Statistics help

[Battlingmaxo]

I was messing around with one of the battle calculators and came across some discrepancies but it may be my lack of statistics acumen. This is the situation:

A 1 rd strafe w/ 3infantry should yield a hit  1/2 times  however the calc says 42%

with 6 inf it should be successful all the time …calc says 67%

12 inf should guarantee 2 hits… calc says 88% for 1 hit

meanwhile if you run these strafes against 1 defending inf the odds of losing an attacking unit are 33% which is correct. The only scenario where you get the obvious predicted rate  for the attacker is in a 1 v 1.  Any thoughts on this.

[HolKann]

Any thoughts on this.

Yes.

You’re confusing “average” and “expected”. The average amount of hits for 6 inf attacking is 1 hit. The expected percentage you’ll get an amount of hits equal or greater to 1 is 67% (=1-(5/6)^6). The difference is in “or greater”. There is a possibility you get 6 hits with 6 inf attacking, but it is very small (1/6^6). Likewise, there is a possibility you get 5 hits, which is a bit bigger, and a possibility you get 4 hits, and so on. On average, with 6 inf attacking, you will get as much hits extra (so greater than 1), as you will get no hits. That’s why it averages out to 1.

The only time “expected” and “average” are the same, is when you’re rolling with 1 unit: then there is only the chance of scoring one hit and the chance of scoring no hits, so no “averaging out”. Hence your experiences: the more units, the less “average == expected”.

[Battlingmaxo]

Thanks. so the way to figure it out is to assume 1 for the event happening  minus the probabilities of the event not happening.

so a 2 man strafe would be 1-(.83x.83)=30.5%

[Guest]

Yes think about it, to expect 100% success with 6 infantry attacking would be like thinking that if you roll 6 dice, one of them will always have a one.

[Guest]

Here is a good explanation of it:

In one roll, the probability of rolling a 1 is 1/6.

For two rolls, there is a 1/6 probability of rolling a one on the first roll.
If this occurs, we’ve satisfied our condition. There is a 5/6 probability
that the first roll is not a 1. In that case, we need to see if the second
roll is a 1. The probability of the second roll being a 1 is 1/6, so our
overall probability is 1/6 + (5/6)*(1/6) = 11/36. Why did I multiply the
second 1/6 by 5/6? Because I only need to consider the 5/6 of the time that
the first roll wasn’t a 1. As you can see the probability is slightly less
than 2/6.

For three rolls, there is a 1/6 probability of rolling a one on the first
roll. There is a 5/6 probability that the first roll is not a 1. In that
case, we need to see if the second roll is a 1. The probability of the second
roll being a 1 is 1/6, giving us a probability of 11/36. There is a 25/36
probability that neither of the first two rolls was a 1. In that case, we
need to see if the third roll is a 1. The probability of the third roll being
a 1 is 1/6, giving us a probability of 1/6 + (5/6)(1/6) + (25/36)(1/6) =
91/216. Again, this is less than 3/6.

The general formula for rolling at least one 1 in n rolls is 1 - (5/6)^n.

Hope this helps.

[Battlingmaxo]

The general formula does not jive with what you wrote in the paragraph before but does agree with the results on the calculator. I interpet the the formula as:

1 minus the probability of the event not occurring = the probability of the event occurring.

Intuitively I follow what you discuss in the preceeding paragraph and that simply put is the way to calculate the event happening.

Thanks for the illustration BD

[Guest]

I see. You want an explanation for the formula. It basically boils down to this:

In the first roll, you have a 5/6 chance of failure. Total probablilty of success is 1 - 5/6 = 1/6.

On second roll, of the 5/6 that failed in the first roll, you have a 5/6 chance of failure again.
5/6 of 5/6 = 5/6*5/6 = 5/6^2 = 25/36. Total probability of success is 1 - 25/36 = 11/36

On the third roll of the 5/6^2 that failed, you have a 5/6 chance of failing yet again.
5/6*5/6^2 = 5/6^3……

I think you get the idea.

[Battlingmaxo]

I get it…sorry to say I was a math major but I’m far removed from those days.  In the not too distant past I would say +1 to you.

[Guest]

@Battlingmaxo:

I get it…sorry to say I was a math major but I’m far removed from those days.  In the not too distant past I would say +1 to you.

We have something in common, I was also a math major, I have always loved math and as a result I still remember quite a lot of it, and I love trying to explain math in a way that others can understand. I find probability particularly fascinating.

[losttribe04]

I love this topic but I have a quick question…

Why doesn’t the math work both ways?  What I mean by that is if you steepen the curve with your attack hits to get the “average” hits instead of the “expected” hits why doesn’t the same rule apply when calculating the enemy defense hits?

That would make your estimate a lot more conservative.

LT

[Wilson2]

@LT04:

I love this topic but I have a quick question…

Why doesn’t the math work both ways?  What I mean by that is if you steepen the curve with your attack hits to get the “average” hits instead of the “expected” hits why doesn’t the same rule apply when calculating the enemy defense hits?

That would make your estimate a lot more conservative.

LT

I does do the same with defense hits. It is just that in the scenario he mentioned there was only 1 unit so the odds are simple - 33%. If there were more units in defense you would have to do the same thing. However, generally there are more attacking units than defending units, because the attacker would not do a battle that he would lose. So, this makes the number of hits the defender gets roughly the average since the attacker almost always has => the defender units.