to calculate the chance for a pod to have at least one player qualify, given some threshold.
Note that
8/216=1/27=1/81*3 (3 players each have 1/81 chance to win 4 straight games)
216/13824=1/64=1/256*4
is identical the first table you have given. Yes, if you go down further it will start to be different for obvious reasons (that is, the probability can not be larger than one so obviously you cannot just multiply that number by 3 or 4), but it is more due to the fact that when there are more than one player qualifying then a naive multiplication counts those instances multiple times. It is not due to the fact that sometimes one sequence can prevent another (for example, when one player scores 1111 the others can't.) Actually when there are mutual exclusiveness the probability will be that number multiply by the number of players.

A better question to ask perhaps is given a threshold what is the expected number of players per pod to qualify. This question can also be answered more easily. And more importantly, the answer to this question is the same as if we treat the record of each player as independent.

The reason is as follows: let us denote the record for a player (ex. something like 1-4-3-2) R. R1 denotes the record of the first player, etc. Now in a 4p game, R1, R2, R3, R4 are four correlated random variables; say if R1=1111 then R2 cannot be 1234. Notice there are conditional probability as well; ex. given R1=1221 the chance for R2 to be 2134 will be different from not knowing R1=1221.

We can encode all this information in the joint probability density function P(R1,R2,R3,R4): P(1111,1234,R3,R4)=0, etc.
Let us also define the threshold function Q(R), basically when a record qualifies it returns 1. So for example if threshold=22, (with 6-4-2-0 scoring) then Q(1111)=Q(1112)=Q(1121)=Q(1211)=Q(2111)=1 and the rest is zero.

The expected number of qualified players is then <Q(R1)+Q(R2)+Q(R3)+Q(R4)>.

Now here is an important observation. When we know nothing about other players, the probability of say P1 to have a certain record is the same as our previous calculation, which let us call it p(R) (That is, assuming equally skilled players, the chance to score 1111 is 1/256, 1112 and the like 4/256, etc, etc.) In equations it reads

sum_(R2,R3,R4)P(R1,R2,R3,R4)=p(R1).

So, even though they are not independent (ie, P(R1,R2,R3,R4)!=p(R1)p(R2)p(R3)p(R4)), the marginal probability is the same. Intuitively, since you know nothing about others, correlations don't matter.

The expected number of player to qualify, then can be calculated as

<Q(R1)+Q(R2)+Q(R3)+Q(R4)>=<Q(R1)>+<Q(R2)>+<Q(R3)>+<Q(R4)>=4 <Q(R)>.

That is, the expected number of qualified player per pod is just the chances you tabulated times the number of players.

Now as we see the difference of the chance between 3p and 4p is oscillating, which means that it does not favor 3p pods in particular. (I do agree if you look in detail at the table there are more spots favoring 3p though; but I think this is coincidental. That is, you can change this by changing the number of games played. And at the very top of course it favors 3p, which is not surprising as it's easier to get to first place in 3p, and at the top you need to get to first place.) There is a factor of 4/3 from the number of players, but this is expected. More players in a pod should generate more qualified players.
 

But the point is, a 3p pod is more likely to produce players at these higher point threshholds, even if any particular player may not be so much favoured. Which is important insofar as the 3p pods increase the threshold more quickly than 4p pods do.

Not sure this is right. As is tabulated 3p pod is more likely to produce 24 pts players but less likely to produce 22 and up. at 20-21 they are almost equal. Anyway this goes back to the same table. If you say that table favors 3p then ok, it favors 3p.

If I add an extra player to the pod, it is more likely that the top player in that pod will score worse, even with the added points injected. Certainly you see this at the top end, which is what is really important.

Not sure how you reach the conclusion with two cancelling factors (harder to win vs. more points at 2nd place), but this comes back to the question that whether the expected score of a player represents his skill well, across different number of players. This scoring method ensures that for an average player it will work. For player at higher end, it depends on the game itself, so yeah, maybe your feeling is right, but I didn't find convincing argument for that.

[timchen]

I didn't find convincing argument for that.

I think I now find one. Just to exaggerate, suppose 4th position in a 4p game is just so bad that no one can win from that position. Now, using the scoring system (6-4-2-0) and (6-3-0), the expected score for a player who can score #1 x of the time and #3 y of the time (x+y<1), when he is not 4p, will score in average
(6x+4(1-x-y)+2y))*3/4 in a 4p game, and 6x+3(1-x-y) in a 3p game.

The former is 1.5(x-y)+3 and the latter is 3(x-y)+3. Notice that (x-y) is a measure of how good the player is. x-y=0 gives an average player. Indeed, if the game is tilted so much that at 4th seat you can't win, then 3p game is favorable for a good player. An average player is never affected. Note that for the analysis to work, the inability to win at 4th seat is the same for everyone.

