So, if you remember from last time, we introduced the concept of combinations of N things taken K at a time, and used this to work out the probability of rolling (for example) a given number of ones on 5d6, which (as anyone who's played the game by now will realise) is useful to know in Chain of Command.
Reproducing our table from last time: the odds of rolling K of a given number on 5 dice:
Useful. Taking this, we can figure out a few handy probabilities:
Chance of getting two phases in a row?
You need 2 or 3 sixes: so that's about 19.5%. Let's call that one chance in 5 for the sake of the next one.
Number of phases before you get two in a row?
If you remember the article on looting rolls in Dux Brit, you should be able to figure this out.
Chance of not getting a second phase = roughly 4/5 each phase.
Chance of failing to get one twice in a row = 4/5 * 4/5 = 16/25 = 64%
Chance of failing to get one THREE times in a row = 64/125 = just over 51%
To get that below 10%, i.e. to be 90% confident of having had an extra phase (which if you remember, is what we discussed in a prior post as maybe being acceptable odds), you need to roll TEN times. Moral? Don't bank on that extra phase, but welcome it when it shows up.
Chance of activating a team?
Obviously enough, you need to roll a 1. Odds of rolling at least 1 one = 1 - (odds of failing to roll any 1s). So just under 60%. Again? Less than our desireable odds, so don't bet the farm on it.
Chance of activating a team with a leader present?
Let's say it's a MG42 team with a Junior Leader. You need either a 1 or a 3, i.e. 1 - (odds of failing to roll any 1s or 3s).
We'll have to calculate this one: we have 4 chances of 6 on any one dice of NOT rolling a 1 or 3, and 5 dice, so that's (2/3 * 2/3 * 2/3 * 2/3 * 2/3), which is 13.2%, giving us odds of 86.8%. Which is probably close enough, and demonstrates why leaders are really useful in Chain of Command.
How long will it take me to get a Chain of Command dice?
For now let's leave out the odd results, and concentrate on rolling 5s. How many 5s, on average, will we roll on 5d6?
Let's calculate the expected value. We have a 40.2% chance of 0, a 40.2% chance of 1... and so one. The expected number is thus (0.402 * 0 + 0.402 * 1 + 0.161 * 2 + 0.032 * 3 + 0.003 * 4 + 0.0001 * 5), or 0.83 fives. To get the 6 fives we need will take on average 6/0.83 turns, or just over 7.
Number of phases before the turn ends?
For this we need three or more sixes. Odds of this on any one turn are about 3.5%. If you really want to know, this makes a turn 50% likely to end after 20 phases. If you want 90% certainty? SIXTY FIVE phases. Don't wait for it, you'll have about NINE Chain of Command dice by then on average :D