## Saturday, 28 September 2013

### Probability for Wargamers 12 - Chain of Command revisited

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:

0: 40.2%
1: 40.2%
2: 16.1%
3: 3.2%
4: 0.3%
5: 0.01%

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

1. But you can add dice together to make up a 2, 3 or 4.

it makes the calculation quite a bit more difficult.

2. This comment has been removed by the author.

3. For example you want to work out the "chance of activating a team with a leader present".

The team itself can be activated if you roll one or more 1s with 5D6.

You can activate a Junior Leader by rolling any combination of dice that add up to three. That's one or more 3s, three rolls of 1, or one or more 2s, plus one or more 1s.

If there's a Senior Leader hanging about you can also activate the team with any combination of dice that add up to four.

It's all horribly complicated.

4. Possibly a little bit of clarification is called for here. You will note that in the posting, it was said that the number of '5's you could expect out of 5 rolls of a D6 (5xD6) was 83.3%, or 5/6 - what you might have guessed , there being 5 rolls at odds each time of 1/6.

But that is not the same as asking: what is the probability of rolling at least one '5' in 5 rolls? The posting tells you that there is a whisker more than 40% chance of getting none - which means your likelihood of rolling at least one five is the same whisker under 60% - not even close to 4 chances in 6, let alone 5.

How do you account for the difference? The clue lies in the phrase "at least one '5'". That you have an almost 20% chance of rolling more than one '5' accounts for the difference between the probability of rolling a '5' and the (statistically) expected number of '5's you will roll.

This can be proved by adding the products of (number of 5s rolled) and (probability of rolling that number). Thus:
0x0.42 + 1x0.42 + 2x0.16 + 3x0.03 + 4x0.003 + 5x0.0001
The first term is zero, of course; and we'll approximate by ignoring the last one as insignificant.
0.42 + 0.32 + 0.09 + 0.01 = 0.83 (the extra 0.00333... is taken up by rounding).

5. Got to love the stats :)
Didn't stop me rolling 5 blanks out of 5 dice (50% chance of a die being blank) 3 times in one game of something last night!

6. I'm one of the five people out of four, who have trouble with stats... :-/

7. You would think that rolling 'hits' at 50-50 propositions 5 times in a row as being 1/32, and doing it three times the cube of this: 1/65536. Actually, it's not so vast. These odds refer to a specific group of three rolls and the 4 following each.

In a given sequence of rolls you will very likely roll 'blanks' close to half the time. Given a lot of rolls, you are pretty near certain to roll 'blanks' at some point. Having rolled a blank, the probability that the next 4 are also blanks is 1/16. Which might suggest, then, that (given enough rolls), the odds of rolling 3 5-blank sequences in the course of an evening would be (1/16)^3 = 1/4096.

Well... Maybe. Roughly half the time you roll a blank, the next roll will also be a blank. Given a long enough sequence - a fist full of dice type game - this could easily happen 3 times during the game. The probability of 3 further 'blanks' to follow is 1/8 each time. That seems to imply that the odds of this happening 3 times is 1/512...

A pattern seems to be emerging here.

Suppose, during the course of an evening's play you rolled 100 dice, all of which had a 50-50 chance of rolling a 'blank'.
Then:
Expected number of blanks = 50
For each blank, there is a 50-50 chance of rolling a blank on the next roll
Expected number of 2 blanks in a row = 25
Expected number of 3 blanks in a row = 12 (truncating)
Expected number of 4 blanks in a row = 6
Expected number of 5 blanks in a row = 3

In any sequence of 100 rolls, then, you can expect a sequence of 5 'blanks' to appear about 3 times.

As Granny Weatherwax was wont to say: 'Million-to-one shots crop up nine times out of ten...'

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