While in practice it isn't so bad, I can imagine it's true that a good player's winning percentage is hurt more than an average player at 4th seat. So this picture should be qualitatively correct.

I guess for some this is trivial. I guess I've started from the wrong perspective.

[WanderingWinder]

An average player is never affected.

Can you clarify what the average player is never affected by? Because it seems obvious to me that an average player benefits from having a never-win player in his pod. And that by having everyone be terrible, he is going to qualify every time. And, moreover, that he's going to benefit more from having one never-win player in a 3p pod than in a 4p pod. So I'm sure I just don't get what you're saying with this claim.

[WanderingWinder]

Okay. Think I'm counting right again now? 3p pod means 216 order-invariant possible result lines, with 8 of these giving perfect scores, for a ~3.7% chance. 4p pod means 13824 lines with 216 of them giving perfect scores, for a rate of only 1.5625%. As the number of possibilities grows, though, it becomes increasingly difficult to count them, because there's LOTS of different cases, especially for the middle ones like 18 points.

So first you are trying to calculate the chance for a pod to have at least one player qualify, given some threshold.
Note that
8/216=1/27=1/81*3 (3 players each have 1/81 chance to win 4 straight games)
216/13824=1/64=1/256*4
is identical the first table you have given. Yes, if you go down further it will start to be different for obvious reasons (that is, the probability can not be larger than one so obviously you cannot just multiply that number by 3 or 4), but it is more due to the fact that when there are more than one player qualifying then a naive multiplication counts those instances multiple times. It is not due to the fact that sometimes one sequence can prevent another (for example, when one player scores 1111 the others can't.) Actually when there are mutual exclusiveness the probability will be that number multiply by the number of players.

A better question to ask perhaps is given a threshold what is the expected number of players per pod to qualify. This question can also be answered more easily. And more importantly, the answer to this question is the same as if we treat the record of each player as independent.

This is absolutely a worse question to ask, because it fails to model what actually happens - i.e. that the threshold to qualify is NOT fixed. Which was the whole point of my other calculation. Indeed, I can see no reason that this is better to ask, other than it is easier to calculate. Of course, this way, things will track pretty well with the chart above, but even so, let's take a look at your math...

The reason is as follows: let us denote the record for a player (ex. something like 1-4-3-2) R. R1 denotes the record of the first player, etc. Now in a 4p game, R1, R2, R3, R4 are four correlated random variables; say if R1=1111 then R2 cannot be 1234. Notice there are conditional probability as well; ex. given R1=1221 the chance for R2 to be 2134 will be different from not knowing R1=1221.

We can encode all this information in the joint probability density function P(R1,R2,R3,R4): P(1111,1234,R3,R4)=0, etc.
Let us also define the threshold function Q(R), basically when a record qualifies it returns 1. So for example if threshold=22, (with 6-4-2-0 scoring) then Q(1111)=Q(1112)=Q(1121)=Q(1211)=Q(2111)=1 and the rest is zero.

The expected number of qualified players is then <Q(R1)+Q(R2)+Q(R3)+Q(R4)>.

Now here is an important observation. When we know nothing about other players, the probability of say P1 to have a certain record is the same as our previous calculation, which let us call it p(R) (That is, assuming equally skilled players, the chance to score 1111 is 1/256, 1112 and the like 4/256, etc, etc.) In equations it reads

sum_(R2,R3,R4)P(R1,R2,R3,R4)=p(R1).

So, even though they are not independent (ie, P(R1,R2,R3,R4)!=p(R1)p(R2)p(R3)p(R4)), the marginal probability is the same. Intuitively, since you know nothing about others, correlations don't matter.

More or less, I am with you here. I maybe wouldn't phrase it like this, but I would certainly not say that this is wrong.

The expected number of player to qualify, then can be calculated as

<Q(R1)+Q(R2)+Q(R3)+Q(R4)>=<Q(R1)>+<Q(R2)>+<Q(R3)>+<Q(R4)>=4 <Q(R)>.

That is, the expected number of qualified player per pod is just the chances you tabulated times the number of players.

But here you lose me. Certainly <Q(R1)+Q(R2)+Q(R3)+Q(R4)>=<Q(R1)>+<Q(R2)>+<Q(R3)>+<Q(R4)>. But this is not the same thing as 4<Q(R)>, because R1, R2, R3, and R4 are not independent! Sure, for R1, you can say that you don't know what the other players did, and thus it is R. But this fails to hold for R2, R3, and R4. Now, quite possibly this is still right somehow - I cannot say that I have definitively calculated it to be wrong. But you're making the claim without showing it sufficiently well to be true, at least for my standards.

