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Two from List 1 and One from List 2, with Overlap

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MatthewF
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Posting this for a friend. I came up with a couple of mechanics for her, as did some friends, but rather than listing any of the potential solutions here and coloring your ideas, I'd like to leave the question open. Here's the deal:

To start a round players need to randomly select two things from among 4, which I'll call A B C and D. These two need to be different.

Then players need to randomly select one thing from among 4, which I'll call C D E and F. This one cannot match either of those selected the first time. That is, if C or D were selected the first time, they should not be selected the second time.

So we're looking for a foolproof, simple method that will be very easy for players to understand (which means charts or other similar inelegant solutions are less desireable). It could use dice or cards or tiles or whatever, though obviously production costs are always an issue, so the cheaper alternative would win out (no electronic randomizing selector!) :)

Nando
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Curious

What is the criteria for an item to appear in both sets? Do you know, at the time the first list is made, which items are carryovers? Must the lists be assembled simultaneously? If not, why not just arbitrarily exclude any selected would-be carryover? Why carryover an item that cannot be chosen?

WHY?! WHY?!

MatthewF
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These are fixed lists, so you

These are fixed lists, so you can literally just work with ABCD and CDEF. They'll never change. As such, yes, you know when creating the solution exactly what carryovers there are.

As to why, hey, I didn't design the game. :P They're categories of questions, so I'm sure there's some sense to it.

MatthewF
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Actually, I will go ahead and

Actually, I will go ahead and include my solution, to see if someone has something more elegant and less likely to confuse players. And that doesn't use too much more in the way of publication resources.

6 tiles (or cards or whatever) as follows:

All white tile with category A
All white tile with category B
Half white and half black tile with category C
Half white and half black tile with category D
All black tile with category E
All black tile with category F

Instructions: Draw tiles until you have two with white and one with black, discarding any you don't need to match that pattern.

Nando
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Two from List 1 and One from List 2, with Overlap

Yes, assuming you could make all your choices simultaneously (which I wasn't sure about), I was going to say that a more concise way to phrase the problem would be: randomly choose 3 options from ABCDEF, but not both E and F.

MatthewF
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Gotcha!

Gotcha!

sedjtroll
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Nando's solution seems

Nando's solution seems concise and good...

Another way (not as good) might be to take cards ABCD and choose 2 randomly, then remove any of A or B that you didn't choose and add E and F and shuffle/choose 1 more.

bluesea
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Ok, hopefully I understand

Ok, hopefully I understand this:

1. Six cards: (A-F). Use A,B,C,D face down to determine first two random *things*. Pick two cards.
2. If C & D are chosen, then AUTOMATICALLY choose one from face down E & F.
3. Else, place E and F face down with C or D or both, depending on what was chosen, then chose final card.

Alternate
1. Place four dice size labeled cubes (A thru D) in a bag. Choose two.
2. Empty bag and place E and F cubes in the bag along with either C or D cubes or both as determined by what was chosen in step 1.
3. Choose last cube from bag.

It would help to know the game a bit to come up with a more thematic solution. Hope I grasped what were asking.

Cheers

Edit: I didn't see sedj's solution before I posted; I think we have the same solution.

Taavet
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Which to choose

I guess my only addition (all the above sounds good), is if a player randomly gets an invalid selection who or what determines which is kept and what is put back.

For instance if I choose E and F which do I put back which do I keep?

MatthewF
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Nando's solution was the one

Nando's solution was the one originally proposed, but it had the "remember to remove A and B before putting in E and F" problem where people could screw it up pretty easily.

sedjtroll
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MatthewF wrote:Nando's

MatthewF wrote:
Nando's solution was the one originally proposed, but it had the "remember to remove A and B before putting in E and F" problem where people could screw it up pretty easily.

Um... no it doesn't. Or did you mean mine and Bluesea's solution?

In Nando's solution just draw the things one at a time and ignore/redraw the second of E or F that you happen to draw. I think that's the best way to go.

Nando
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Two from List 1 and One from List 2, with Overlap

Please note that Nando did not propose a solution. Nando merely rephrased the problem.

Desprez
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Are these items drawn and

Are these items drawn and then kept? Or the process of selecting indicates some effect?

If you don't mind replacing some cards between drawing rounds, you can use 4 cards, split into top and bottom halves, and during the first round you look at the top half, and the second round you look at the bottom half.

A / E
B / F
C
D

You can simply place the cards face down and draw randomly, but you have to replace the split cards for the second round, and not replace the non-split C and D. This difference in manipulation may or may not be elegant enough.

