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 ava17
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#61791
Does the F :dblline: K mean that one of F and K must be in the out group?
 Adam Tyson
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#61805
That's exactly right, ava17! The same goes for K and M - at least one of them must always be out. That's a good way to read the double-not arrows: "at least one of these two variables is out, maybe both are." We also often discuss it in terms of what is in, saying "these cannot both be in." Both are correct statements, just looking at the relationship from two different angles. Well done!

And TB, you're almost right on the game type. It's actually overloaded, not underfunded, because there are more variables (9) than can be selected to fill the spaces (5). An underfunded game would be where there are not enough variables to fill the slots, such that either some variable must repeat or else some slots go unfilled. There are only the two numeric distributions. Not sure I would pursue templates on this one, and I am a bit of a template junkie. If there are more than 4, then it's probably not even worth chasing them down, especially since there are only 5 questions.
 k100
  • Posts: 10
  • Joined: Nov 21, 2019
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#77430
Hello,

Why can't 1-1-3 be possible? Given that if more than one botanist is selected :arrow: at most one zoologist is selected, if exactly one botanist is selected, shouldn't 3 zoologists be allowed? Thank you!
 Adam Tyson
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#77742
That's correct, k100! When we write a distribution like 3-1-1, it is what we call an "unfixed" distribution, which means that it is not assigning those number to any particular group. It just means that one of the groups has three members while the other two have one each, without specifying which one is the group of three. It could be either the botanists or the zoologists. So your 1-1-3 is what we would call a "fixed" distribution, assigning the 3 to the zoologists, while ours covers that possibility but also covers other possibilities. Good work!
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 Mikelsat
  • Posts: 3
  • Joined: Jul 08, 2021
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#89402
Why would

K :dblline: M
and
M :arrow: P & R

not constitute K :dblline: P and R :dblline: K inferences? I came about it as 'if not k then M, if M then P and R"

Thanks!
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 Mikelsat
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#89403
Mikelsat wrote: Thu Aug 05, 2021 5:44 pm Why would

K :dblline: M
and
M :arrow: P & R

not constitute K :dblline: P and R :dblline: K inferences? I came about it as 'if not k then M, if M then P and R"

Thanks!
Wait a minute... is it because "if x then not y" means both cannot be chosen AND both do not have to be chosen, while "if not x then y" means if not one, the other must be chosen? Therefore my initial interpretation of the inference I was trying to force is invalid
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 Beatrice Brown
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#89501
Hi Mike! Thanks for your great question :)

You're correct that those two inferences you tried to draw are invalid. It's helpful to think about what exactly the double-not arrow means to see why these inferences are invalid:

K :dblline: M is the same as the combination of the following two conditionals:
K :arrow: not M, and
M :arrow: not K.
Taking the two rules you mentioned together, this would just tell us that if M is in, then K is not in but P and R are in. We can't read across the double-not arrow to infer anything about the necessary condition in the rule that if M is in, then P and R are in.

If K is not in, all we know is that M could be in. And if M is not in, all we know is that K could be in, but this doesn't prevent P and R from being selected (inferring that P and R couldn't be selected would be a Mistaken Negation). As such, we don't know whether P and R must be in, but they could be in, and it is not necessary for either of them to be out if K is in (as your two inferences would suggest). It is possible for K, P, and R to all be in, because P and R can be in even if M is not in; all M :arrow: P and R tells us is that P and R must be in when M is in.

I hope this helps, and let me know if you have any other questions!
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 amb8ds
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  • Joined: Aug 03, 2022
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#96510
Can't we go a step further and say M -> P+R -> L/Q since K is not available and we already accounted for the F/G/H choice?
 Rachael Wilkenfeld
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#96541
Hi amb8ds,

Yes! That's a completely valid inference. I wouldn't do templates on this game, but I did draw out what happens when M is in. We know 3/5 spots from that single rule when M is in, and it's worth it to think about who the other two spots are. It only takes a few seconds to do and is potentially helpful in this game. I think that the cost in time is worth the benefit in clarity here, but it wouldn't be wrong to go into the game without that.

Hope that helps!

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