Now as we see the difference of the chance between 3p and 4p is oscillating, which means that it does not favor 3p pods in particular. (I do agree if you look in detail at the table there are more spots favoring 3p though; but I think this is coincidental. That is, you can change this by changing the number of games played. And at the very top of course it favors 3p, which is not surprising as it's easier to get to first place in 3p, and at the top you need to get to first place.) There is a factor of 4/3 from the number of players, but this is expected. More players in a pod should generate more qualified players.
 

But the point is, a 3p pod is more likely to produce players at these higher point threshholds, even if any particular player may not be so much favoured. Which is important insofar as the 3p pods increase the threshold more quickly than 4p pods do.

Not sure this is right. As is tabulated 3p pod is more likely to produce 24 pts players but less likely to produce 22 and up. at 20-21 they are almost equal. Anyway this goes back to the same table. If you say that table favors 3p then ok, it favors 3p.

I wasn't saying that. My whole points were based on things you can't get from the table. However, now that I do look at the table, it's pretty clear to me that by-and-large, it does favour 3p. 4p gets a slight bump on exactly 16 and 22, but everywhere else, 3p has an advantage, and in some of these cases, quite a significant one.

[theory]

I moved all this discussion over into this topic, by the way.

[timchen]

An average player is never affected.

Can you clarify what the average player is never affected by? Because it seems obvious to me that an average player benefits from having a never-win player in his pod. And that by having everyone be terrible, he is going to qualify every time. And, moreover, that he's going to benefit more from having one never-win player in a 3p pod than in a 4p pod. So I'm sure I just don't get what you're saying with this claim.

That average player also cannot win when he is in 4th seat. So if you play a 4-game series with each player in each seat once he's score will not be affected. To put it more clearly, he's average score will be 3 per game (average to 4 when not at 4th seat, 0 when at 4 seat), which is the same as if there is no disadvantage at 4th seat.

[WanderingWinder]

An average player is never affected.

Can you clarify what the average player is never affected by? Because it seems obvious to me that an average player benefits from having a never-win player in his pod. And that by having everyone be terrible, he is going to qualify every time. And, moreover, that he's going to benefit more from having one never-win player in a 3p pod than in a 4p pod. So I'm sure I just don't get what you're saying with this claim.

That average player also cannot win when he is in 4th seat. So if you play a 4-game series with each player in each seat once he's score will not be affected. To put it more clearly, he's average score will be 3 per game (average to 4 when not at 4th seat, 0 when at 4 seat), which is the same as if there is no disadvantage at 4th seat.

Oh, you're talking about fourth SEAT. I thought you were talking about having one player who was just really terrible no matter where he was.
Seating order will tend to bring everyone closer to average over time, no? (which obviously doesn't affect Mr. Average)

[timchen]

But here you lose me. Certainly <Q(R1)+Q(R2)+Q(R3)+Q(R4)>=<Q(R1)>+<Q(R2)>+<Q(R3)>+<Q(R4)>. But this is not the same thing as 4<Q(R)>, because R1, R2, R3, and R4 are not independent! Sure, for R1, you can say that you don't know what the other players did, and thus it is R. But this fails to hold for R2, R3, and R4.

Your statement is kinda of funny. Suppose you accept it for R1. Note that when you calculate the expectation value for Q(R2), that term DOES NOT depend on R1, R3 and R4, in the same way that Q(R1) does not depend on the rest. This is why the expectation value only cares about the marginal probability. The correlation is encoded in the pdf P(R1,R2,R3,R4), but we have argued that the marginal probability for R1-R4 will all be the same as p(R). Technically, basically you plug in the above equality and then use this formula which you say you accept:

sum_(R2,R3,R4)P(R1,R2,R3,R4)=p(R1)
(and its permuted version)

 you then get the answer.

(To be perfectly clear I'll just calculate for example, <Q(R2)> here, the calculation is the same for every R)
<Q(R2)>=sum_(R1,R2,R3,R4)(Q(R2)P(R1,R2,R3,R4))
             =sum_(R2)Q(R2)sum_(R1,R3,R4)P(R1,R2,R3,R4)
             =sum_(R2)Q(R2)p(R2)
             =<Q(R)>.

This is absolutely a worse question to ask, because it fails to model what actually happens - i.e. that the threshold to qualify is NOT fixed. Which was the whole point of my other calculation. Indeed, I can see no reason that this is better to ask, other than it is easier to calculate. Of course, this way, things will track pretty well with the chart above, but even so, let's take a look at your math...

You have to do things in steps. What you were calculating IS the probability of qualification given a threshold so I don't see why you are complaining. You have to know this first(or actually, know what I said is better to calculate first) then you can use the size of the tournament and the slots for the qualification along with this table to calculate the expected threshold and see if it favors 3p or 4p.

I wasn't saying that. My whole points were based on things you can't get from the table. However, now that I do look at the table, it's pretty clear to me that by-and-large, it does favour 3p. 4p gets a slight bump on exactly 16 and 22, but everywhere else, 3p has an advantage, and in some of these cases, quite a significant one.

