This is the rule as stated:
Your opponent guards for 2 pass, and you have a Vanguard that does triple drive. If your first drive check is a trigger, you should all-in, because the chance of netting an additional trigger on 2nd or 3rd drive check is high.
Imagine a common scenario: You smile to yourself as you declare an attack with your stride unit, knowing your opponent has to guard your attack. (extremely evil on-hit effect or extremely evil attack when your opponent is at 4 damage) Well, your smile starts to fade a little as you notice your opponent's eyebrows starting to knit. I mean, you didn't even bother to announce the power level, knowing the effort would go to waste if your opponent smacks a perfect guard on the table.
But alas, your opponent starts putting down 10k shields with contemplation, hesitating a little. (At this point, your own eyebrows start to knit) Finally, your opponent says, "two to pass". Taking a deep breath, you reveal the top card of your deck.
It's a trigger. What now?
And so, the question arises: should you pass the trigger effect to your Vanguard, knowing that you have two more chances to check a trigger?
It's the kind of thing that could spark off some friendly debate among friends or in forum discussions, an idea which taunts at the nerd within you. Intuitively, you "feel" that it's right because in many games, you opt for the safe route, and then regret it as soon as you check a second trigger that could have had devastating (and extremely evil) effects.
What, then, are the actual probabilities? If you're interested to see the calculations, scroll down a bit more. For now, let's focus on the results.
Okay, so your first check is a trigger. Let's just say we want a probability of at least 50% to check another trigger in the next 2 drive checks. In other words, we are interested in cases where we can succeed most of the time.
If it's a serious game, you'd better start counting your triggers, because that may affect your decision. It's pretty obvious that if you have more triggers in your deck, the chances of success will be higher. And of course, you'd fare much better with a "thin" deck that contains higher trigger density.
So, let's consider the probabilities when we have, say, 30 cards left in our deck:
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| Scenario: Triple Drive! First check is a trigger! y-axis: Probability of second and third check containing 1 or 2 triggers x-axis: Number of triggers in a deck of 30 cards |
Here is the full list of the minimum number of triggers required for a 50% success rate for different amounts of cards left in your deck:
For 10 cards in deck: at least 3 triggers left
For 11 - 14 cards in deck: at least 4 triggers left
For 15 - 17 cards in deck: at least 5 triggers left
For 18 - 20 cards in deck: at least 6 triggers left
For 21 - 24 cards in deck: at least 7 triggers left
For 25 - 27 cards in deck: at least 8 triggers left
For 28 - 31 cards in deck: at least 9 triggers left
For 32 - 34 cards in deck: at least 10 triggers left
For 35 - 38 cards in deck: at least 11 triggers left
For 39 cards in deck: at least 12 triggers left
Do you notice something about the trigger density required?
A general rule of thumb can be seen from this: you only require a deck of at least one-third triggers to have a better-than-average chance of benefiting from Stride Rule #1.
Considering that the initial proportion of triggers in a deck is 16 out of 49, or about a third, a well-shuffled deck with an even distribution of triggers should always contain about one-third triggers. This means that Stride Rule #1 would work most of the time for a well-shuffled deck!
I hope that this post has given a somewhat legitimate support to a rule that we mentioned in passing. It may seem a little bit crazy to be so rigorous on a game meant for fun, but it's always good to reassure yourself with extravagant details, because... why not? Good luck and have fun in your games!
Details of calculations
In my post, I only consider raw probabilities of drawing X number of triggers from a deck of N cards (without replacement). This can be done rather conveniently through the hypergeometric distribution function.There are 4 inputs to this function: N (number of cards), X (number of triggers), n (number of cards drawn) and x (number of triggers drawn). It calculates the probability that, from a deck of N cards containing X triggers, after drawing (or drive checking, in this case) n cards, we get exactly x triggers.
In MATLAB, the calculations can be performed by the in-built function hygecdf(x,N,X,n). For instance, in the case where there are 30 cards left in a deck of 9 triggers, we want to calculate the probability when we drive check 2 cards (n = 2) and get 1 or 2 triggers. Because this is a cumulative distribution function, it sums up the probabilities of all cases where we get x triggers or below. So, taking hygecdf(1,30,9,2) will give us the probability of checking 1 or 0 triggers. When we enter hygecdf(0,30,9,2) we get the probability of checking no triggers. Therefore, one way to calculate our desired probability is to take 1 - hygecdf(0,30,9,2), which gives us 0.5172.
This calculation can then be repeated for varying numbers of triggers (varying X) in a deck of 30 cards, to generate the bar chart shown. The same set of calculations was then performed with varying numbers of cards left in deck (varying N). From the set of probabilities generated in these calculations, we are only interested in the minimum value of X that would cause the probability to rise above 0.5.
One should note that such calculations are general and can, in fact, apply to a twin drive instead. It does show that you are indeed more than likely to check at least one trigger in a twin drive, given that your deck contains one-third triggers. This is by no means the only way to analyse the situation, and there may be some other way that could provide a more comprehensive outlook.
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