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I Found A Mathematical Paradox In My Mechanic!

8 replies [Last post]
Ark1t3kt's picture
Joined: 09/12/2017

The Mechanic:

Every player has a strip of cardboard on which there are slots numbered left to right from 6 to 0. Players take turns drafting tiles from a general pool. Each tile has a number on it. When you draft a tile, you place it under the slot corresponding to the number on said tile. Then, you slide all of your tiles down one slot. Whenever a tile hits 0, its action is triggered and then it is placed back under the slot corresponding to its number. For example, a number 1 tile will trigger every turn, whereas a number 4 tile will trigger every four turns. Of course, the higher numbered tiles will have more powerful effects.

The Paradox:

I was playing around with this idea a little bit last night, and started recording how many turns it would take for two differently-numbered tiles placed in the same slot to arrive back in the same slot. For example, it takes 15 turns for a 5 and 3 tile originally placed in the same slot to re-align.

The pattern seemed to be very simple at first: You simply multiply the two numbers on the tiles, and this gives you the number of turns.

However, there are exactly four exceptions:
- 2 and 4 will realign in 4 turns
- 2 and 6 will realign in 6 turns
- 3 and 6 will realign in 6 turns
- 4 and 6 will realign in 12 turns

The Question:

Can anyone make sense of why this is the case?

Joined: 09/06/2017
They arrive back at the same

They arrive back at the same slot at least common multiple between the two events.

Joined: 01/27/2017
Least Common Multiple

The consequence of this is that 5 will be a special case. It will never sync up reliably with a non-5 action, but a player can get different 5s to stack or fire off in sequence more or less at will.

Ark1t3kt's picture
Joined: 09/12/2017
Of Course!

Hahaha, lol I can't believe I didn't figure this out... evidently math isn't my strong suit!

Yeah, the case with the number 5 is certainly interesting... I don't think it will be an issue though if I design number 5 tiles with this in mind...

Thanks for your fresh pair of eyes!

Ark1t3kt's picture
Joined: 09/12/2017

Hmm... now that I think about it, how do I design number 5 tiles with this in mind?

Anyone have any ideas?

X3M's picture
Joined: 10/28/2013
Consider this

5 is a prime number.
6 is called a perfect number. (1×2×3, in short 2×3)

To get rid of overlap. Get rid of perfect numbers.

If you need more prime numbers. And skip the problem with 6. Replace it with 7.


Also. When players have 1 all to 6. The whole system will repeat itself.
Every 60 turns. (2×2×3×5)

If you replace the 6 with 7. Then the whole system repeats itself evey 420 turns. (2×2×3×5×7)

Good luck with your game.

Ark1t3kt's picture
Joined: 09/12/2017
Thanks for doing the

Thanks for doing the math!

Hmm, interesting ideas for sure.

I'm not too worried about the the whole system repeating itself every 60 turns, though... I think the game will typically last about 20 to 30 turns.

X3M's picture
Joined: 10/28/2013
Just some more idea's.

The 50% of 60 is 30. That means that if a game is 30 turns. There is a 50% chance that all 6 are at the same time.

It is even faster, waaay faster. With 5 slots being active. Simply said, the 5 itself has a great influence. And perhaps, you don't need to change much of the system. Simply make the 2 and 3 weaker than usual. So, this is what you can do. First we look at what happens.

This is the list of occurence when all 6 slots are used:

1: 1
2: 1,2
3: 1,3
4: 1,2,4
5: 1,5
6: 1,2,3,6
7: 1
8: 1,2,4
9: 1,3
10: 1,2,5
11: 1
12: 1,2,3,4,6

This means that every 12 rounds, you have 5 events active. Now, there is some build up as well here. And it is even possible that 5 events can happen within 6 rounds.

Solution 1:
Higher tier cards are even more "expensive" to place. Let's say, you may place a card after a certain round.
1 remains 1.
2 could cost 3 (rounds).
3 could cost 6.
4 could cost 10.
5 could cost 15.
6 could cost 21.
So, as early as 5 event occuring. You have played at least 15 rounds. And have to wait 5 more to the maximum event occuring.

Solution 2:
You don't do solution 1. But instead. The list that I posted. You could say that 2 events do not exceed a certain weight. Thus you calculate backwards how heavy a certain card should be.

1: 1 = 1
2: 1,2 = 3
3: 1,3 = 3, not 4
4: 1,2,4 = 6, not 7
5: 1,5 = 3, not 6
6: 1,2,3,6 = 10, not 12
7: 1 = 1
8: 1,2,4 = 6, not 7
9: 1,3 = 3, not 4
10: 1,2,5 = 6, not 8
11: 1 = 1
12: 1,2,3,4,6 = 15, not 16

And all 6 could weigh 21.

What I mean with this is that card 2 and 3 are both equal in weight, namely 2. 3 is simply occuring less. Now we look at round 4. The 4 card should weigh 3.
Round 5 says that the 5 card whould weigh 2 as well.
Round 6 has the 6 card, but this one can cost 5!.
Round 10 will actually be worth 5, not 6, not 8. This is because we said that the 5 card is going to weigh 2.
Round 12 will actually be worth 13, not 15, not 16.

So, 1, 2, 2, 3, 2, 5 That are the weights for the cards in design. And all 6 will have a total weight of 15, not 21.

Solution 3:
And perhaps the best.
Make all cards equal in their main event.
But higher tier cards will amplify the effects of lower tier cards.
For example, the tier 1 card will rot 2 fruit. The tier 2 card will rot 2 fruit as well. But also will amplify the tier 1 card with +1. So, you have 2 fruits rotten. Then 5 with both cards.

Ark1t3kt's picture
Joined: 09/12/2017
Hey, I really appreciate your

Hey, I really appreciate your extensive feedback.

Working the math out like this is really useful, and I will definitely take all three solutions into consideration...

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