I am not disagreeing with you on this particular table seemingly favoring 3p. It depends really on where your threshold will sit. But the point is there is no reason for the threshold to prefer sitting on places that favors 3p, on the contrary of what you have said. Also, while this particular table has more 3p favoring thresholds I don't see this as a general trend. You can try to make a table of say 12 game-series I'll say it will be different, but still oscillating.

The only thing that is beyond this table is when you weigh in the skills of the player. That is you have to consider players that win more often. As I've shown earlier with an example, as long as the expected score is the same for the player in question in 3p or 4p games, he will not be particularly favored. But if not, all this calculations are redundant. Obviously, when the expected score is different, he will in average score more in the format which he has a higher expected score. With a common threshold then of course he is going to have an advantage over there.

[timchen]

Oh, you're talking about fourth SEAT. I thought you were talking about having one player who was just really terrible no matter where he was.
Seating order will tend to bring everyone closer to average over time, no? (which obviously doesn't affect Mr. Average)

Yeah, I guess it is sort of obvious. Sorry if I phrase it in a confusing way. But the point is that this effect indeed brings down the expected score for a good player, in this scoring scheme.

Actually, in principle one way to compensate this effect is to make it something like (6420) and (5 3 1) for 4p and 3p. (average not changing so the average player again is not affected. Good player will score less in a 3p game comparatively.) But somehow it is a terrible idea for a 4 game series. With threshold at 20 now in 3p games the player has to win all games to qualify!

[WanderingWinder]

But here you lose me. Certainly <Q(R1)+Q(R2)+Q(R3)+Q(R4)>=<Q(R1)>+<Q(R2)>+<Q(R3)>+<Q(R4)>. But this is not the same thing as 4<Q(R)>, because R1, R2, R3, and R4 are not independent! Sure, for R1, you can say that you don't know what the other players did, and thus it is R. But this fails to hold for R2, R3, and R4.

Your statement is kinda of funny. Suppose you accept it for R1. Note that when you calculate the expectation value for Q(R2), that term DOES NOT depend on R1, R3 and R4, in the same way that Q(R1) does not depend on the rest. This is why the expectation value only cares about the marginal probability. The correlation is encoded in the pdf P(R1,R2,R3,R4), but we have argued that the marginal probability for R1-R4 will all be the same as p(R).

I believe this is what I fail to accept. Perhaps I said I accepted it above, but if that's the case, it was because I misunderstood what you were saying. My argument here is that R1, R2, R3, and R4 are not equal to each other, and in fact they can't be. Now, I'm getting lost a little bit inn your I'm-trying-to-write-math-using-only-plain-text, especially as I have not used multivariate probabilities in some years. I do NOT work in quantum physics as you do. I never liked using mathematical equations to describe things anyway - it is always much simpler to have some kind of example to explain your phenomenon, some concrete thing. It is just so difficult to keep so many abstractions in your head. Or at least it is for me.
So I will present a clear (clear to me, anyway) explanation of why R1, R2, R3, and R4 seem to me as they should not be equivalent to each other. (Obviously, if they are equivalent, then the Q will be equivalent as well - I am with you there).
So R1 is the first player's results. They can be anything, since we so far have no constraints (other, of course, than the fact that you can't do better than 1st place or worse than 4th, and the artificial no-ties constraint we've been using). But now, you go and pick R2. But now we have constraints. For every one of the four games, there is one result for the second player which he cannot achieve, because the first player already achieved it. And for the third player, there are two results he cannot achieve. The fourth player is entirely deterministic. If we are calculating the expectation value of R2...
Okay, I think I just got our problem. You are calculating the expectation value for each individual player? This is not what I am calculating. That is absolutely pointless - it's trivially obvious that every player will expect to score the same number of points, if we assume them to be of equal skill going in, and all games independent. I fail to see the need to use any math to prove this, and moreover,  I fail to see its importance. IS that what you're calculating?

This is absolutely a worse question to ask, because it fails to model what actually happens - i.e. that the threshold to qualify is NOT fixed. Which was the whole point of my other calculation. Indeed, I can see no reason that this is better to ask, other than it is easier to calculate. Of course, this way, things will track pretty well with the chart above, but even so, let's take a look at your math...

You have to do things in steps. What you were calculating IS the probability of qualification given a threshold

No, that's what I did in the chart. What I was really trying to calculate though, was a function for what the threshold WILL BE. Which is what you need if you actually want to calculate the chance of qualifying, which is what we care about.

so I don't see why you are complaining. You have to know this first(or actually, know what I said is better to calculate first) then you can use the size of the tournament and the slots for the qualification along with this table to calculate the expected threshold and see if it favors 3p or 4p.

Well, yeah. But that's already done. I was moving on to the next step. I think we're totally just misunderstanding each other here.

