Rule 32B3 Question

[Steve_Immitt]

@jwiewel

@BobMessenger

@jwiewel

Splitting the two U2400 prizes three ways gives the U2400 players more than splitting four prizes four ways.

Actually, it doesn’t. ($1379 + $766) / 3 = $715. ($536 + $306 + $1379 + $766) / 4 = $746.75.

So we have the following options mentioned so far:
A) 2495 gets 536, others get (1379+766+306)/3
B) 2495 gets (536+306)/2, others get (1379 + 766 + (536+306)/2) /3
C) All four get (1379+766+536+306)/4
D) 2495 gets 536, others get (1379+766)/3

I think C is technically correct (the U2400 players are NOT getting less than they would splitting only the U2400 prizes - so a shift from an equal split is not triggered) while A is probably what people were aiming for when this rule about place prizes was first written (I vaguely remember hearing that it was based on a US Open case where one expert tied overall with some prize-winning GMs and after the prize division he ended up getting an amount less the first place expert prize).

D is wrong with the actual prize structure but would have been technically correct if it was 456 and 256 instead of 536 and 306 (with the 2495 getting 456 and the 256 being available for the 3.5 players).

A comes across as feeling fairer, and the 2495 would get the same amount he would have gotten if he was sole third. If I had my preference, I’d want to simply avoid the issue by having a different prize structure. From a practical standpoint, four prizes were used for the four players and the 2495 (the only player who would receive more money under C) received the maximum he could have if he was the only one with his score. So an appeal seems a bit unlikely.

It’s probably only just a matter of time before an appeal reaches the Rules Committee and TDCC.

[EricHammond]

@Robgetty

I’m sticking with B.

I also support B. It is technically correct, as explained by Harold S.

[BobMessenger]

@wilecoyote

@BobMessenger

What’s different where the Under 2400 prizes are $1379 and $766 is that the four-way split prize would be $746.75, which is bigger than any single prize that the 2495 can win.

Your observation is very interesting, and a bit troubling. I think the principle you are trying to apply is that a player should not take from the pooled prize money an amount greater than any of the prizes for which he is eligible.

That may indeed have been the intention of rule 32B, but I can not find the wording to justify that reasoning. Specifically, rule 32B1 states that “the award may be one full cash prize if a clear winner, or parts of two or more cash prizes if tied with others.” There is no mention of a limit on the sum of the parts in rule 32B1.

Rule 32B1 doesn’t prevent a player tied with others from winning an undivided prize if the other players in the tie can win more money by splitting prizes for which the first player is ineligible. For example, if instead of a 4th prize of $306 and a 5th prize of $230 there had been a 3rd Under 2400 prize of $306, I think you’d agree that the three players rated under 2400 would each win $817, leaving the undivided 3rd prize of $536 to be won by the player rated 2495.

Although rule 32B1 doesn’t explictly mention a limit on the sum of the parts in rule 32B1, it does say that players are limited to “one cash prize per player”. Going back to the actual prize structure, as I see it this means that the 2495 is eligible for either the 3rd prize or the 4th prize but not both, which implies that the maximum amount the player can win is $536. “A clear winner of more than one cash prize must be awarded the most valuable prize.” Although the 2495 is not initially a clear winner, he/she becomes a clear winner if the other players in the tie divide the remaining prizes (not counting the 5th prize, because four players can win only four prizes.) Using your terminology, 3rd prize is “blue currency” and 4th prize plus the two U2400 prizes are “pink currency”.

At any rate, that’s how I interpret rule 32. If/when the Rules Committee decides otherwise I will revise my understanding of the rules accordingly. In any case I think the rule should be revised/clarified, preferably with one or more fairly complex examples.

[BobMessenger]

@Harold S.

Let’s look at it this way. What if there were only 2 players tied for third and neither was a class player. It’s simple to see that they would each get half of the sum of the 2 prizes.

Now we also have class prizes here and the decision was to give the only player above the class limit the higher of the 2 place prizes. Doesn’t this mean that when the class prizes were calculated that only the 4th place prize was added to the pool for them while third went to the player over the class cap? That’s what it looks like to me.

Yes, that’s correct.

@Harold S.

I see no justification for taking the lower of the 2 place prizes instead of half of their sum.

I gave my justification in my most recent reply to Ken Ballou, citing a sentence from rule 32B1:

“A clear winner of more than one cash prize must be awarded the most valuable prize.” Although the 2495 is not initially a clear winner, he/she becomes a clear winner if the other players in the tie divide the remaining prizes (not counting the 5th prize, because four players can win only four prizes.)

