So this is something I'm working on and I'm generally interested in feedback and mistakes and supersticion. I'll also have to look for a way to make it prettier however, I'm working with a keyboard so what can I do?
If it doesn't interest you I recommend just not reading it. It's pretty lenghty, lol.
As described this is going to be the first part of a three parter. I have several times attemted to write a guide about startegy but alsways realised that it is futile to anyone who doesn't get the odds. So this is my take on helping people to get the basics who well... don't get the basics.



A Bloodbowl guide by Wreckage
This piece is designed to teach you bloodbowl quickly and to play it to a very high level.

It will consist of 3 parts that are most cruitial to success in the game.

Part 1: Probabilities
Part 2: In Game Strategy
Offense/Defense/Re-Rolls/Time Management
Part 3: Events
Set Up/Kick-Off-Events/Inducements/Team Management


Part 1
Bloodbowl can be played simply on emphasis and a coach with plenty of experience can obviously get pretty good at it. However the idea of teaching something is the idea to circumvent the process of trial and error and to let you succeed right away without trying first.
To be able to do that and also become a really good coach understanding probabilities is critical.
Because this is a guide and this is probably nowhere else explained like this, I’ll try to teach you the actual math behind bloodbowl teach you the concept of probabilities altogether.
If you do know about probability but have difficulties to apply them to bloodbowl I recommend reading this aswell.

Very basic
Most of the odds you are dealing with are pretty simple, related to d6.
In dices we assume the outcome of every roll to be equally likely. In case of a d6 the possible outcomes are 1, 2, 3, 4, 5, 6. There are six possible outcomes each has a chance of 1/6.

Basic
The same applies to the block dice, they have Skull, Both down, Push, Push, Defender Stumbles, Defender Down. Each has a probability of 1/6. Obviously the probability to get either of the two push results is 2/6 wich equates 1/3.

The reason why I write 2/6 instead of 1/3 is because we want probabilities we can compare, we don’t really want to calculate the least common multiple but rather use a mid high multiple that is easy to calculate and often used in bloodbowl.
Because the majority of more complex actions requires 2 dice, I like to calculate everything in a 36 divident, rather than a 6 divident. For the case of single die actions that would mean we have to multiply everything with 6 to get a 36 divident. For example 1/6 = 6/36. Or: 2/6=12/36. So a push result on a d6 equates a 12/36 chance.

2 Dice Basics
When dealing with 1d6 rolls you may have noticed how I just added the probabilities of the two 1/6 push chances together to get a 2/6 chance but you can’t just add the odds you get from 2 different dice, because they don’t exist in a universe where only 6 possible outcomes exist., there are now 36 possible outcomes. It’s an entirely different reality and you have to be aware of that reality if you want to calculate odds properly.
It’s actually pretty simple, you get the foundation of every probability when you just lay the possible outcomes down. For 2 d6 those are:
11, 12, 13, 14, 15, 16,
21, 22, 23, 24, 25, 26,
31, 32, 33, 34, 35, 36,
41, 42, 43, 44, 45, 46,
51, 52, 53, 54, 55, 56,
61, 62, 63, 64, 65, 66

In other words it’s six times six possible outcomes now. The designers made a conscious choice to make us add the eyes of the rolls together. Some different eyes added together will produce the same result. For example 46 equates 64 equates 55.
What is important here is that you understand that you don’t live in a world of 12 possible outcomes but in a world of 36.
Because a very high result with 2 dice requires for both dice to be very high while a middle result could have one of the two dice to be low and one of the two dice to be high in different variations a very high result is very unlikely. The same applies to a very low result. For instance getting an added result of 12 can only be archieved by getting the 66 on the scale of possible outcomes. Ie: The probability to roll a 12 is 1/36.
For the added result of 11 there are two variants: The 56 result and the 65 result. Two results equate a 2/36 chance.
For the added result of 10 there are 3 variants as mentioned above. That would equate a 3/36 chance.
For the added result of 9 there are 4 variants with 45, 54, 63 and 36.
8 has 5 variants and 7 the most middle added number finally climaxes with 6 possible variations in 16, 61, 25,52,34,43. The 7 is special because it is also the one number on a 2dice roll that is equally likely to get to any roll made with a single d6.
Unfortunatly when we move away from the middle there aren’t that many variations anymore that can form the lower numbers mirroring the effects of high numbers, the added number of 6 has only 5 possible variants, 5 has 4 variants, 4 has 3, 3 has 2 variants and 2 has only 1 variant, the double 1.

