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Shifted balance issue [Math problem?]

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X3M's picture
Joined: 10/28/2013

Seems I have caused a shifted balance issue within one of my games. I do not understand how this happened. I don't understand how this shift works exactly. But it clearly occurs when I add something "special"; like a split up cannon.

Here is the problem:
In my game, we have some tanks fighting.
Each tank has it's own weaponry system.

Now the case is regarding 2 tanks.
Tank A, is a classic tank doing 25 damage and has 75 health.
Tank B, is a special tank doing 9 damage and 16 damage, and also has a health of 75.

A cannon can; miss, hit, or hit twice.
This means that Tank A (3 options) does damage; 0, 25 or 50.
Tank B (9 options) does damage; 0, 9, 16, 18, 25, 32, 34, 41 or 50.

Obviously, having 3 hits means that the other tank dies.
And obviously, in the first round, no tank dies.

However, (after building a probability tree) after the second round.
There is a chance of 27 out of 81 that Tank A beats Tank B. Yet Tank B has only a chance of 12 out of 81 for beating Tank A. (Correct?)

This surprized me. Clearly I am forgetting something, etc. etc. Blablabla.

What happens after round 3?
What am I missing? Maybe it turns around after a couple of rounds?
Or should I consider the total damage over X rounds?
Can someone help me get a complete list of dying chances, after X rounds?
Or can someone explain to me why there is such a difference?

MarkJindra's picture
Joined: 01/24/2014
Odds of hitting?

Are the odds an even chance of miss, hit, or hit twice for each cannon?


X3M's picture
Joined: 10/28/2013
Yes they are. 1/3th

Yes they are. 1/3th each.
That is why the tank with the double weapon has 9 outcomes each round.

Tomorrow, I will draw out the tree much further for observation.
In theory, both tanks should be equally effective against targets that have 75 health.

But then again, perhaps I have made a huge mistake in my calculations.

If someone can tell me what a good program is to use for these kind of tests?

debiant's picture
Joined: 05/03/2015
Have you tried multiplying

Have you tried multiplying the health by 10 or 100 and seeing if that effects the model at all? Just on a gut reaction, it seems that the 2nd tank would be more prone to having average attacks because of the additional randomization its attacks undergo.

MarkJindra's picture
Joined: 01/24/2014
I could be wrong but aren't

I could be wrong but aren't the odds of damage something like this

Tank 1
Less than 25: 33%
Equal to 25: 33%
Greater than 25: 33%

Tank 2
Less than 25: 44%
Equal to 25: 12%
Greater than 25: 44%

Tank 1 would have a 66% chance in 3 attacks of destroying Tank 2
Tank 2 would also have a 56% chance in 3 attacks to destroy Tank 1

You might be able to use google spreadsheets to put together a rudimentary random number generator

Joined: 08/02/2014
Let's discuss 2-round death

Let's discuss 2-round death because that I think is your major concern.

I'll start by confirming your statistical analysis:

For tank A, the possible damage for each round are 0, 25, 50 with equal chances (1/3).
If you want to kill tank B in 2 rounds, there are 3 options: 25+50, 50+25, 50+50. So your chance is 3/9=33.33%

For tank B, the possible damage for each round are 0, 9, 16, 18, 25, 32, 34, 41 or 50 with equal chances (1/9). If you want to kill tank A in 2 rounds, there are 11 options: 25+50, 50+25, 32+50, 50+32, 34+50, 50+34, 34+41, 41+34, 41+50, 50+41, 50+50. So your chance is 11/81=13.58%

Now, the reason for such a major difference is that tank B is more likely to perform an "average" damage than a "critical" hit. And because of your health and damage value, it still relies heavily on scoring a 50 during one attack in order to achieve 2-round death. As you can see from the above analysis of tank B, the only way to avoid needing a 50 damage is 34+41. Yet the chance of scoring 50 damage for tank B is only 1/9 in comparison to tank A's 1/3.

My suggestion is that you raise the health to 80 or change the split damage to 10+18. Otherwise, tank B wouldn't really have much of a chance to see round X (X>=3).

X3M's picture
Joined: 10/28/2013
Who would have thought

Who would have thought, that splitting up a cannon. Would have this side effect. I have taken some time in calculating other effects. And summarized the most important parts. (No need for a TLDR 12 page calculations, right?)

The big picture:
In other situations (dealing with smaller units with health 3, 12, 27 and 48 (tier 1 to 4)), Tank B would be a better option.
While Tank A would be the best option when dealing with 147 health or more (tier 7 and beyond).
There are other Tanks for dealing with other tiers in a most optimal way.

However, these 2 tanks should be on equal terms if there was no randomness. I think we all agree on that.
Thus expanding to infinite health for both tanks. Both tanks are equal. Tank A simply has luck on his side when dealing with tier 5. And I know that having a weapon that has more than one projectile, is better in the long run. Tank B has better weapons when you can spend XP on both tanks. And as a plus bonus, tank B also can kill smaller units in just 1 round.