@Harold S.

I believe the correct way to distribute is to give half of 3rd + 4th to the player rated over the cap and add the other half to the 2 class prizes and divide by 3.

I think that would be contrary to rule 32B3, which says that “If winners of different prizes tie with each other, all the cash prizes involved shall be summed and divided equally among the tied winners unless any of the winners would receive more money by winning or dividing only a particular prize for which others in the tie are ineligible.” Either all four prizes are split among the four players, including the 2495, or the players rated under 2400 split the prizes for which the 2495 is ineligible.

@Harold S.

This is because class money may not travel up to players who are only eligible for place prizes.

To look at this a different way, what if there were two players over the class cap and two under? It would be agreed that neither place player is entitled to clear third. How would changing the rating of a player to be under the cap entitle a player over the cap to win more money?

Look at Example 3 on page 183 of the 5th edition rulebook. If player 6 was an expert instead of a B player player 3 would have won $93.75, a four-way split of 3rd, 4th and the two Class A prizes. But since player 6 is a B player, player 3 wins $100, a four-way split of 3rd, 4th, the 1st Class A prize and the Class B prize. Using your terminology, $6.25 in Class B money “travelled up” to a player rated over 2000.

Many players apparently believe in the principle you’re citing, that class money may not travel up to players who are only eligible for place prizes, but I don’t think there’s any basis for it in the rulebook. Players only eligible for place prizes can win more money by being tied with players eligible for class prizes, as long as the class players win more money by sharing all the prizes into a single prize pool than they would by splitting just the prizes for which the higher rated players are ineligible.

[BobMessenger]

@BobMessenger

Although rule 32B1 doesn’t explictly mention a limit on the sum of the parts in rule 32B1, it does say that players are limited to “one cash prize per player”. Going back to the actual prize structure, as I see it this means that the 2495 is eligible for either the 3rd prize or the 4th prize but not both, which implies that the maximum amount the player can win is $536.

This implied rule that a player in a tie can’t win more money than he/she would win as a clear winner is important in another situation: where one player in the tie has a limited prize. As an example, the prizes in the Under 1400 section of a 5 round tournament are:

1st: $400
2nd: $200
Unrated players can’t win more than $100.

The top three players are:

Player 1, rated 1380, 4.5 points
Player 2, unrated, 4.5 points
Player 3, rated 1250, 4 points

How should the prizes be awarded?

A. Player 1 wins $400, Player 2 wins $100, Player 3 wins $100.

B. Player 1 wins $500, Player 2 wins $100, Player 3 wins nothing.

C. Other

If Player 1 is limited to a maximum prize of $400 under rule 32B1 then I believe the correct answer is A. But if there is no limit on the sum of the parts of a shared prize the correct answer appears to be B, following rule 32C6 in the Rulebook Changes document, the remaining prize money to be paid “Within the point group in which the limit was awarded.” That would mean that if Players 1 and 2 were paired in the last round, Player 1 would win more money by drawing the game ($500) than by winning it and achieving a perfect score ($400).

[jwiewel]

@BobMessenger

@BobMessenger

Although rule 32B1 doesn’t explictly mention a limit on the sum of the parts in rule 32B1, it does say that players are limited to “one cash prize per player”. Going back to the actual prize structure, as I see it this means that the 2495 is eligible for either the 3rd prize or the 4th prize but not both, which implies that the maximum amount the player can win is $536.

This implied rule that a player in a tie can’t win more money than he/she would win as a clear winner is important in another situation: where one player in the tie has a limited prize. As an example, the prizes in the Under 1400 section of a 5 round tournament are:

1st: $400
2nd: $200
Unrated players can’t win more than $100.

The top three players are:

Player 1, rated 1380, 4.5 points
Player 2, unrated, 4.5 points
Player 3, rated 1250, 4 points

How should the prizes be awarded?

A. Player 1 wins $400, Player 2 wins $100, Player 3 wins $100.

B. Player 1 wins $500, Player 2 wins $100, Player 3 wins nothing.

C. Other

If Player 1 is limited to a maximum prize of $400 under rule 32B1 then I believe the correct answer is A. But if there is no limit on the sum of the parts of a shared prize the correct answer appears to be B, following rule 32C6 in the Rulebook Changes document, the remaining prize money to be paid “Within the point group in which the limit was awarded.” That would mean that if Players 1 and 2 were paired in the last round, Player 1 would win more money by drawing the game ($500) than by winning it and achieving a perfect score ($400).

