Charts for Blood bowl

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Brownie_Monstar

Charts for Blood bowl

Post by Brownie_Monstar »

Hi,

After several years out of BB I'm going to start again.
As I recall it there is "a lot" of number crunshing and I would like to know if anyone got an overview chart with chances for Succes/Failure when you are blocking, tackling, passing and so forth with different skills including team reroll!

Blocking Example:
1 Block D, offence no skills, defence no skills: succes 67%, failure 33%.
1 Block D, offence Block, defence no skills: succes 83%, failure 17%.

Thanx in advance

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voyagers_uk
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Post by voyagers_uk »

Welcome,

glad to see you are going to start up again, which version did you play with?

As for Number Crunching you will be wwanting to speak with Galak, Mestari or one of the other guys who are good at that.

what is your favourite team, most succesful season etc?

again, Welcome... :smoking:

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Vesticle
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Post by Vesticle »

If you want, I can make one for you if you aren't good at numbers yourself. I've never made one yet, personally, because I tend to know the more common ones in my head, or I do a quick calculation of a specific situation on the spot, since there are so many different situations that can occur anyway.

Additionally, though I'm all for helping other people out, I do think that coaches should do some of the calculations of odds and such on their own. I think it gives more of a 'hands on' kind of feel to coaching, and makes them more familiar with their own team, and I think it makes them better coaches to work with the numbers themselves - I think the game loses some appeal and becomes a bit more impersonal if you're reading numbers of someone else's chart.

For the most part, the numbers are one of 3 things: a combination of D6 rolls, a combination of 2d6 rolls, or a combination of D8 rolls. Every number on a D6 is ~16.7% (6=16.7%, 5=33.3%, etc.). Block dice rolls I just consider D6 rolls, and translate 6 = Pow, 3,4 = pushback, etc. 2d6 rolls work in a "pyramid" format, the total denominator being 36, and the end numbers being worth 1, #7 being worth 6. So 2 = 1/36, 3 = (1+2)/36, 4 = (1+2+3)/36, etc. Then to combine results, say 1d6 with 1 re-roll or 2 block dice, the chance of *both* dice being successful (i.e. 2 dice, defender's choice) is the chance of 1 being successful squared (e.g. 4/6 * 4/6 = 16/36 or 44.4%) The chance of at least one dice being successful (i.e. catch with the catch skill) is 1 - the chance of failure, squared (e.g. 3+ to catch, 2/6 to fail once, 2/6 * 2/6, 1/9 to fail twice, 8/9 to catch the ball).

Anyway, hope that helps.

David

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Brownie_Monstar

Post by Brownie_Monstar »

Hi voyagers_uk,

here's a little about my BB history:

I used to play Imperial, the team formely known as Human, but I'll start with Orc's for several reasons.
A) I got the minatures, I dont want to spend my hard earned money if I don't want to play BB anayway
B) It's a hardhitting league
C) And i want to bash some heads :)

I played in a local league for 6 or 7 (?-93) seasons with the 2nd Edition (incl. Star Players and Companion rules) and was Champion all the season. Lost 1 game and 1 tie in about 45 games

In 94 I started to play 3rd ed. and then turn overs were introduced. Then I started to lose. But the game was more fun and way more balanced but not as bloody

But the league dissolved in 94 and I havent played on a regularly basis ever since

Thanks for the advice

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Brownie_Monstar

Post by Brownie_Monstar »

Hi Vesticle,

Though I havent got problems solving the math and the statistics I see your point
But I see no point in reinventing a chart somebody already has made - if he wants to share it.

The only problem - for me -is all the skills which affect each other one way or another which complicate such a chart. 'Cause there's come a lot of skills since I played last and the rules has also changed.

I'll do it the hardway :)

Thanks for the advice

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Mestari
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Probability counting

Post by Mestari »


Counting the outcome probabilities of single actions is elementary.
To be truly strong, One has to learn to count the probabilities for the different scenarios for his entire Turn so One can choose a likely triumphant scenario, or if need be, knowing the odds choose a more daring one.

