This is for those who are statistically gifted.

When you are running Longshot, is it better to have two equal numbers of Army guys, or is it better to have more of one than the other? Basically, the question is that if one of the characters is a virtual guarantee on every Longshot, does that improve your chances even if the other character is slightly less common?

It seems like it would be obvious that equal is better than weighted, but the more I think about it the more I wonder if there's some weird math thing that gives slightly improved odds if one card is more common than the other. I don't know the math I need to figure this out and I was wondering if someone more statistically inclined than I can help me out.

Example: you have 35 slots for your Army guys. Will you get better Longshots with a 18-17 distribution, or will you hit more often with 21-14? Granted, the amount of army guys you run depends on how many you need, what drops you need them at, etc, but as pure math, which is better and what are the odds?

If mike can wash a car in 6 minutes and bob can wash the same car in 8 minutes how long will it take them to wash it together ?

it will take them less time

I am going to assume when you name longshot, you are naming the two army guys. If that is the case, it is better to have half and half. Reason being, need to hit both army characters on every draw, so you need to improve your chances of hitting both. If one is weighted, you will hit the weighted one more often, but that does not increase your chances of a success, because longshot requires you to hit both, not just one.

One the other hand, if you are going to name 1 army guy + 1 random card (either the other army guy or a plot twist), then, of course, it is better to weight towards the army guy.

I have always wondered the same thing, but I am an artist not a mathematician.

In practice, I like 24/12 alot since you can start calling a card you need with the 24 copy character.

Is it 3 Min and 25.5 seconds?


As for the main question, why not just run Night Vision or Base of Operation and take all the guess work(except when you see the main army card on top) out of it?

mike does 1/6th of a car in a minute, bob does 1/8th of a car in a minute. Assuming they keep going at the same rate they will do 7/24th of a car together in a minute so...

3 minutes and a little under 26 seconds.

As for the longshot math, the closer you get to even the better.

The math goes something like... if you have 18 copies of a card, the odds of hitting it w/ longshot is equal to 1-(the odds of missing it entirely).

the odds of missing it entirely is (42/60)*(41/59)*(40/58)*(39/57).

Now you can't just do that twice since the probabilities are conditional (in other words, if you hit your wild sentinel then the odds you hit your mark 4 go down).

So you have to assume you hit the wild sentinel and then see how often you'll hit the mark 4 so w/ 17 copies it would be...
1-(42/59)(41/58)(40/57).

Anyway... the math works out to this:

D1 D1% D2% LS%
18 0.77 0.65 0.50
19 0.79 0.62 0.49
20 0.81 0.59 0.48
21 0.83 0.56 0.47
22 0.85 0.53 0.45
23 0.86 0.50 0.43
24 0.88 0.47 0.41
25 0.89 0.43 0.39
26 0.90 0.40 0.36
27 0.92 0.36 0.33
28 0.93 0.32 0.30
29 0.94 0.28 0.26
30 0.94 0.24 0.22
31 0.95 0.19 0.18
32 0.96 0.15 0.14
33 0.96 0.10 0.10
34 0.97 0.05 0.05

Damn, that was easy, I did it the hard way.
Stupid Decimal system.

I think the reason you run more of one than the other is to naturally draw into that card. Vomit isnt much if you cant hit the wild sentinels early game and the mark 4s late game. So its better to run half and half; but will having the same amount of one half in your hand, in the early game, help? Mark 4s really dont do much turn 1-3.

The best method is to find a number of each you are comfortable with ... and then up the number of the one you want to be weighted.

The thing to remember with Longshot is that you'll sometimes hit more than 2 cards ... and in those cases, it will often be the weighted card you'll get more of.

Once you get into REALLY high numbers of characters, you'd rather 'stop' at a reasonable number for the 'enabling' character while you up the 'main' character ... otherwise you'd be losing too many support cards.

If, hypothetically ... 16 of each is 'good enough' ... 16 and 20 is better. You might be more consistent at 18 a piece ... but getting extra copies of say ... Mark IV won't be as helpful as extra copies of Wild Sentinel.

So assuming 18-17, you've got a 50% chance of hitting, but at 21-14 you only have a 47% chance. Interesting. Thanks Dalton.

Well, Mike's figures assume you are drawing into a 60 card pool with your starting value (18/17) of Army dudes.

This will not be the case. At the very maximum you will be drawing into a 54 card pool and your starting values for army characters will depend on how many you have drawn already.

(You can't assume that the remaining pool has the same % of army characters either as, the very first time you are succesful with LS, the weighting from random draws diminishes, and the first time you fail, the weighting increases again. Plus you have the (potential) knowledge of the mulligan)

So you can't take the figures he presents as accurate. However the conclusion he draws (that balanced numbers give a better % chance) is accurate, as you will always be presented with a bell curve with the median weighting giving a better chance.