But now you are running into the no more than one prize per player limit. $500 would be all of first and half of second for one player. $400/$100/$100 gives player 1 the entirety of first. The other initial poster’s situation would have had each player getting 1/4 of each of four prizes.

P.S. B was the option used at the National Open where this was applicable.

[BobMessenger]

@jwiewel

But now you are running into the no more than one prize per player limit. $500 would be all of first and half of second for one player. $400/$100/$100 gives player 1 the entirety of first. The other initial poster’s situation would have had each player getting 1/4 of each of four prizes.

I think it’s a similar situation. If Player 2 was rated Players 1 and 2 would each get $300. Since Player 2 has a $100 prize limit, there is $200 available for distribution. Rule 32C6 says the money should be distributed within the point group in which the limit was awarded, so in the absence of a rule limiting Player 1’s prize to $400, Player 1 would get the entire $200 in addition to the original $300, for a a total of $500.

@jwiewel

P.S. B was the option used at the National Open where this was applicable.

To me, that’s just wrong.

[jwiewel]

@BobMessenger

@jwiewel

But now you are running into the no more than one prize per player limit. $500 would be all of first and half of second for one player. $400/$100/$100 gives player 1 the entirety of first. The other initial poster’s situation would have had each player getting 1/4 of each of four prizes.

I think it’s a similar situation. If Player 2 was rated Players 1 and 2 would each get $300. Since Player 2 has a $100 prize limit, there is $200 available for distribution. Rule 32C6 says the money should be distributed within the point group in which the limit was awarded, so in the absence of a rule limiting Player 1’s prize to $400, Player 1 would get the entire $200 in addition to the original $300, for a a total of $500.

@jwiewel

P.S. B was the option used at the National Open where this was applicable.

To me, that’s just wrong.

Point group is fourth in 32C6. In this particular case it is trumped by 32B1.

[BobMessenger]

@jwiewel

Point group is fourth in 32C6. In this particular case it is trumped by 32B1.

So you think 32B1 imposes a prize limit in this case but not in the original post?

[jwiewel]

@BobMessenger

@jwiewel

Point group is fourth in 32C6. In this particular case it is trumped by 32B1.

So you think 32B1 imposes a prize limit in this case but not in the original post?

In the original post it looked like each of four players should get one fourth of each of four prizes. Four fourths of prizes = one prize so no individual player gets more than one prize and 32B1 is not triggered.
In your new example your are suggesting that a player gets all of first plus 1/2 of second while another player gets the other half of second. In total there are two prizes for two players (so far, so good) but one of those is getting one and a half prizes, which is more than one prize and which triggers 32B1.

[BobMessenger]

@jwiewel

In your new example your are suggesting that a player gets all of first plus 1/2 of second while another player gets the other half of second. In total there are two prizes for two players (so far, so good) but one of those is getting one and a half prizes, which is more than one prize and which triggers 32B1.

How do you know that Player 1 is getting all of the 1st prize and $100 of the 2nd prize? Maybe he’s getting $350 of the 1st prize and $150 of the 2nd prize. As Ken Ballou said, “rule 32B1 states that ‘the award may be one full cash prize if a clear winner, or parts of two or more cash prizes if tied with others.’ There is no mention of a limit on the sum of the parts in rule 32B1.”

[jwiewel]

@BobMessenger

@jwiewel

In your new example your are suggesting that a player gets all of first plus 1/2 of second while another player gets the other half of second. In total there are two prizes for two players (so far, so good) but one of those is getting one and a half prizes, which is more than one prize and which triggers 32B1.

How do you know that Player 1 is getting all of the 1st prize and $100 of the 2nd prize? Maybe he’s getting $350 of the 1st prize and $150 of the 2nd prize. As Ken Ballou said, “rule 32B1 states that ‘the award may be one full cash prize if a clear winner, or parts of two or more cash prizes if tied with others.’ There is no mention of a limit on the sum of the parts in rule 32B1.”

Then $350 would be 7/8 of first and $150 would be 3/4 of second for a total of 1.625 prizes for one person.

[BobMessenger]

@jwiewel

@BobMessenger

@jwiewel

In your new example your are suggesting that a player gets all of first plus 1/2 of second while another player gets the other half of second. In total there are two prizes for two players (so far, so good) but one of those is getting one and a half prizes, which is more than one prize and which triggers 32B1.

