I Found A Mathematical Paradox In My Mechanic!

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

https://en.wikipedia.org/wiki/Least_common_multiple

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.

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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.

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.