Have a logic puzzle!
Jim, AJ, Sam and Dana are playing EDH. Between them, they have all five colours in their decks.
Two of them are playing red.
Two of them are playing white.
One is playing blue, one is playing green, and one is playing black.
None of the decks have more than two colours.
Jim's deck has blue in it.
AJ's deck has only one colour in it.
Sam's deck is the only one with an allied colour pair.
WHO IS PLAYING WHAT COLOUR???
also, rules question: someone casts Rune-Tail, Kitsune Ascendant with 40 life. Is it on the field as a creature long enough to Swords to Plowshares it?
Lemme take a crack at this.
[spoiler]We have two reds, two whites, one blue, one green, and one black, among four players, and we know nobody is using more than two colors. So we have one single-color deck and three two-color decks. We know that AJ's deck is the single-color one, and that Jim's using blue, so AJ is either a red, a white, a green, or a black. We know that Sam's deck is the only one with an allied color pair, which means that both Jim and Dana are using enemy color pairs - so Jim is either Blue/Green or Blue/Red; Sam is either Red/Green, Red/Black, or White/Green; and Dana is either Red/White or White/Black - all of Sam's solutions use up either the only Green or the only Black, so Dana can't be Green/Black.
From here, we play with it. Two possibilities for Jim, let's see what happens if we assume he's Blue/Green. That means nobody else can be Blue or Green, so Sam must be Red/Black, and since we don't have two Blacks, Dana must thus be Red/White. This leaves AJ with mono-White. Two Reds, two Whites, one Green, one Blue, and one Black. So that's certainly
a possible solution, but it's based on an assumption, so unless we can show that assuming otherwise is impossible, we can't say whether it's the
only possible solution.
So, what if Jim isn't Blue/Green, and is instead Blue/Red? Well, that tells us very little conclusively - his being Red doesn't rule out anyone else being Red, and we knew nobody else was Blue. Dana could be Red/White, or could be White/Black - let's assume Red/White. That's both Reds, so Sam must be White/Green, and that's both Whites. That leaves AJ to be using Black. Two Reds, two Whites, one Green, one Blue, one Black. Another possible solution.
Go back a step, say Jim is Blue/Red and Dana is White/Black instead. Now, we still can't conclude anything - Sam can't be Red/Black, but he could be either White/Green or Red/Green, leaving AJ with either the leftover Red or leftover White. And these are all viable solutions.[/spoiler]
So, the answer is [spoiler]that there is not enough information presented to draw a conclusion. Even if you assume that the "One is playing blue, one is playing green, and one is playing black" line means three different people, so no Blue/Green, Green/Black, or Blue/Black combinations, there would still be multiple possible solutions. Unless the fact that they're playing EDH specifically, instead of just multiplayer Magic, imposes some kind of restriction - I don't know how EDH is played so if that rules out some of the possibilities, that would change the answer.[/spoiler]