I wasn't saying that. My whole points were based on things you can't get from the table. However, now that I do look at the table, it's pretty clear to me that by-and-large, it does favour 3p. 4p gets a slight bump on exactly 16 and 22, but everywhere else, 3p has an advantage, and in some of these cases, quite a significant one.

I am not disagreeing with you on this particular table seemingly favoring 3p. It depends really on where your threshold will sit. But the point is there is no reason for the threshold to prefer sitting on places that favors 3p, on the contrary of what you have said. Also, while this particular table has more 3p favoring thresholds I don't see this as a general trend. You can try to make a table of say 12 game-series I'll say it will be different, but still oscillating.

The only thing that is beyond this table is when you weigh in the skills of the player. That is you have to consider players that win more often. As I've shown earlier with an example, as long as the expected score is the same for the player in question in 3p or 4p games, he will not be particularly favored. But if not, all this calculations are redundant. Obviously, when the expected score is different, he will in average score more in the format which he has a higher expected score. With a common threshold then of course he is going to have an advantage over there.

I still believe that, on top of the inherent gameplay factors and flaws in our assumptions, basically ALL of which help the players in 3p pods, there will be a general trend toward the favorability of 3p. Of course it depends on the thresholds, but what I am trying to argue above is that it is most likely for a threshold to sit somewhere which favors 3p. Of course you won't accept that, so fine, whatever. Don't think we're going to make progress on that. But at least, a 12-game series isn't what we're talking about HERE. Here we're talking about 4s

[WanderingWinder]

Oh, you're talking about fourth SEAT. I thought you were talking about having one player who was just really terrible no matter where he was.
Seating order will tend to bring everyone closer to average over time, no? (which obviously doesn't affect Mr. Average)

Yeah, I guess it is sort of obvious. Sorry if I phrase it in a confusing way. But the point is that this effect indeed brings down the expected score for a good player, in this scoring scheme.

Actually, in principle one way to compensate this effect is to make it something like (6420) and (5 3 1) for 4p and 3p. (average not changing so the average player again is not affected. Good player will score less in a 3p game comparatively.) But somehow it is a terrible idea for a 4 game series. With threshold at 20 now in 3p games the player has to win all games to qualify!

Right. My entire point throughout all of this is that it's impossible to make a fair system.

[timchen]

So R1 is the first player's results. They can be anything, since we so far have no constraints (other, of course, than the fact that you can't do better than 1st place or worse than 4th, and the artificial no-ties constraint we've been using). But now, you go and pick R2. But now we have constraints. For every one of the four games, there is one result for the second player which he cannot achieve, because the first player already achieved it. And for the third player, there are two results he cannot achieve. The fourth player is entirely deterministic. If we are calculating the expectation value of R2...
Okay, I think I just got our problem. You are calculating the expectation value for each individual player? This is not what I am calculating. That is absolutely pointless - it's trivially obvious that every player will expect to score the same number of points, if we assume them to be of equal skill going in, and all games independent. I fail to see the need to use any math to prove this, and moreover,  I fail to see its importance. IS that what you're calculating?

Yeah, I can see your problem here. That is why sometimes you do need some kind of abstraction.

What you are saying is completely true in the first paragraph. HOWEVER, as I said, I am calculating the expected number of players to qualify from a 4p prod, including the effect you listed in the first paragraph. So the point I am trying to show is that the correlation between the record of individual players does not change the expected number of players to qualify.

To give a concrete example, maybe I should just use numbers. If threshold=24:
(i) assume no correlations, each player has 1/256 chance to score 4 wins. The possible number of qualified players ranges from 0 to 4 though. The expected number of players to qualify from this 4 player prod. is just

1*4*(1/256)*(255/256)^3+2*6*(1/256)^2*(255/256)^2+3*4*(1/256)^3(255/256)+4*(1/256)^4=4*(1/256)=1/64.

The first half is just to count the number of qualified players times the chance for that to happen. The first equality you can either just calculate or prove by binomial expansions. Or you can use the physical argument that since four players are independent, the average number of players to qualify should be equal to the sum of the chance of each player to qualify.

(ii) now assume correlations. So only 1 out of 4 can qualify IN ANY GIVEN INSTANCE. You have calculated this, the chance for there is any player to acheive 4 wins is 216/13824=1/64. So the expected number of players to qualify is
 
1*(1/64)+2*0+3*0+4*0=1/64.

See, the point is that with or without the constraint between records, the expected number of players to qualify is the same. While I only use 24 as threshold and assume each player has equal skill, this statement is general, and works for any threshold and any player skill. The proof is what I have written there.

So the point that confuses you is I think, that the constraint is always there in any given instance. But when you average over all instances and see the average number of players to qualify from the pod, the constraint has no effect.

