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Subject: Deduction Numbers. rss

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John A. White
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I have a deduction game idea. Like Clue.
Lets work with 18 cards.

I would assume one column of 18 unique type would be the hardest to track.
What if there are 2 rows, Color/type. Now what is the optimum for creating difficulty?
(A) 2,9
(B) 3,6
(C) 4,4 (16)

If they get a match in one column they are told.
in figure (A) Let's assume they figured out row with 9 types. Now the other row is Boolean (no value).

With that said it STILL might be the 2,9.. so I am here asking.

I also assume a 3rd row creates faster solving.
 
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Jeremy Lennert
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It's not entirely clear to me from this post how you are assuming the cards will be used. My guess from your description is that you divide the cards into groups (columns), analogous to suspect/weapon/room cards in Clue, and select 1 card from each group as a "solution". Then players guess combinations of 1 card from each group (all at once), and get told how many of their guesses are correct (but not which ones).

If that's the case, the number of possible solutions is the product of the sizes of all groups:
1 group of 18 = 18 possible solutions
2 groups of 9 = 9^2 = 81 possible solutions
3 groups of 6 = 6^3 = 216 possible solutions
...
6 groups of 3 = 3^6 = 729 possible solutions
9 groups of 2 = 2^9 = 512 possible solutions
(notice NxM and MxN give different results)

The number of possible solutions should correlate to the difficulty if players are basically guessing randomly. To maximize the number of possible solutions with 18 cards, you probably want 6 groups of 3.

However, because you tell the player how many groups are correct, the players also get more information the more groups you have (there are more possible answers to the question "how many groups were correct?", which means higher Shannon entropy). This might move the point of highest difficulty somewhat in the direction of having fewer groups, but quantifying it would require a complex analysis involving assumptions about what guessing strategy the players use.
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Jeremy Lennert
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Gave this some further thought. If you wanted to do a more complete analysis, here's the strategy I'd probably use as a starting point:

1. Guess a random solution, and see how many groups are correct
2. For your next guess, change your answer in ONE group, while keeping all other groups the same
- If the number of correct groups increased, your answer in that group just changed from wrong to right; lock that group on the right answer
- If the number of correct groups decreased, your answer in that group just changed from right to wrong; lock that group on the right answer
- If the number of correct groups remained the same, then both of your answers in that group were wrong, and can be eliminated

Using this strategy, you spend 1 guess to establish a "baseline", and after that you are effectively only making guesses in one group at a time. That means the overall complexity should closely correlate with the total number of cards, regardless of the number of groups they are divided into.

That strategy is not necessarily optimal, but it's pretty simple and will likely occur to many of your players (it's also easy to copy once you see someone else do it). Since this strategy makes the difficulty almost independent of the number of groups, I would guess that the number of groups you choose will actually not have much bearing on the difficulty of the game. (But larger numbers of groups may be slightly easier.) Total number of cards will be more important.

(Unless, of course, I have made an incorrect assumption about your mechanics.)
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John A. White
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Antistone wrote:
Gave this some further thought. If you wanted to do a more complete analysis, here's the strategy I'd probably use as a starting point:

1. Guess a random solution, and see how many groups are correct
2. For your next guess, change your answer in ONE group, while keeping all other groups the same
- If the number of correct groups increased, your answer in that group just changed from wrong to right; lock that group on the right answer
- If the number of correct groups decreased, your answer in that group just changed from right to wrong; lock that group on the right answer
- If the number of correct groups remained the same, then both of your answers in that group were wrong, and can be eliminated

Using this strategy, you spend 1 guess to establish a "baseline", and after that you are effectively only making guesses in one group at a time. That means the overall complexity should closely correlate with the total number of cards, regardless of the number of groups they are divided into.

That strategy is not necessarily optimal, but it's pretty simple and will likely occur to many of your players (it's also easy to copy once you see someone else do it). Since this strategy makes the difficulty almost independent of the number of groups, I would guess that the number of groups you choose will actually not have much bearing on the difficulty of the game. (But larger numbers of groups may be slightly easier.) Total number of cards will be more important.

(Unless, of course, I have made an incorrect assumption about your mechanics.)


This is it... I am seeking probability numbers.

I did not tell you that the 18 cards are tied..
6 parent colors, Each with 3 child types... or another set of numbers if that is harder.
(inverse=3 repeat color, 6 repeat type)


 
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Jeremy Lennert
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aVoidGames wrote:
This is it... I am seeking probability numbers.

I did not tell you that the 18 cards are tied..
6 parent colors, Each with 3 child types... or another set of numbers if that is harder.
(inverse=3 repeat color, 6 repeat type)

I followed this even less than your initial explanation (and I'm still not sure if I got that right).
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Jan Åke Hansson
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What are you trying to identify - one specific card?

Then if you have it as 1x18 it takes a maximum of 17 guesses to identify which it is.

With 2x9 the max needed guesses are 9, for 3x6 it can take 7 guesses and for 4x4 it's 6.


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