Oh yea I forgot about that. In comparison, Hearthstone will let you pick up to 2 of each card except for legendaries of which you can only have 1.

Oh yea I forgot about that. In comparison, Hearthstone will let you pick up to 2 of each card except for legendaries of which you can only have 1.

One thing I need to add is that these game (Eternal, Magic as well I think) allow you to have up to four copies of the same card in your deck.

I know nothing about Magic or similar card games, but I saw that

**mathematics**was one of the topics assigned and I was intrigued by the questionWhen does it make sense to have more than 75 cards?

It sometimes helps to mess around with a smaller or simpler problem and then apply the rule(s) discovered to the actual problem.

Letβs say you have three cards labeled π¦, π, π

*I could label them 1,2,3 or a,b,c but I like using emojis.*If the synergy comes from dealing the π¦ card first followed by the π card, then there is 1 out of 6 possibilities that you will get dealt that sequence:

π¦π

π¦π

ππ¦

ππ

ππ¦

ππ

So 1:6

Another way to look at it is that you had 3 possibilities for the first card you drew and two possibilities for the second card you drew.

And 3 times 2 equals 6 combinations. If you had 4 cards, youβd have 4 possibilities for the first card times 3 possibilities for the second card, which would equal 12 combinations.

So 76 cards would equal 76*75=5,700 possible combinations.

Back to the 3 cards. What if the sequence dealt didnβt matter for synergy, that is

π¦π has the same value as

ππ¦

Then in a 3 card deck there are 2 successful draws out of 6 possible combinations. So 2:6 or 1:3.

**Your chances are doubled if the order doesnβt matter.**What if your 3 card deck had two of the same card? That is your 3 cards are π¦ππ.

π¦π

π¦π

ππ¦

ππ

ππ¦

ππ

If order doesnβt matter, you have 4 successful chances out of 6 or 2:3.

So adding another πcard increases your chances of a synergy event.

If there are π cards that can be combined with a number of different cards

π¦π

ππ¦

ππ

to create synergies, then the value of adding the π card goes up even more.

Hope that was both understandable and somewhat helpful.

- FactotumAndreas
@apm I think yours is a good example to show the general problem we're talking about, if expanded slightly.

In that example

> our minimum deck size is three

> we're drawing a hand of two cards

> drawing {duck, pizza} is a good event, drawing {X, cake} in our initial hand is bad, but we still want to draw cake at some point.If our deck consists of just

**{duck, pizza, cake}**, then as you point out we have a probability of 1/3 to get the good event. We have a 2/3 probability of immediately drawing {X, cake} - but we only want cake later.Replacing cake with another pizza (deck is

**{duck, pizza, pizza}**), we increase the "good draw" probability to 2/3, but at the same time we decrease our probability of drawing cake to 0 - and we*want*to eventually draw cake.So, if instead of replacing cake with pizza, we add another pizza on top of that - making our deck

**{duck, pizza, pizza, cake}**- we will still have a 1/3 probability of drawing {duck, pizza}, while our probability of drawing cake early is reduced to 1/2 while still being available*at some point*.Adding another duck for

**{duck, duck, pizza, pizza, cake}**increases our "good draw" probability to 2/5, while decreasing the probability for an early cake draw to also 2/5. On the other hand, the fact that we've added another card also means that there's a 1/5 probability of cake being the fifth card in our deck. If we're only ever drawing four cards, this can mean that we're not drawing cake at allSo, in this toy example, depending on what value we want to give to the events "initial good draw", "eventual cake draw" and "no cake draw at all", it might be sensible (or not) to play with a deck consisting of four or five cards instead of just three.

At the same time though, we're making pretty big assumptions again and stay very vague in some regards, so I'm not convinced that this outcome can just be generalized to any deck size and any value for synergistic effects between cards.

- FactotumAndreas
@Eric Looking at this and squinting hard, it looks as if the mathematically proper way to deal with this would be to do a discrete sum over all possible deck permutations and positions ("position" meaning number of cards drawn over time) using the value you suggest, then dividing by some factor that depends on deck size, to get a closed formula for something like an

**Average Deck Value (ADV)**.Then, manipulating the inequality

**ADV(deck_small) < ADV(deck_large)**(basically stating that we're looking for a situation where the larger deck is better than the smaller deck) to remove constant factors etc. might lead to some more clues under which circumstances this can be true, if at all. @Factotum I was thinking about trying to have a small card set example. Suppose there is no

**synergy**and you have 6 cards with their**card value**equal to their number, 1, 2, ... 6. You could create a deck with 5 cards**D1**, and you could create a deck with 6 cards**D2**. So the logical way to build**D1**is to put cards 2 through 6 in the deck, and leave out card 1 which has the lowest value.In this hypothetical situation we start with 0 cards in our hand. Then the draw potential for each deck would be:

D1 draw potential = (2 + 3 + 4 + 5 + 6) / 5 =

**4**D2 draw potential = (1 + 2 + 3 + 4 + 5 + 6) / 6 =**3.5**If you could generalize this to where you are comparing a deck of size n vs n+1, where the deck of n+1 has all of the exact same cards, plus one more which is guaranteed to be smaller, we want to prove something like:

**sum(n):**the sum of value in deck 1 (avg value * n)**c:**an n + 1 card that is lower value than the lowest value in the deck

of n cards.**prove:**draw potential for the larger deck is worse than the smaller deck.1: sum(n) / n > (sum(n) + c) / (n + 1) 2: (n + 1) * sum(n) > (sum(n) + c) * n 3: n * sum (n) + sum(n) > n * sum(n) + c * n 4: sum(n) > c * n 5: sum(n) / n > c

Seen in this form, we can see that as long as the card

**c**has a value less than the average value of the smaller deck, then it reduces the average value of the larger deck. Meaning lower draw potential.I think this can be generalized to include synergy as well, it just makes things more complicated.

Edit: Math

- FactotumAndreas
@Eric Thanks. I think this is similar to the original reasoning of just cutting the "worst" card from your oversized deck. This is definitely correct for card games without synergy.

In the meantime, I've thought about your last post and my response to it. One thing led to another, and now I have this crazy formula for the "Average Deck Value"... I think I've got some explaining to do! :D