How do you know that Player 1 is getting all of the 1st prize and $100 of the 2nd prize? Maybe he’s getting $350 of the 1st prize and $150 of the 2nd prize. As Ken Ballou said, “rule 32B1 states that ‘the award may be one full cash prize if a clear winner, or parts of two or more cash prizes if tied with others.’ There is no mention of a limit on the sum of the parts in rule 32B1.”

Then $350 would be 7/8 of first and $150 would be 3/4 of second for a total of 1.625 prizes for one person.

True, but that’s also the case with the SwissSys/Ballou prize payout in the original tournament. The 2495 gets $746.75. 3rd prize is $536 and 4th prize is $306. You might say that the $746.75 is $500 from the $536 prize and $246.75 from the $306 prize. That’s 0.933 of the 3rd prize and 0.806 of the 4th prize, for a total of 1.739 prizes for one person.

What I’m worried about is that if the Rules Committee endorses the SwissSys/Ballou prize payout in the original tournament, I don’t see how you can avoid giving $500 to Player 1 in my limited prize example. I really don’t want to do that.

[jwiewel]

@BobMessenger

…True, but that’s also the case with the SwissSys/Ballou prize payout in the original tournament. The 2495 gets $746.75. 3rd prize is $536 and 4th prize is $306. You might say that the $746.75 is $500 from the $536 prize and $246.75 from the $306 prize. That’s 0.933 of the 3rd prize and 0.806 of the 4th prize, for a total of 1.739 prizes for one person.

What I’m worried about is that if the Rules Committee endorses the SwissSys/Ballou prize payout in the original tournament, I don’t see how you can avoid giving $500 to Player 1 in my limited prize example. I really don’t want to do that.

In the initial example, the four players would each get 25% of the four prizes (3rd, 4th, 1st u2400, 2nd U2400). 4 x .25 is still only 1.00 prize. The payment method Ken suggested (blue and pink money) is technically not the prize proportions actually given, but rather was just a way of making sure that the U2400s got at least as much as they would have by splitting only U2400 money.

32B1 still says that “No winner shall receive more than one cash award. The award may be one full cash prize if a clear winner, or parts of two or more cash prizes if tied with others.”

[BobMessenger]

@jwiewel

@BobMessenger

…True, but that’s also the case with the SwissSys/Ballou prize payout in the original tournament. The 2495 gets $746.75. 3rd prize is $536 and 4th prize is $306. You might say that the $746.75 is $500 from the $536 prize and $246.75 from the $306 prize. That’s 0.933 of the 3rd prize and 0.806 of the 4th prize, for a total of 1.739 prizes for one person.

What I’m worried about is that if the Rules Committee endorses the SwissSys/Ballou prize payout in the original tournament, I don’t see how you can avoid giving $500 to Player 1 in my limited prize example. I really don’t want to do that.

In the initial example, the four players would each get 25% of the four prizes (3rd, 4th, 1st u2400, 2nd U2400). 4 x .25 is still only 1.00 prize. The payment method Ken suggested (blue and pink money) is technically not the prize proportions actually given, but rather was just a way of making sure that the U2400s got at least as much as they would have by splitting only U2400 money.

32B1 still says that “No winner shall receive more than one cash award. The award may be one full cash prize if a clear winner, or parts of two or more cash prizes if tied with others.”

I guess that’s what it comes to, if you assume that in rule 32B1, the “parts of two or more cash prizes” can’t add up to more than a single prize. To justify giving $746.75 to the 2495 player without violating 32B1 you have to say that the 2495 is receiving Under 2400 prize money.

I still prefer my interpretation: that because the 2495 can only win one prize, he/she gets the entire 3rd prize and is therefore ineligible for the 4th prize, which is available to be divided among the players rated under 2400. But if the Rules Committee finds otherwise you’ve at least given a rational basis for limiting Player 1 in my limited prize example to a prize of $400.

[jwiewel]

@BobMessenger

…I guess that’s what it comes to, if you assume that in rule 32B1, the “parts of two or more cash prizes” can’t add up to more than a single prize. To justify giving $746.75 to the 2495 player without violating 32B1 you have to say that the 2495 is receiving Under 2400 prize money.

I still prefer my interpretation: that because the 2495 can only win one prize, he/she gets the entire 3rd prize and is therefore ineligible for the 4th prize, which is available to be divided among the players rated under 2400. But if the Rules Committee finds otherwise you’ve at least given a rational basis for limiting Player 1 in my limited prize example to a prize of $400.

