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 Paul Marsh
PowerScore Staff
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#74530
Hi tfab!
tfab wrote:Basically, due to the 3rd and 4th rules, "U" cannot be out at the same time that R is in the 2nd slot?
Not quite. In fact, U must be out if R is in the 2nd slot. Our combination of the 3rd and 4th rules gives us the biconditional R=2 :dbl: U just mean that we've got two possibilities for R and U:

- First possibility: R is in the 2nd slot, and U is out
OR
- Second possibility: U is in, and R is not in the 2nd slot (so R can be 1st, 3rd, 4th, or he can be out).

That's all it means! We're always either in the first possibility or the second. Hope that helps.
 The Stig
  • Posts: 3
  • Joined: Jul 13, 2020
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#78039
Overall, my setup for this game conformed to what's been written above. I saw that the out group is extremely limited, with just one available space. I understood the biconditional interaction between rules three and four, and I got the inference that Q cannot be in the fourth race. However, two of the earlier posts by Powerscore employees Adam and David seem to imply that there's a another inference to be had regarding the placement of S.
Adam Tyson wrote: ...Just be sure to test inferences, though - the rules might not say it explicitly, but maybe the way they interact would prevent S from running the 1st or 3rd race, or maybe he has to run one of the races. You can't read that rule as meaning that he is DEFINITELY allowed to run in 1st or 3rd or that he could DEFINITELY be left out, just that the rules don't explicitly prohibit those things.
David Boyle wrote: ...Do some of the rules interconnect, e.g., if S can't be 2 or 4, does that relate at all to the last two rules? But you already have some valuable observations!
Annoyingly, I can't seem to derive the inference that they are hinting at in these posts. Could anyone help point me in the right direction? Thank you in advance for your insightful responses!
 haileymarkt
  • Posts: 13
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#78056
I know that if U is out, we know R is second and that if U is in R is not second. But if R is not second, can we make any inferences?

I'm really struggling with understanding these rules!
 Rachael Wilkenfeld
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#78156
Hi Steig and Hailey,

Steig, there's no additional inference that I'd write down here, but I would note the role that S's placement plays with the QT block. S can only go 1, 3, or out. If S is 3, that means QT block has to be 1 and 2, or Q is out. There's more flexibility if S is 1. If S is out, that means that R can't go in 2 (because that would put U out as well). I wouldn't write any of this as part of my diagram, but I'd consider it part of my mental notes when working through the game.

Hailey, we can absolutely make an inference about when R is not in second by taking the contrapositive of rule 3. Rule 3 says that u :arrow: R2. Our contrapositive would be R2 :arrow: U. So we can draw the inference that if R is not in 2, U must be in. That's just a straight inference from the contrapositive.

Hope that helps!
Rachael
 angelsfan0055
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#86446
Is it correct to make the inference that either R is in 2nd or U is in the race (but not both, per the last rule, so R can't be in second while U is in) has to happen in the game? This helped me make diagrams quicker and I've noticed a pattern with this type of rule
 Adam Tyson
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#86539
That's correct, angelsfan0055! The biconditional nature of the combination of the 3rd and 4th rules means that there are only two possible outcomes regarding R and U:

1) R runs 2nd and U is out
2) R does not run 2nd and U is in

It's a "both or neither" situation, where either both of those things occur (scenario 1) or neither of them does (scenario 2).

Note that it may be possible for both U and R to be in, so long as R is not 2nd, or R may be out if U is in.
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 bebeg3168
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#96494
I was trying to make heads or tails of the 3rd and 4th rule. You said exactly what I was thinking but I questioned myself and missed 3 questions on this game. Great lesson.
Paul Marsh wrote: Fri Mar 27, 2020 7:16 pm Hi tfab!
tfab wrote:Basically, due to the 3rd and 4th rules, "U" cannot be out at the same time that R is in the 2nd slot?
Not quite. In fact, U must be out if R is in the 2nd slot. Our combination of the 3rd and 4th rules gives us the biconditional R=2 :dbl: U just mean that we've got two possibilities for R and U:

- First possibility: R is in the 2nd slot, and U is out
OR
- Second possibility: U is in, and R is not in the 2nd slot (so R can be 1st, 3rd, 4th, or he can be out).

That's all it means! We're always either in the first possibility or the second. Hope that helps.

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