Unity of Command Forums

I recently ran into a situation where the game showed me the enemy unit had a 100% retreat chance. I attacked it, but it didn't retreat(even though it had retreat options). Which made me have the following question:

When the game shows retreat percentages, does the game take into account the projected chances of retreat on an average roll only? -And does that damage roll affect retreat chances - (retreat rolls are perhaps done after damage rolls)? In other words if you get bad luck on your damage roll and/or de-entrenching roll can it be that the actual retreat chances are lower - or higher on a lucky hit - than projected in the estimates screen due to the bad attack?

Thanks!

NightPhoenix

To quote the manual (section 3.3. Combat - Retreat and Overrun):

To calculate the retreat probability, the odds number is modified by the above, retreat-specific shifts and again randomized. The retreat probability is then looked up from the retreat table and a retreat roll is taken.

That means that 100% retreat does not really mean 100% - it's just the central value.

What are you saying, that this feels misleading?

We've never had a problem with that before, but with UoC 2 rules on cumulative retreat shifts, these "100% retreat" predictions appear much more often than in UoC1.

Hey, thanks for the response. Don't really mind or think it's unfair. You don't really notice if you have a number below 100% projected either. It's just a hidden number. But at 100%, you'll have this "what just happened?" kind of moment. Just wanted to have that cleared up since that never happened to me before. And i'd like to be 100% clear on the rules. I'm just curious then in how far the randomization go and how i'm supposed to read these odds.

Taking the retreat table -> say that i'm on retreat shift 7, if the randomisation can be max +1/-1 it would mean: The enemy retreats but there is a 33% chance the shift will be -1 -> leading to a 90% retreat chance, thus it's a 3,33% it won't. The actual retreat chance is only 96,66% then or rounded up 97%. The same can happen the other way around too then if there is no projected retreat - shift 0 - there is still a 33% chance the shift goes up - and then you have a 10% chance for a retreat.

If the game were to tell me the retreat chance is 10% (shift 1) The odds are actually more in my favor: 33% of getting 0% - 33% of 10% and 33% - of 30% - So rather 13%

Am i understanding it correctly?

Yes, except the randomization is Gaussian, i.e. it can be literally any value, not just +/-1.

We are using a sigma of 1, so there is 68.3% probability of the randomized result being within +/-1 and 95.5% of being withing +/-2.

I see, thanks for taking the time to clearing this up!

This is somewhat counterintuitive. In a general probabilistic model, if you have a 100% odd of something happening, that means that whatever random number you draw, you should have that outcome. If there are random draws which make it such that the unit will not withdraw, then you can't have 100% odds. The actual probability that the unit will retreat should be 1 less all random draws that make it not retreat. I guess the issue comes from the fact that you are showing the odds of the unrandomized roll despite the fact that your probability distribution's E[x] does not equal 100%. And it's screwing with our simple monkey brains

People can understand that a +3 shift in your favour means you should kill on average 3 steps plus or minus X depending on the roll (numbers are for illustration purposes only). So showing the outcome as 3 steps destroyed if the randomized roll doesn't apply any change is alright. However, in situations with binary outcomes (retreat/not retreat), you should really show the true probability of that event (i.e. E[x]). Otherwise, when you are on the edges (0% or 100% at your current shift pre-randomization), you will get non-intuitive outcome given that most people can't really understand that it's 100% plus or minus X% (especially since 100% plus X% doesn't mean anything)

Not that it's a priority, but I do like my odds to be the right ones

I think the simplest solution is that we change the topmost three retreat columns to retreat chances that are something like 95%, 98% and 99% (instead of 100%, 100%, 100%).

I think exact chances can be worked out based on the gaussian, but just by virtue of these numbers never being exactly 100% you won't get into the "broken" situation where the probability is 100% yet the defending unit does not retreat.