Alternately, you can also have front and back sides. You use the side you first see, and, again, during the first round you look at the top half, and the second round you look at the bottom half.

front     back
A / E     A / F
B / E     B / F
C / E     C / C
D / F     D / D

Pro: Less manipulation, you don't have to replace cards during drawing rounds.
Con: Drawing a single (C or D) in round 1, leaves an elevated chance to draw (F or E) in round 2 (this may or may not matter in your game). Also, keeping the cards hidden while drawing, and keeping players honest about the specific side drawn might cause logistic issues.

Desprez
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Another idea. There's only 18

Another idea.
There's only 18 possible combinations.
You could have a deck of 18 cards with all 3 picks on each card. So you just draw a single card.
Or you could use a custom 20-sided die, with 2 re-roll spaces, or some other creative event.

EDIT:
A custom 20-side might be expensive. You could use a normal d20 and consult a chart, though you specifically said you didn't like charts.
Alternately, you could use 3 custom 6-siders. Dunno if that would be cheaper, but the spaces are bigger for the three icons or whatnot. So you'd blindly pick one of the dice from a bucket and roll that.
You could even cleverly distribute the choices among the dice sides and let the player choose one, each die perhaps giving slightly different + or - chance to roll certain yields. e.g. The red die has a very good chance to roll C or D, but can't roll A, while the blue die has a good chance for A and B, but rarely gives C and D and never together. Or something.

Taavet
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The 18 Option

With 18 possibilities I would agree with Desprez that either a stack of 18 cards or 3 6 sided dice and a chart should be used.

These would be the easiest for players to use although component wise they might be more expensive, but only slightly so.

So the 18 possibilites could be presented as:
-cards to draw
-tiles which are flipped
-chits to pull out of a bag
-a spinner with 18 slots
-a die(dice) that correlate to a chart with 18 results, or custom with the faces showing the results.
-small player board with 18 spaces, players progress along this board together or separately and get results according to the space they are on

Other alternatives could be to have an opponent select A or B, C or D, and E or F. The players could bid for results or purchase them with some form of ingame exchange system.

That's pretty much all I could come up with. Hopefully, one of those works for you, or your 'friend'. I love that part in Spiderman 2 when he goes to see the doctor and says 'I'm Spiderman' 'Well, in my dream I'm Spiderman' 'Actually its not even my dream its a friend of mine'.

So if you must disguise 'your' game as your 'friends' idea we won't hold it against you, even a superhero like Spiderman needs anonymity sometimes.

kungfugeek
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half- thought out solution

Have a track with 6 spaces numbered 1-6.
Put the tiles on the track in order: ABCDEF
Roll a d4. Remove the tile on that number from the track. Slide the remaining tiles back to close the gap (thus leaving the number 6 slot empty).
Roll a d3. Remove the tile on that number from the track. Slide the remaining tiles down.

Remove A and/or B, if they're still there. Slide the remaining tiles down.

Roll either a d3 (if there are 3 tiles left) or a d2 (if there are only 2 tiles left).

For the d3's and d2's you can use d6 and divide the result by 2 or 3 accordingly, or have custom d6's for that purpose.

Actually a d12 will handle all those rolls, but you'd have to divide by the right number every time.
Or, you can use a d4 for everything and reroll on the 4 if you're going for a d3.

Ok, now that I have it written down, it's not as elegant as I thought. Oh, well.

Katherine
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Your own solution is the

Your own solution is the easiest for me to read & understand.

I guess it depends on how much information needs to be displayed & what is done with the components after they have been chosen.

MatthewF
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Thanks for all the good

Thanks for all the good suggestions. It really is a friend, I promise. I'm more than happy to ask questions for my own needs. :D

SiddGames
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Three Columns

If I understand this right, you actually have three columns here, not two: AB, CD, EF. You must make 1 choice from any two columns.

As for doing that randomly, I suppose you could roll a die to decide which column isn't chosen or something similar.

EDIT: okay, wait, I missed the part about 2 choices from ABCD and only 1 from CDEF. Nevermind :)

kungfugeek
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MatthewF wrote: Instructions:

MatthewF wrote:

Instructions: Draw tiles until you have two with white and one with black, discarding any you don't need to match that pattern.

Wouldn't that force a player to stop after only 2 tiles, if they drew both C&D?

You could use 4 dice, roll them all at once.
Die 1 is a d4 with these sides: a green C, a blue D, a red E, and a red F (can use little symbols instead of colors for the color blind, but that's probably more expensive and would still have to correspond to the three d6's)
Die 2 is a red d6 with these sides: AB, AC, AD, BC, BD and CD
Die 3 is a green d6 with these sides: AB, AD, BD (each side appears twice)
Die 4 is a blue d6 with these sides: AB, AC, BC (each side appears twice)

Roll them all at once and look at the d4 and the d6 that matches the color of the letter on the d4. The letters on the d4 plus the ones on the matching d6 are your categories.

Hm, I wonder if the probabilities would skew to certain combinations?

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