Wednesday, 26 September 2012

Probability for Wargamers 3 - Loot that Farm

Let's go back a couple of weeks to the last Dux Britanniarum battle Andy and I fought. Chris Stoesen commented on the battle report that it was a shame I didn't get the loot, and I was prompted to wonder, just how hard is it to get the loot?

Basically, to get the loot, you need to roll a 6, so how many rolls is it typically going to take? Let's rephrase that as 'how many times do you have to roll before you have a better than even chance of having succeeded?'

Obviously, on the first roll, you have a 1/6 chance of success, and therefore a 5/6, or 83.33%, chance of failure. What's our chance of success on exactly the second roll?

Actually, that's pretty easy. We need to fail on the first roll, and succeed on the second. That tell-tale "and" hints that we should multiply, so that's 5/6 * 1/6, or 5/36, which works out to 13.89%. Our chance of succeeding in no more than two rolls, then, is the chance of succeeding on the first roll, PLUS the chance of succeeding on the second having failed on the first, as these are mutually exclusive events (yes they are - think about it). That's 1/6 + 5/36, or 11/36, which is 30.56%.

But actually, there's an easier way, because we know the odds of failing twice - that's the odds of failing on the first roll AND the second. 5/6 * 5/6, or 25/36, or 69.44%. So the odds of not failing twice (i.e. succeeding on one of the first two rolls) is one minus that - 30.56%, as before. By extension, the odds of succeeding in three rolls or less is one minus the odds of failing all three, or (1 - 5/6*5/6*5/6).

I have to admit, at this point I bunged it in a spreadsheet, and it turns out, after filling down a few rows, that for a chance of succeeding more than half the time, you only need to make four rolls - 51.77%, to be exact.

Right - homework for this post. The rules actually say that if you roll a 1, there is no loot to find, and you can stop rolling dice. How does that change the odds? How many dice do you need to roll for a better than 50% chance of success now? The best method of attack might be to work out the chance of getting to roll again next time, and proceeding from there.

The answer will be in the next post, and may surprise you. I suspect it may surprise Mr. Clarke, too :D

5 comments:

  1. Well, the way I worked it out, you can never have even a 50% chance of success no matter how many rolls you make (short of infinity, anyway).

    Actually it's worse than that, because your best shot is your first roll and even that's only a 1/6 chance. After that it's downhill all the way.

    I think you'd better break this gently to Richard ...

    Tede

    ReplyDelete
  2. First of all let's take the (hypothetical!) situation in which you have an unlimited number of tries. If on a 1 you keep rolling, and you stop only with a success, then the chance of success approaches certainty (As the number of throws tends to infinity, the probability of success - rolling a 6 - tends to 1).

    Adding the rider that rolling a 1 means automatic failure (so to speak) changes this completely. The Probability of success and the probability of failure are exactly equal (think about it) and hence precisely 50-50 or one-half, again taken over an indefinite number of rolls.

    The chances of rolling a 6 in a limited number of rolls is changed because you can just about ignore the 1s
    The probability of rolling a 6 in 4 rolls, say, falls below 50%:
    P(S4) = P(roll 6 on first go) + P(roll 6 on 2nd go GIVEN YOU GET A 2ND GO) + P(roll 6 on 3rd go GIVEN you get a 3rd go) + P(roll 4 on 4th go GIVEN you get a 4th go)
    = 1/6 + 1/6.4/6 + 1/6.4/6.4/6 + 1/6.4/6.4/6.4/6
    = (216 + 144 + 96 + 64)/1296
    = 520/1296
    = 40%.
    The probability that you discover there is no treasure to be found (in 4 tries) is the same: 520/1296.
    So the chances that you are still looking after 4 die rolls is (4/6)^4 = 256/1296. This is a useful check to determine the validity of our results: 520+520+256 = 1296.
    Cheers,
    Ion

    ReplyDelete
  3. Just answer the question explicitly: how many rolls to ensure more than a 50% chance of success? The chances will never be greater than 50-50, and the even chance itself assumes an infinite number of tries. Given a finite number of rolls, the chances are less than 50-50 of finding any treasure.

    Looking at this from the practicalities of scenario design, the addition of automatic failure on a 1 is not a very good feature in my view. What it means is that there is less than a 50-50 chance that ANYONE will find the treasure. That's the sort of thing that can really get up the nose of a wargamer...

    ReplyDelete
  4. Me again... After spending the minute or two's thought I ought to have put in BEFORE writing that last paragraph, I realise I may have mis-spoken.

    My discussion assumed there is but one treasure to be found. It also assumes that a roll of 1 means there is no treasure to be found at all, everyone can stop looking. But if the 1 is 'player dependent' i.e. tells the player he is never going to find anything, but another player might, then what I said in my first comment applies to each individually.

    What one learns from this is just how tricky probability problems can become.

    ReplyDelete
  5. Quite so. Using the sum of a GP we can show that p(finding loot) → 0.5 as n(umber of tries) → infinity.

    For a finite number of tries n, p(finding loot) = 1/2 * (1 - (4/6)^n)). This is of course the same as p(there is no loot).

    ReplyDelete

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