And I’d be perfectly happy with an ADM that limits a person’s prize money to no more than the largest amount the person could get if the other players in the tie had not scored as well (that would be the gist of it, the wording would need to be tightened up). That would then easily support the actual prizes awarded.

I would also guess that it was the original intent of many of those who pushed for 32B3 in the first place, but they didn’t take into account unusual situations where most of the winners of the overall prizes were also eligible for higher value under prizes.

[wilecoyote]

@jwiewel

And I’d be perfectly happy with an ADM that limits a person’s prize money to no more than the largest amount the person could get if the other players in the tie had not scored as well (that would be the gist of it, the wording would need to be tightened up). That would then easily support the actual prizes awarded.

I would also guess that it was the original intent of many of those who pushed for 32B3 in the first place, but they didn’t take into account unusual situations where most of the winners of the overall prizes were also eligible for higher value under prizes.

I’ve tried to start a discussion of this example among the rules committee members, but I have been rather unsuccessful. I changed the prize amounts to “nice” round numbers that still showed the problem. I presented three different prize distributions: the one I come up with strictly following 32B3 (call that “#1”), Steve Immitt’s and Bob Messenger’s solution (call that “#2”), and Harold Stenzel’s solution (award the 2495 half of the sum of third and fourth place prizes, and put the rest in the pool to divide among the three players rated under 2400) (call that “#3”). I asked specific questions about which prize distribution was correct and how to justify that decision.

I’ve basically gotten three answers:

1. A two character reply: “2.”
2. An answer that “#2” was correct and that a justification of “I think the rules as stated are clear on this.” (There were also some interesting and humorous comments about the advisability of such a prize fund, although I specifically asked not to discuss this.)
3. An assertion that #1 was correct under 32B3.

Well, the rules as stated may be clear, but I’m afraid I am a Bear of Very Little Brain and the combination of 32B1 and 32B3 is not clear to me. There were two interesting sentence in a further discussion of the second reply: “2495 is not eligible for the U2400 prize. He is eligible for 4th but cannot win two prizes, therefore can receive only the 3rd place prize.”

Personally, my thinking on this question has evolved. While 32B3 seems to govern in this situation, I’m persuaded it is simply wrong that the 2495 should win more prize money when there is a tie than he would win if he were the only player in the 4.0 score group. Stated differently, a tie should decrease your prize (or leave it unchanged), not increase it. Also, I am uncomfortable saying the rules are clear when three NTDs and an ANTD divide a prize fund and come up with three different answers. I am especially uncomfortable considering the (completely deserved) reputations of the other three TDs involved in the discussion. (Look, let’s be honest … compared to them, I’m the “clueless newbie” of the bunch!)

If I can rebuild some of the rules karma I spent this year , I will propose to the committee adding words such as “provided no player receives more than the largest prize for which he would be eligible if there were no tie” to 32B3. On the other hand, if the rules really are clear and I’m just missing something obvious, then maybe cluttering up 32B3 is not a good idea.

[BobMessenger]

Thanks for the update, Ken. I do wish that the Rules Committee would make a ruling on this so Steve Immitt can pay the remaining prize in that section.

[wilecoyote]

@BobMessenger

Thanks for the update, Ken. I do wish that the Rules Committee would make a ruling on this so Steve Immitt can pay the remaining prize in that section.

Technically, I didn’t do anything that would trigger the rules committee to issue a ruling. I simply sent e-mail to the committee stating that an interesting thread had been started in the forum and presenting a version of the question with slightly modified prizes (round numbers).

[wilecoyote]

I’ve studied one of the responses I received more carefully (the one explaining that the rules were clear in this case). I think I just managed to wrap my mind around the explanation. It hurts.

Anyway, if we just apply 32B3 to this situation, we would compare the result of summing the four prizes in question and dividing evenly and the result of dividing the sum of the two under 2400 prizes among three players. The former is greater, suggesting a four-way division.

However, there is an alternate analysis. Under rule 32B1, a player may not win more than one prize. Rule 32B1 could be interpreted as meaning that the 2495 is not eligible for the both the third and fourth place prizes. So, to apply 32B3 properly, we should actually compare the four-way division with the result of dividing the sum of the fourth place prize plus the two under 2400 prizes three ways. The latter is greater and thus represents the correct prize distribution.

I’m still a little uncomfortable with this definition of “ineligible,” but I think I understand the reply better now.