Question #1675

The power of distributions: the way I solved this game.

Started by Brett-Lindsay · started 2020-07-31 13:22 · last activity 2020-08-27 03:35 · 2 replies

I'm not an instructor, and I'm pretty new to LG, but I used some of the stuff on Mehran's video that wasn't really mentioned in the solutions for this game, and it seemed to make a lot of sense to me, at least. The first thing I did was notice the distribution from the first two rules. There were three possible games: (3-1-1-1) and (2-1-1-1-1) and (1-1-1-1-1-1) I wrote out the rules exactly the same as Naz in the video. Then, I made some inferences, some the same and some different to hers: I noticed that if we had G, then we had a (1-1-1-1-1-1) game: G --> G + F + [P -- T] + [H -- L] I noticed that if we didn't have G but had F, we'd have a (2-1-1-1-1) game: not G and F --> F + [P -- T] + [H -- L] + X (an unknown repeat of F or P or T or H or L) I also noticed that if we didn't have G or F, then we'd have a (3-1-1-1) distribution: not G and not F --> P + T + [H -- L] + XX (an unknown repeat of P or T or H or L) For the first question (Q45), I did it the same as Naz For the second question, I realized that for the first and last message to be the only messages left by the same person, we must have a (2-1-1-1-1) distribution. Therefore, we CANNOT have G, but MUST have F. If we have F, then we have the following constraints: [P -- T] + [H -- L], which means that none of those 4 people could possible leave both the first and last messages. F was the only person left who had no constraints. For the third question, as Greta left a message, we had a (1-1-1-1-1-1) distribution, which meant that everybody only left 1 message, and all the constraints were in place, most importantly [P -- T]. Therefore, T could not be first. For the fourth question, I used the distributions. D CAN be true, but that's only the (2-1-1-1-1) distribution. We also have the (3-1-1-1) distribution and the (1-1-1-1-1-1) distribution, so it doesn't have to be true. For the fifth question, I used trial and error, similar to Naz, and it took ages. I certainly didn't come up with any of these methods - I just remembered Mehran showing them in one of the LR videos.

Replies

  1. lexiw · 2020-08-25 03:22

    Hi Brett, I know that you are not a tutor but I was wondering if you could possibly explain rule one and two to me? During the explanations it seems that every person that left a message 2+ times they were put directly next to each other. For example: I don't understand why we cannot have: L L T H P L It did not specify that it had to be consecutively. Thank you!!

  2. Brett-Lindsay · 2020-08-27 03:35

    Sure @lexiw, There were 6 people and 6 messages. Rule 1: At most one person left more than one message. This just means that only 1 person could leave more than one message. There is nothing mentioned about order in this rule, but order does come into play in later rules. I'll mention why LLTHPL doesn't work soon. Rule 2: No person left more than three messages. This just means that the maximum number of messages any one person left was 3. Combined with the first rule, we know that there are three possible distributions: 1-1-1-1-1-1 (everybody left 1 message) 2-1-1-1-1 (one person left 2 messages, everyone else left 1) 3-1-1-1 (one person left 3 messages, everyone else left 1) Remember that distributions such as these are not related to actual sequence. It would be equally correct (although perhaps less clear) to write them with variables: 1-1-1-1-1-1 --> A-B-C-D-E-F 2-1-1-1-1 --> A-A-B-C-D-E (or with a random sequence: B-A-D-A-E-C) I think we just write them with the highest number first because it's systematic and clearer. LLTHPL cannot work because of the final rule: If Pasquale left any message, Hildy and Liam did also, all of Hildy's preceding any of Liam's. This means that if Liam left 3 messages, all of his messages would have to be after Hildy's. Hope that helped somewhat.

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