meandme wrote:Hey BP

I did PT9 Game 2. I got 5/6 but I made 4 scenarios. Which I know is wrong because it took me awhile. And I didn't understand question 9, 10, and 13. Please explain those to me.

Thanks

God bless

Going 5/6 is pretty solid.

Alright, so I've got a pretty straight-forward In and Out, stable grouping game; 4 are selected, 3 aren't.

First two rules are awesome: Exactly 1 of J and K are is selected; Exactly 1 of N and P is selected. That means, of my 4 In slots, 2 are reserved. Of my 3 Out slots, 2 are reserved. Then, I get two straightforward tag-along rules (If X-> Y).

So where to go from here?

There are three ways. I think you went with the first way.

The first way is to create 4 scenarios: 1) J and N are in; 2) J and P are in; 3) K and N are in; 4) K and P are in. This would be perfectly fine, and it shouldn't have taken you a lot of time (it took me about 20 seconds to go through all 4 of them). If you went this way, you should have easily gotten through the questions.

The second way (probably the fastest, and the one I would recommend) is to instead of creating scenarios, just notice what's left. I've placed 4 of my 7 players as linked options. LMQ are left; there are 2 slots in, 1 slot out. So 2 of LMQ will be selected. That deduction is honestly enough to get me through the game.

However, the third way (probably the slowest) would be to create 3 scenarios from this deduction: 1) L and M are in/Q is out; 2) L and Q are in/M is out; 3) M and Q are in/L is out.

Why did I go through all of this? To show you that as long as you make logical deductions, there are many ways to arrive at the same answers. Some are faster than others, but many can be correct.

So question time.

9) So I'm looking for 2 people who are in an 'at least one' relationship. That's a conditional statement with a negated sufficient condition (~X->Y is the same as saying 'at least one of X and Y).

Well, there's my big deduction: between LMQ, I have to select 2. That means if I pick any two of those players, at least one of them has to be in (since I can't have 2 from that group out). B gives me L or M, so it's the right answer.

If you did scenarios, just go through each answer choice. If there's a scenario where both can be out, then it's not the right answer.

10) This is asking me to find a 'can't be both' relationship. A can't be both relationship is one of my big four grouping relationships (just like the 'at least one' relationship from the last question), and it's when I negate the necessary condition (X->~Y is the same as saying I can't have both X and Y).

Hopefully, you spent some time making deductions with your 4 conditional rules. If you did, you'd get the following chain: J->~K->~Q. That's the only conditional rule I can combine (except for N->~P and L, but that's not saying anything new). Since the game straight up gave me 'can't have both J and K', I'm expecting my deduction (can't have both J and Q) to be the answer. And it is.

13) The ever-popular 'what let's me say everything for sure!' question. These aren't particularly hard, just time consuming; especially if you just start trying out every option. If you do that, you'll get to the right answer (when you try one scenario out and it becomes completely filled up), but it might take you 3 minutes.

Instead, let's think about our rules. If I'm trying to fill slots up, I want to use my tag along relationships, because if I select one of those people, I get someone else in the mix. My two tag-along sufficient conditions are N and Q, so I want to start with answer choices that have one of those two in them. But should I start with N, or with Q?

I'd say Q. Why? Because with N, I know P is out and L is in. With Q, I know K is in and J is out, but I also know that either L or M is out. Q tells me a little bit more.

Now, I can do one of 2 things: I can try out both answer choice D and E, and that would get me the right answer. Or, I can think about which one is more likely to give me the correct answer. If L is in, it tells me M is out; nothing else. However, if M is in, it tells me L is out, which tells me N is out, which tells me P is in. Hmm, that sounds pretty complete. Let's try E first.

(Try E).

You'll see that everything is set if M and Q are in - L must be out (because I can only have 2 of LMQ), K must be in (because Q is in); L being out means N is out; K being in means J is out; N being out means P is in. That's the whole shebang. There's my answer.

If you did out the 4 scenarios as described above, this question is even easier - you don't have to think or do any work. Just start with A and see if there is only one scenario where both can be in. The only combo that only shows up once is (E) M and Q.