Is 'snecko eye' a bad choice without 2-cost cards?

Solution 1:

This assumption is actually mostly wrong. While indeed the average card costs 1.5, the cards are randomly distributed. By increasing your hand size to 7, it is likely to see multiple zero- and one-cost cards. This little piece of C# code tries to simulate how many cards are playable with 3 energy, on average:

int numCards = 7;
int[] costs = new int[7];
Random rand = new Random();
int totalPlayed = 0;
for(int tries = 0; tries < 1000000; ++tries) { 
    for (int i = 0; i < numCards; ++i) {
        costs[i] = rand.Next() % 4;
    }
    int avE = 3;
    int cardsPlayed = 0;
    Array.Sort(costs);
    for (int i = 0; i < numCards; ++i) {
        avE -= costs[i];
        if (avE < 0) {
            break;
        } else {
            cardsPlayed++;
        }
    }
    totalPlayed += cardsPlayed;
}
Decimal avgPlayed = (Decimal) totalPlayed / (Decimal) 1000000;
System.Console.WriteLine("AVG cpt: " + avgPlayed);

One million hands are simulated. Try running the code: It should print out a figure of about 3.83.

This means that, if you don't draw any additional cards, Snecko eye allows you to play 3.83 cards out of the 7. If your average card cost used to be 1, then that is a distinct improvement over the 3 cards you could play before.

A pure energy relic would up the number to '4' cards. The break-even card cost (where Snecko eye does not increase or reduce) is thus at an average cost of around 3/3.83 ~ 0.78. For a mix of 20 1- and 0-cost cards, this would mean at least 5 zero-cost cards.

Here's what the probability distribution histogram of each number of playable cards looks like: