# Flipping Green

**"Two simple means to get your greens."**

βPharma, Chief Medical Officer at Delphi(

*missing reference*)

Featured in two of the top 8 decklists at the first Energon Invitational, the Air Strike Patrol made a grand entrance into the competitive scene, showing its potential in both defensive and offensive strategies. Tailwind's remarkable ability comes at the very inexpensive cost of playing the green and blue/orange cards, most of which possess a good power level with effects in line with the strategies favored by their battle icon colors.

The aim of these notes is to quantify our rate of success in triggering abilities like Tailwind's, i.e. flipping at least one green icon while battling. Our focus will be on green icons because of the relevance of the Air Strike Patrol in the *Siege* metagame. But there's nothing specific to the green icon playing any role in the following notes. Therefore, the same approach could be used to calculate the probability of flipping at least one orange, blue, or black icon.

Let's start with the result, and leave some more explanatory comments for later on in these notes.

**Figure 1** shows our minimum and maximum success rates as a function of the number of green cards we play. By green cards we mean every card featuring a green icon, which includes **π
πΆ **cards**.** Where exactly our probability lies between the minimum and maximum values depends on how many white (non-green) cards we play. A good approximation for the minimum is given by

P_{min} = ( 2 β P_{G }) P_{G} ,

where P_{G} is the fraction of green cards in our deck. For example, with 10 green cards in a 40-card deck, P_{G} = 0.25, and P_{min} = 44%.

We can increase our success rate by playing white cards, so that our final probability takes the form of a product between this minimum probability and a **boost factor**:

P = boost factor Γ P_{min} ,

where

boost factor = 1 + ( 2 β P_{W }β 2P_{G }) P_{W} ,

**Figure 1**

and P_{W} is the fraction of white, non-green, cards in our deck. In our example, 6 white (non-green) cards would correspond to P_{W} = 0.15 and a boost factor of 1.2. Therefore, we would flip at least one green card with probability P = 1.2 Γ 44% = 53%.

For easier reference, the minimum and maximum probabilities shown in **Figure 1** are also listed in the following table:

This is everything we need to know. The remainder of these notes just provides some additional details about our approximation, and the factorization of the probability that we've employed.

## Appendix

We know already that with no Bold or Tough, probabilities with replacement are a good approximation of exact formulae. In this approximation, our probability of flipping at least one green card with our initial two flips is simply ( 2 β P_{G }) P_{G} . Alternatively, we might flip at least a white (non-green) card, and have a second chance with our next two flips. Our chances of flipping just white is ( 2 β P_{W }β 2P_{G} ) P_{W}. Therefore, our final probability of flipping at least **πΆ** is

P = [1 + ( 2 β P_{W }β 2P_{G }) P_{W} ] ( 2 β P_{G }) P_{G} .

Notice that **π
πΆ cards are counted in** P_{G} **and not in** P_{W}, because flipping two additional cards is irrelevant if we've already flipped a **π
πΆ** card. For example, A deck with 3 **π
πΆ**, 15 **πΆ** and 7 **π
** cards gives us the same success rate as a deck with the simpler composition 18 **πΆ** and 7 **π
**.

Of course we haven't tried to maximize P in these notes, as that's simply achieved by playing as many green cards as we can. And we can easily check that our formula gives the right limits, P = 0% when P_{G} = 0%, and P = 100% when P_{G} = 100%. Instead, it has been more useful to find the minimum and maximum probability we can achieve, given the number of green cards we play. As an intuitive way of tackling this problem we have factorized P into the product of two terms: a **base probability**, ( 2 β P_{G }) P_{G} , multiplied by a **boost factor**, 1 + ( 2 β P_{W }β 2P_{G }) P_{W} . The boost factor includes the effect of white cards, and increases with their number. It starts from an initial value of 1, corresponding to P_{W }= 0%, and reaches its maximum, 1 + ( 1 β P_{G })^{2}, when we play as many white cards as we can, so that P_{W } = 1 β P_{G}. Therefore, the lowest success rate corresponds to the case in which we don't play any white (non-green) card, and it's just equal to our base probability. Our maximum rate of success corresponds instead to the highest boost factor, and is equal to

P_{max} = [ 1 + ( 1 β P_{G })^{2} ] ( 2 β P_{G }) P_{G} .

These minimum and maximum rates are the black, dashed lines in **Figure 1**, where they are compared with their exact counterparts (green, continuous lines). The two sets of values are hardly distinguishable from each other in this plot, and our approximation ends up being remarkably good.