This point is indeed not intuitive, and that is why you need a little bit of abstraction to see it. Mathematically this is quite obvious though; it is just that the expectation value is the first moment so it does not care about correlations. If you try to calculate the variance of the number of qualified players in a pod given a threshold, then it will depend on this constraint.

Well, yeah. But that's already done. I was moving on to the next step. I think we're totally just misunderstanding each other here.

Ok, maybe you are trying to move to the next step. My point is that the calculation you've done midway (getting 216/13824 and other things) does not actually give new information, or move you to the next step. And what you say about the next step are pure speculations.

I still believe that, on top of the inherent gameplay factors and flaws in our assumptions, basically ALL of which help the players in 3p pods, there will be a general trend toward the favorability of 3p. Of course it depends on the thresholds, but what I am trying to argue above is that it is most likely for a threshold to sit somewhere which favors 3p. Of course you won't accept that, so fine, whatever. Don't think we're going to make progress on that. But at least, a 12-game series isn't what we're talking about HERE. Here we're talking about 4s

1. I am trying to convince you your opinion is wrong. I give numbers and reasons behind it. It is not that I am not accepting any thing you say, it is just that you did not provide anything to support it. You know, your argument I feel is like this:
If it requires all wins to qualify, then 3p pods is definitely more favorable. And this generalizes.
The first part I agree. Sure you can also generalize this to situation with a fixed number of qualifying slots and a very large playing pool, since in this case it may require you to win it all. But it's not really true for the general case. I dunno how to tell you this in a different way except listing the probabilities/expected number to qualify as a function of the threshold. There is really no lock in effect for the threshold to always stay somewhere which favors 3p.

Well, I dunno if this will help, but the relation of fixing the threshold and fixing the number of qualified slots is pretty much the same as fixing the density or fixing the chemical potential. ie, you can calculate things using canonical ensemble or grand canonical ensemble, but the result in average will not change.

Right. My entire point throughout all of this is that it's impossible to make a fair system.

I thought your entire point is that for any fair system (to average players) 3p is generally going to be favorable for skilled players. I am saying this is not true, and there is no general trend toward 3p pods except at the very very top (all wins to proceed).
Heck, I don't even agree it is impossible to make a fair system. I think it is entirely possible. The problem we are facing here is some discretization error which is a technical problem instead of a fundamental problem. (Well it is funamental and inherit but I mean, that kind of thing is always there, can only be solved by a larger number of games. There is nothing fundamentally wrong to compare results from 3p and 4p games using some scale, if we disregard the inherit difference between 3p and 4p games and just consider it from how-many-people-can-win perspective, as we are doing here. And we can even compensate some of the inherit difference between the games by changing our point scale.
And we don't even need it to be perfectly fair. As long as the unfairness is quite a bit smaller than the intrinsic random factor, we are in good shape.

[WanderingWinder]

So R1 is the first player's results. They can be anything, since we so far have no constraints (other, of course, than the fact that you can't do better than 1st place or worse than 4th, and the artificial no-ties constraint we've been using). But now, you go and pick R2. But now we have constraints. For every one of the four games, there is one result for the second player which he cannot achieve, because the first player already achieved it. And for the third player, there are two results he cannot achieve. The fourth player is entirely deterministic. If we are calculating the expectation value of R2...
Okay, I think I just got our problem. You are calculating the expectation value for each individual player? This is not what I am calculating. That is absolutely pointless - it's trivially obvious that every player will expect to score the same number of points, if we assume them to be of equal skill going in, and all games independent. I fail to see the need to use any math to prove this, and moreover,  I fail to see its importance. IS that what you're calculating?

Yeah, I can see your problem here. That is why sometimes you do need some kind of abstraction.

No, I don't think so. The entire problem is BECAUSE you're using abstractions. BECAUSE of that, I don't know what you're talking about. Which leads to us misunderstanding each other. Which is the crux of our problems.

What you are saying is completely true in the first paragraph. HOWEVER, as I said, I am calculating the expected number of players to qualify from a 4p prod, including the effect you listed in the first paragraph. So the point I am trying to show is that the correlation between the record of individual players does not change the expected number of players to qualify.

To give a concrete example, maybe I should just use numbers. If threshold=24:
(i) assume no correlations, each player has 1/256 chance to score 4 wins. The possible number of qualified players ranges from 0 to 4 though. The expected number of players to qualify from this 4 player prod. is just
1*4*(1/256)*(255/256)^3+2*6*(1/256)^2*(255/256)^2+3*4*(1/256)^3(255/256)+4*(1/256)^4=4*(1/256)=1/64.
The first half is just to count the number of qualified players times the chance for that to happen. The first equality you can either just calculate or prove by binomial expansions. Or you can use the physical argument that since four players are independent, the average number of players to qualify should be equal to the sum of the chance of each player to qualify.
(ii) now assume correlations. So only 1 out of 4 can qualify IN ANY GIVEN INSTANCE. You have calculated this, the chance for there is any player to acheive 4 wins is 216/13824=1/64. So the expected number of players to qualify is
1*(1/64)+2*0+3*0+4*0=1/64.
See, the point is that with or without the constraint between records, the expected number of players to qualify is the same. While I only use 24 as threshold and assume each player has equal skill, this statement is general, and works for any threshold and any player skill. The proof is what I have written there.

