Suppose you have one action with your (all stats 4) Striker, and need to basically run, pick up the ball, run, pass, catch the pass with another Striker and score on that action. What

*are*the odds?

Let's step through it.

To pick up the ball you need two successes (4,5,6) on four dice (remembering that 6's explode)... as it's late and I'm lazy, I'm availing myself of some superb work by BalrogBond on the Mantic Forums which you may need to read.

From the above, P(2 successes) = 29.1%

Now we have the ball: we run and throw with the free action we just got. It'll be a three dice roll assuming we got to short range - +1 dice for being a Striker, and -1 for having run. How many successes?

- Well, we have a 12.5% (1/2 * 1/2 * 1/2) chance of none at all.
- For each dice we have a 41.7% chance of one success, so that's 3 * (50% * 50% * 41.7%, or just a hair over 31.2% chance of one on three dice.
- for two it's a little hairer, as we can do it either by:
- one success on each of two dice (3 * 50% * 41.7% * 41.7%) = 26% PLUS
- two successes on one dice (3 * 50% * 50% * 6.9%) = 5.2%
**making a total of**31.2%- for three successes there are several ways,
- one success on each dice = 7.2%
- three successes on one dice, 3 different ways = 3 * 1.2% * 50% * 50% = 0.9%
- two on one, one on another, six different ways = 6 * 6.9% * 41.7% * 50% = 8.6%
- f
**or a total of**16.7% - four? (my brain is starting to hurt).
- 4 on one dice = 0.2% * 50% * 50% = 0.05%
- 3 one one dice, one on another, six ways = 1.5% (look, just trust me!)
- 2 on two dice, 3 different ways = 0.7 %
- 2 on one, 1 on two, 3 different ways, = 1.1%
**for a total of**just over 3.3%

Let's stop there, as we have a long tail of 5+ successes.

The Dreadball throw/catch rules say we get as many dice to catch as we made throwing successes. We NEED two successes to get the final throw. because we need to double to get the free action.

- Two successes on one dice? = 9.3%
- on two dice? the easy way is (1 - the odds on one or none) = 33%
- on three dice? 56.3%
- four? 77.8%

So the odds of making the catch are (31.2% * 9.3% + 31.2% * 33% + 16.7% * 56.3% + 3.3% * 77.8%) = a smidgeon over 25% not counting the 5+ successes odds.

And our final throw is a three dice throw (+1 Striker, -1 small target). We only need one success, so that's an 87.5% chance ( 1 - the odds of none).

OK. Let's tot those up.

Final odds of scoring = 29.1% (pickup) * 25.1% (pass + catch) * 87.5% (strike).

About 6.4%. Wow. My dice were on yesterday!

So. Here's the question.

You have one coaching dice. Where do you use it?

I admit to some curiosity here (Sheldon Cooper moment, sorry). Supposing (I suspect incorrectly), that to test the success of an action, you have to roll 4 dice, of which there must be at least two 'hits' (scoring 4 or better) for it to be carried out (an interesting game mechanic). The Probablility of each number of 'hits' is:

ReplyDeleteP[0] = 6.25%

P[1] = 25.00%

P[2] = 37.50%

P[3] = 25.00%

P[4] = 6.25%

[Total the percentages, and you'll get 100].

The probability of scoring at least two =

P[2] + P[3] + P[4] = 68.75% - a little better than 2-to-1 in favour.

But you mentioned '6s explode, which seems to mean the catch is successful, but the ball explodes in your face. That does sound as though you really want only 4s and 5s. In that case (approximating to the nearest whole percentage number) your chances become:

P[0] = 20%

P[1] = 40%

P[2] = 30%

P[3] = 10%

P[4] = 1%

Your chance of a successful 'catch' are, then, about 41%.

Setting this aside, it is my belief that in a game like this, in which, having a certain allowance of action, you carry on until the allowance runs out, the probabilities become a little different. I'm not talking mathematically, here, but 'game-ically', which brings in psychology and decision-making.

Let's take a hypothetical situation in which every action you take has a 50-50 chance of coming off (except run). We are dicing for pickup, complete pass, successful catch on the return pass, touchdown.

At the start of the action the ball is loose on the ground, close to Blitzer Blotz, and it seems to us desirable that the ball be in our hands. Blotz is the closest player, we try him first, and he successfully retrieves the ball. Had he not we would have tried the next closest player.

Ball possession is a positive gain, and if the circumstances dictated we stop here, we would probably be reasonably satisfied. But we have plenty of allowance left, so we push our luck, run clear of close-by opponents and let rip a pass.

We are risking loss of possession here, but, aside from 'nothing venture nothing win' - and a successful enemy tackle will compromise possession, in any case - that the ball is so much closer to the opposing end zone is still of benefit to us, and our catcher might yet be able to pick up the loose ball anyhow. Success, then, will benefit us greatly; failure won't hurt us much. Of course, we'll go for it.

And so it goes. Each successive action will gain us a great deal; a failure will interrupt the sequence, but otherwise not harm us much except (possibly) to compromise our possession of the ball.

So a series of actions that have only a 6% chance of overall success is well worth undertaking if the benefit is great and the cost of failure negligible. It is especially true if once the first action is successfully complete, whatever happens improves the situation that existed at the beginning of the turn.

Having said all that, it is most satisfying when you pull off something heroically unlikely. I still remember, from over twenty years ago, my Orc Bloodbowl team, with just 5 guys still standing and time running out, to receive the kickoff against a full team that included a large monster... and scoring a touchdown with a single running play. We lost the game 2-1, but, man, seeing my star player, the running blitzer Adul Fitla score in the corner: that was just glorious!

Then the designers went and stuffed up the game...

Just and explanation of "sixes explode".

ReplyDeleteA six is a success, and allows another dice to be rolled - for an additional success, another six = another dice etc.

So in theory one six in the first throw can result in an infinite number of successful throws - 6 to 6 to 6 to 6 etc!! (Obviously at very low odds!!).

For me less of a Sheldon cooper moment and more of a Hari Sheldon moment!

DeleteThe war games equivalent of psycho-history? What would you call that, I wonder? Game Theory...?

Delete