Answers.Body: <p>The expected score for a player with a rating advantage of e is (1+10^(e/400))(^-1) The expected score over seven boards is seven times this, and for the trick to work (ie not to lose) this must be at least 4. The break-even point comes when e=400*ln_{10}(3/4)=49.95.</p>
  <p>Many years back, in the British Chess Federation team competitions, I quizzed my players in advance how strongly they objected to being Black. Those with the least objection were put on the odd boards. If we won the toss, we took Black on those boards. So permutation can be advantageous.</p>
  
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Answers.Text: Does an expected score of 4 translate into a 50% chance of actually winning the match, though? In a 3 board match, assuming no draws, the cheating side would win 9/16 of the time if we balance it so they get 50% of the points. (They can never win all the boards but the honest side can.)
Answers.UserDisplayName: D M
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Answers.CreationDate: 2024-03-20T12:52:54.343Z
Answers.Text: While a very exact computation would have to consider the binomial distribution of results and average over it, I'm OK with it, especially as it so nicely matches my Fermi calculation. :-)
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Answers.UserDisplayName: Philip Roe
Body: <p>To my best knowlegde mere permutation of team order is pointless (at least by German rules here): everyone playing too low loses by default, everyone too high is a weaker opponent than in the &quot;regular&quot; order.</p>
  <p>But browsing my old club zines, I just unearthed a titanic scandal that can only be measured in megalewinskys: The opponent (of our youth team) for the climb into the higher league re-registered a dummy for board 1 (which of course defaulted, so to be not shown) and everyone else could play one board lower.</p>
  <p>I would be interested how effective this shenanigan is. Assume we have 8 players (also standard here), and the teams are equally strong with the board 1 Elo is 2000 (the actual value shouldn't matter) and the value decreases by e for each lower board (a random but rather reasonable assumption).</p>
  <p>If e is zero, clearly it is pointless, they &quot;earn&quot; just a default on board 1 so the outcome would be 4.5:3.5 for us in average. If e is gigantic, all lower boards win resulting in a 1:7. Thus, some e must exists where the 4:4 line is crossed in favor of the cheater.</p>
  <p>At the age of the incident, Germany still had Ingo, which would make it even easier to compute since the system is strictly linear - e must be of the order 50/8 in that system (since 50 Ingo points difference=100% win), and since the factor between the two systems is about 8, e should be around 50 Elo points.</p>
  <p>Can you compute e mathematically by chaining some probabilities?</p>
  <p>(Honk if I should take this to a math board.)</p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2024-03-20T02:42:34.233Z
Comments.Text: Can you explain why "everyone playing too low loses by default", please?
Comments.UserDisplayName: Laska
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2024-03-20T05:30:06.717Z
Comments.Text: @Laska It's a rule. Your highest rated player must play board 1. If they play a lower board they lose by default.
Comments.UserDisplayName: D M
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2024-03-20T12:44:12.023Z
Comments.Text: Even more precise, everyone playing a lower board than given in the (fixed) order before the new series starts loses by default. (Someone playing higher is unaffected. The rule is easy to implement even with the computer system we have for gathering results.)
Comments.UserDisplayName: Hauke Reddmann
Tags: <elo><cheating><team>
Title: Cheating with team order
ThreadText: # Cheating with team order
  
  (Asked by: username_0 on 2024-03-19)
  
  To my best knowlegde mere permutation of team order is pointless (at least by German rules here): everyone playing too low loses by default, everyone too high is a weaker opponent than in the "regular" order.
  
  But browsing my old club zines, I just unearthed a titanic scandal that can only be measured in megalewinskys: The opponent (of our youth team) for the climb into the higher league re-registered a dummy for board 1 (which of course defaulted, so to be not shown) and everyone else could play one board lower.
  
  I would be interested how effective this shenanigan is. Assume we have 8 players (also standard here), and the teams are equally strong with the board 1 Elo is 2000 (the actual value shouldn't matter) and the value decreases by e for each lower board (a random but rather reasonable assumption).
  
  If e is zero, clearly it is pointless, they "earn" just a default on board 1 so the outcome would be 4.5:3.5 for us in average. If e is gigantic, all lower boards win resulting in a 1:7. Thus, some e must exists where the 4:4 line is crossed in favor of the cheater.
  
  At the age of the incident, Germany still had Ingo, which would make it even easier to compute since the system is strictly linear - e must be of the order 50/8 in that system (since 50 Ingo points difference=100% win), and since the factor between the two systems is about 8, e should be around 50 Elo points.
  
  Can you compute e mathematically by chaining some probabilities?
  
  (Honk if I should take this to a math board.)
  
  **Tags:** elo, cheating, team
  
  > **Comments:**
  > **username_1**: Can you explain why "everyone playing too low loses by default", please?  
  > **username_2**: @username_1 It's a rule. Your highest rated player must play board 1. If they play a lower board they lose by default.  
  > **username_0**: Even more precise, everyone playing a lower board than given in the (fixed) order before the new series starts loses by default. (Someone playing higher is unaffected. The rule is easy to implement even with the computer system we have for gathering results.)  
  
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  ## 1 answers
  
  ### Answer by username_3 on 2024-03-19. Score: 3 | Accepted answer
  
  The expected score for a player with a rating advantage of e is (1+10^(e/400))(^-1) The expected score over seven boards is seven times this, and for the trick to work (ie not to lose) this must be at least 4. The break-even point comes when e=400\*ln\_{10}(3/4)=49.95.
  
  Many years back, in the British Chess Federation team competitions, I quizzed my players in advance how strongly they objected to being Black. Those with the least objection were put on the odd boards. If we won the toss, we took Black on those boards. So permutation can be advantageous.
  
  > **Comments:**
  > **username_2**: Does an expected score of 4 translate into a 50% chance of actually winning the match, though? In a 3 board match, assuming no draws, the cheating side would win 9/16 of the time if we balance it so they get 50% of the points. (They can never win all the boards but the honest side can.)  
  > **username_0**: While a very exact computation would have to consider the binomial distribution of results and average over it, I'm OK with it, especially as it so nicely matches my Fermi calculation. :-)  
  
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