So the point that confuses you is I think, that the constraint is always there in any given instance. But when you average over all instances and see the average number of players to qualify from the pod, the constraint has no effect.

This point is indeed not intuitive, and that is why you need a little bit of abstraction to see it. Mathematically this is quite obvious though; it is just that the expectation value is the first moment so it does not care about correlations. If you try to calculate the variance of the number of qualified players in a pod given a threshold, then it will depend on this constraint.

Well, yeah. But that's already done. I was moving on to the next step. I think we're totally just misunderstanding each other here.

Ok, maybe you are trying to move to the next step. My point is that the calculation you've done midway (getting 216/13824 and other things) does not actually give new information, or move you to the next step. And what you say about the next step are pure speculations.

I still believe that, on top of the inherent gameplay factors and flaws in our assumptions, basically ALL of which help the players in 3p pods, there will be a general trend toward the favorability of 3p. Of course it depends on the thresholds, but what I am trying to argue above is that it is most likely for a threshold to sit somewhere which favors 3p. Of course you won't accept that, so fine, whatever. Don't think we're going to make progress on that. But at least, a 12-game series isn't what we're talking about HERE. Here we're talking about 4s

1. I am trying to convince you your opinion is wrong. I give numbers and reasons behind it. It is not that I am not accepting any thing you say, it is just that you did not provide anything to support it. You know, your argument I feel is like this:
If it requires all wins to qualify, then 3p pods is definitely more favorable. And this generalizes.
The first part I agree. Sure you can also generalize this to situation with a fixed number of qualifying slots and a very large playing pool, since in this case it may require you to win it all. But it's not really true for the general case. I dunno how to tell you this in a different way except listing the probabilities/expected number to qualify as a function of the threshold. There is really no lock in effect for the threshold to always stay somewhere which favors 3p.

Well, I dunno if this will help, but the relation of fixing the threshold and fixing the number of qualified slots is pretty much the same as fixing the density or fixing the chemical potential. ie, you can calculate things using canonical ensemble or grand canonical ensemble, but the result in average will not change.

I had this long response typed up, but then I realized, we're just making no progress here. We're each saying the same things over and over, and it's going nowhere. So there's not much point, is there?

Right. My entire point throughout all of this is that it's impossible to make a fair system.

I thought your entire point is that for any fair system (to average players) 3p is generally going to be favorable for skilled players. I am saying this is not true, and there is no general trend toward 3p pods except at the very very top (all wins to proceed).
Heck, I don't even agree it is impossible to make a fair system. I think it is entirely possible. The problem we are facing here is some discretization error which is a technical problem instead of a fundamental problem. (Well it is funamental and inherit but I mean, that kind of thing is always there, can only be solved by a larger number of games. There is nothing fundamentally wrong to compare results from 3p and 4p games using some scale, if we disregard the inherit difference between 3p and 4p games and just consider it from how-many-people-can-win perspective, as we are doing here. And we can even compensate some of the inherit difference between the games by changing our point scale.
And we don't even need it to be perfectly fair. As long as the unfairness is quite a bit smaller than the intrinsic random factor, we are in good shape.

Disagree with you entirely there.

[timchen]

I don't think I am saying the same thing. Isn't there an example for you to directly see what I am saying? How is that not concrete enough? Abstraction only helps when getting the proof. The example there is not abstract at all.

BTW, I don't think I misunderstand you. And I don't think you misunderstand me (in any consistent way anyway); it's more like you don't understand me.

[zxcvbn2]

Just as an FYI to theory and/or my opponents, I can't currently log in w/ my gmail account (This is the what happens when I try: http://imgur.com/uHa2y). If this issue doesn't resolve itself, like it has in the past, I'll be there under a different name (My Second Account).

[WanderingWinder]

Your literal name is "My Second Account"?
Also, I'm guessing the man you really need to let know this is rrenaud.

[theory]

Just as an FYI to theory and/or my opponents, I can't currently log in w/ my gmail account (This is the what happens when I try: http://imgur.com/uHa2y). If this issue doesn't resolve itself, like it has in the past, I'll be there under a different name (My Second Account).

I don't understand what you mean.  Can you rephrase in a way that includes several unbroken paragraphs of mathematical arguments?

[zxcvbn2]

Your literal name is "My Second Account"?
Also, I'm guessing the man you really need to let know this is rrenaud.