In short:
A vs B, A wins.
After XP assigning when both have killed other units.
B wins.
Vs Smaller, B wins (faster than usual)
Vs Bigger, A wins (in a normal rate)

These are the in game value's for both tanks:
Tank A vs;
3 health = 1
12 health = 4
27 health = 9
48 health = 16
75 health = 25
108 health = 36
147 health = 49
192 health and beyond = 49

Tank B vs;
3 health = 1+1
12 health = 4+4
27 health = 9+9
48 health = 9+16
75 health and beyond = 9+16

If you calculate for both tanks dealing with a tier 4 unit. You will see that Tank B is victorious.
And tier 3 or lower, B will have 1 round victories enabled.


A comparable situation would be:
Trooper A with 12 health does 4 damage
Trooper B with 12 health does 1+1+1+1 damage

Only here, there is only 1 lower tier to look at.


After doing some calculations.
When playing 4 rounds or more. Fighting targets that do not fight back, with 75 health. The difference between A and B is only 0,1%.

Further, XP spending for A is 75% higher then B. While the bonus is only 5,4% better.

I am once again satisfied. And both Tanks may remain in the game. A valid strategy is keeping B alive, such that it will be a better option than A after 3 rounds.

Of course there is a balance between the 2. For this we need "infinite" rounds to consider (simulation). Where of course dying means, not being able to die in the next round. Therefore, the chance of dying in the third round is depending on the survival of the opponent in the second round.

Who can make a simulation for this? Or explain the spreadsheet? Perhaps I can make one for other situations as well.

Joined: 05/05/2014
If it is balanced through

If it is balanced through more variables as experience then it maybe ok.

Now, by example, I will see (trough testing) if using the 2 extreme strategies is mandatory... I will see how much freedom is still in the game to have more than 2 strategies.

But this is complicated. It seems that you need to play the game A LOT to notice these things, surely they are more variables that we dont know.

But, as I said, it seems ok by now in the way that you describe the process.

P.S.: yeah, dont kill me please by my "personal" version of english xD

Joined: 03/02/2014
X3M wrote:Who would have

X3M wrote:
Who would have thought, ...

Well, me, for one. :)

You're making the bell curve taller and narrower, and then the only thing you're measuring is how often they land in the leading edge of the curve. Certainly the one with the shorter, wider curve will have more results at +2 sigmas or higher. If you measure two teams fighting each other to the finish, you'll probably see it even out quite a bit.

There's one effect you usually see from splitting up the attacks (for the same total) is flexibility -- less wasted points. (That is, assuming the one with the split attacks is allowed to retarget between them if his first attack killed one opponent.) However, with the hit points exactly 75, and the attacks at 0-25-50, they never waste points if they are allowed to split their 2-hit turns. However, I'm assuming that's not the case, that it is a single hit for double points, but even so, I suspect that they waste less points overall from the team that almost always is wasting points on every kill.

You'd really see this if you give all tanks 76 hit points instead of 75. Suddenly, the side where the tanks get two smaller attacks would consistently crush the one with one larger attack. The larger attack would waste 24 or 49 points on every kill, whereas the team with the split attacks would rarely waste more then 17.

X3M's picture
Joined: 10/28/2013
I already played a lot

Its just that this comparisson is relatively new to us. And a surprize :D (I love surprizes)

Yeah, a double hit would mean that 1 hit goes to the next target if possible.

There is also the risk for overkill.
Let's say, we did 9+9+9+9+9+9+16=70 damage already, what remains is 5 health. 9 and 16 are both overkill on this one. With 9 and 16, this overkill is abundand too.

Sure there are more factors that are concerned. But for this little test. Speed, range and other units, where ruled out. Just to see the difference between 9+16 and 25.


If you measure two teams fighting each other to the finish, you'll probably see it even out quite a bit.

I have tested against a supply of walls. Yes, the difference is very very small after a while.

But this happens in a fight:
Tank A is Lucky in the first 2 rounds, then it evens out. But, Tank B could be dead by then. Thus Tank B cannot even out with counter luck.
With more units however, this could be evened out for a few rounds. Then you clearly see that Tank A starts to get the upper hand.

Once an unit gets the upper hand. It is over, unless you break through to another unit type. This other unit type is not the case in this test.

What I need is to learn to make simulation programs. Now, I am prety good with my TI92+. But it is damn slow when using random. So, can any one suggest something similar, but on the pc?

I want to simulate a battle over and over. Only then we get accurate numbers without me getting a table of (6units x 3x3random x 6units x 3random)^Rounds
Each round has 972 outcomes with a 6 vs 6 battle.

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