Also, if One is able to position Ones players so that he can lead Ones Enemy to execute his Turn in a predictable fashion, One could calculate the odds of next Turns also...

And if One has a mind like Gari Kasparov's and One can predict his opponent exactly, One could count the probability of a drives outcome when the teams have been set up.


:wink:

A little more seriously:

A word of advise: In a game, probability counting is easy when you do it like this:
The probabilities are rational numbers, id est: X/Y. The denominator (Y in the previous example) in Blood Bowl probabilities is usually some power of 6. When counting probabilities of several consecutive dice rolls, you multiply all the numerators together, and all the denominators together. This is easier to count than if you change them instantly to percentages and use them.

Example:
Roll 1: X1/Y1 Roll2: X2/Y2 Roll 3:X3/Y3
Result: Probability is (X1*X2*X3) / (Y1*Y2*Y3)
In BB this would usually mean (X1*X2*X3) / 6^3

What I'm trying to say is: learn the powers of six by heart! 6, 36, 216, 1296, 7776 etc.
Then you'll always know the denominator. Then you'll only have to count the numerator, and the probability is counted a lot faster.

What about rerolls?
If a certain roll can be rerolled, count the probability as above, but invert the probability of the particular roll (i.e. use the probability that it would fail) and multiply the result again with the non-inverted probability to get the possibility of success in the event that you have to use a reroll. Then add this probability to the probability of success without using a reroll.

BB example: catch (3+), dodge(3+, Reroll), two sprints(2+).

Success without rerolls: 4*4*5*5 / 6^4 = 400/1296
Success with dodge reroll: 4*3*4*5*5 / 6^5 = 1200/7776

Add these together (remember that to add two rational numbers together they must have the same denominator, so multiply 400 by 6) to get
(2400 +1200)/7776 = 3600/7776
Even without doing the division, we see that the probability is slightly less than one half, division gives us the exact number of about 46.3% probability of success.

If you're not good at multiplying in your head, then you might want to divide such scenarios into parts, calculate the odds for the parts, round to a comfortable number and then calculate the total result. For example above we get
3+ 3+(RR) = 4*4/6^2 + 4*3*4/6^3 =144/216 = 2/3
2+ 2+ = 5*5/6^2 = 25/36 = about 7/10

So the total is 2*7 / 3*10 = 14/30 = 46.7%
Due to rounding, that's not an exact number, but it's close and perhaps easier to count. But if you use this, be aware of the fact that the more you round the more it effects the outcome.

What about if the counting gets complex: for example the reroll possibility of a later roll depends on whether you used a reroll earlier, you have multiple rerolls etc.?

Then (if you want to find out the probabilities) you can use what is called a probability tree:

You start drawing from the left. Every die roll is a node that spans two lines: success and failure. On the lines you mark the possibility of that happening (Line1prob+Line2prob = 1/1). Every line ends into an another die roll node or into an end node (that outcome leads to a TO for example). When you've drawn all the scenarios, you should have a tree starting from the left, dividing up into several branches with several die rolls in each. Note that the same roll usually appears in several branches.
Now you look at the endings of the branches and decide which of them are favourable. Then for each such branch, you take the route that leads up to it, and multiply all the probabilities on that route together.
When you've counted all the favourable routes, add them together to get the total probability for a favourable outcome.
Sounds complicated?
It isn't, but I'd need to draw it to make it appear simpler. The way we calculated reroll probabilities above is a simple example of using a probability tree.

And, as a student of mathematics at the university, it might be that even if I don't consider something complicated, it still might be :P

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Mestari
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Post by Mestari »

And one more thing:

I wouldn't consider charts too useful: after all - what comes to the probabilities of a single die roll, you'll soon remember them by heart. I'm more interested in the probabilities of entire actions and turns - and there wouldn't be much point for trying to devise a chart for such things...

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