That's correct. I'm sitting in the tournament lobby right now. I'll PM rrenaud, thanks.

[zxcvbn2]

So I have just a few questions about the Semifinals/ Finals. I hope I'm not pestering you too much.

First of all, how will the semifinal matches be decided? Will they be seeded, random, by qualifying day?

Second, what is the actual format of the match? Is it just one game winner take all?

And lastly, what kind of timeframe do we have to complete those games?I know it was very soon, but when exactly are the games to be completed?

Thanks in advance!

[theory]

Semifinals will likely be seeded in some form, though we will probably rotate the games.

I think we'll do something similar: everyone plays 4 games.  Can be persuaded otherwise though.

Finally, it must finish between June 30 and July 2.  No exceptions.

Let's also discuss 3p v 4p.  On the one hand, we had two 24pt qualifiers from 3p pods.  On the other, what more can you ask of them than winning all their games?  Also, dghunter went 1-1-1-3 but couldn't advance because he was in a 3p pod.

Finally, let's discuss the points system.  Do we keep it in place for Day 3 - Day 4 or revert back to wins?

[Captain_Frisk]

Semifinals will likely be seeded in some form, though we will probably rotate the games.

I think we'll do something similar: everyone plays 4 games.  Can be persuaded otherwise though.

Finally, it must finish between June 30 and July 2.  No exceptions.

Let's also discuss 3p v 4p.  On the one hand, we had two 24pt qualifiers from 3p pods.  On the other, what more can you ask of them than winning all their games?  Also, dghunter went 1-1-1-3 but couldn't advance because he was in a 3p pod.

Finally, let's discuss the points system.  Do we keep it in place for Day 3 - Day 4 or revert back to wins?

I'm in favor of points, but I tend to favor more reward for winning.  1st place - 2nd place > 3rd place - 4th place.  5-3-1-0

As for 4p vs 3p... I strongly encourage drafting warm bodies out of the general lobby to round it out.

If all players had equal skill and there was no seating advantage: A player has a .39% chance of getting a perfect score from a 4p pod.    Same conditions - a player has a 1.23% chance of getting a perfect score from a 3p pod... that's over 3 times as good - not to mention that the 3p games are going to be more skill based - so it will be easier for a higher skilled player to advance well out of there.

[WanderingWinder]

I still think 3p is a massive advantage over 4p, I think the established math bears that out, and would even moreso in a more generalized case, though I guess there are some who disagree with that - and I don't think anyone's going to actually go do all that math.

On the other hand, I think a much bigger effect will be in place when you take into account the different skill levels of the players. It is much easier for a highly skilled player to demonstrate superiority in a small series of 3p matches than in a small series of 4p matches. You have a lot more control over what happens in every individual game, and the games also last longer. I think this is further compounded by the seating order effect being a good bit larger in 4p than in 3p. Overall, it is not clear to me that it is easier to 1-1-1-2 in a 4p than to 1-1-1-1 in a 3p (in fact, I suspect the opposite is true), or even to 1-1-1-1 in a 3p than to 1-1-2-2 in a 4p (which seem close to me; of course this is subjective, as it's extremely difficult to measure this - impossible on the data we have).
So I would try to avoid different pod sizes as much as possible.

I like points in general, and apparently this is what's happening at nationals/worlds? I'd suggest that format, if possible, because this qualifies for that. Of course, my preferred 4p point system would be something along the lines of 6/3/1/0.

Of course, if you do just wins, this gives an even huger advantage to any 3p guys....


Edit: cleaned up a stray line at the end.

[Powerman]

Could you add an extra field to the signup sheet that says "indicate preferred choice: 3p or 4p" and then place people that way, with the number of qualifiers proportional to the number of players in each?  (IE: if it's half and half, 2 qualifiers from 3P and 2 qualifiers from 4P)  You might have to juggle around a few people, but assuming you can either get it to be half-half or 1:3 (3:1) this could alleviate any problems of competing between 4P and 3P.

[theory]

Of course, my preferred 4p point system would be something along the lines of 6/3/1/0.

Of course, if you do just wins, this gives an even huger advantage to any 3p guys....
If we're doing

Isn't this a contradiction?

Could you add an extra field to the signup sheet that says "indicate preferred choice: 3p or 4p" and then place people that way, with the number of qualifiers proportional to the number of players in each?  (IE: if it's half and half, 2 qualifiers from 3P and 2 qualifiers from 4P)  You might have to juggle around a few people, but assuming you can either get it to be half-half or 1:3 (3:1) this could alleviate any problems of competing between 4P and 3P.

I'd prefer not to do that, since that makes dropouts and no-shows even worse.

[theory]

Here is my concern about how we're addressing the 3p-4p unfairness: are we trying to make it "fair" not by removing advantage/disadvantage, but by giving a 50/50 chance of a big advantage and a 50/50 chance of a big disadvantage?