Applying 2d6 to bloodbowl
Fortunatly there is a pattern for us that is very easy to spot. With every number closer to the middle the next roll is more likely by +1. This helps us greatly if we wan’t to compare the effects a change in the armor or the injury roll has for example.
Because the injury roll is static lets have a look at what odds we are dealing with here:
To cause an injury we need to roll a 10+ on the injury roll. That equates the possible outcomes to get the added result of either a 12, 11 or 10.
All we have to do now is to add the odds for those different outcomes together. Because they all are happening as part of the same possible 36 outcomes we look at it and get:
1(result 12)+2(result 11)+3(result 10) = 6 out of 36 outcomes. In other words there are 6/36 variants to get a 10+.
Now it’ pretty simple from here to figure out the chance for a KO, a 8+ roll. We just have to continue adding numbers down the line: 1+2+3+4+5= 15. (1+2+3 represents here the chance for a 10+ result, the outcome 9 is represented by adding 4 more variants, the outcome 8 is represented by adding the +5 more variants in the end. )
So there are 15 out of 36 results to get you on 8 or above equating a 15/36 chance. In comparison a 50% chance would equate 18/36.
Rolling a 7+ has 21/36 variants but be carefull from there because now you have to add lower numbers to the equation again.
With 6+ 26/36, 5+ 30/36, 4+ 33/36 and so on.

How do single dice rolls with a reroll equate to 2 dice rolls?
Those are pretty much the same. You could take the chart of 36 different outcomes and just count the outcomes that would make you succeed a roll and you would have an accurate result.
But it’s easier to go on about it differently.
Generally we want for our brain as little work as possible. That’s why when we have 2 dice we will only look at the first dice result and see if it gives us the desired outcome we want. A 2+ roll has 5 possible outcomes that make it succeed. a 2, 3, 4, 5, 6 equating 5/6 or also 30/36 if you multiply it with six on both sides.
So when I have to make a 2+ roll I will first calculate my chance for making the first roll. Wich is 5/6 and tell myself: Well, if this roll succeeds everything will be fine. It doesn’t matter how the second roll goes, or to be precise in 100% of the possible results of the second roll the entire result will still be successful.
However there is also a chance I won’t succeed the first time So I first look at the likeliness for this to happen, wich is 1/6, ie. What’s left on the d6 after the other 5/6 are used up. Now I look at my second roll and ask myself: How likely is it to succeed this time? It’ s 5/6 again but only for the case of the 1/6 failure in the first roll.
So I multiply those odds: 5/6*1/6= 5/36 (I multiply the divisor and the dividend separately but you really should know that.)
Now I have the scenario of 30/36 outcomes to succeed on the first roll and the scenario out of the remaining six outcomes with 5 give me a success.
I add them together and get a total chance of 35/36., without having to count everything on the formular sheet.
Now those 35/36 are a lot better odds than 30/36 as you can probably guess but in cases like these there is an even easier way to figure the odds out. Rather then looking at the single roll you need to succeed at it’s always easier to look at a row of actions wich ALL need to succeed/fail to get a desired outcome. In other words in this case where I have to succeed either 2+ roll to make it, I have conversely to fail every single time to fail it altogether.
Now that’s simple. The chance for the first roll to fail is 1/6 AND the chance for the second roll to fail is 1/6. A little mind helper says: AND means you multiply, OR means you add. This applies to consequtive dice rolls you have to make anyways.
Making 1/6 AND 1/6 would be then 1/6*1/6 = 1/36. Ie: The first roll needs to fail and in this case the second roll needs to fail and that’s the only case there is. There is no other case where it would turn out like this too.
If you know the fail chance you always also know the success chance (unless some sort of tie is possible). Ie. If 1/36 is the failchance, 35/36 has to be the success chance or to write it mathematically:1 is 100% or 36/36. 36/36 – 1/36= 35/36.
Because the reroll in bloodbowl is generally the same as the initial roll you usually can just square them to get a fail chance.
For example:
2+ reroll  1/6 squared = 1/36 -
3+ reroll -> 2/6 squared = 4/36
4+ reroll -> 3/6 squared = 9/36
5+ reroll -> 4/6 squared = 16/36
6+ reroll -> 5/6 squared = 25/36
The counterchance in those cases you get by simply subtracting those numbers from 36/36.

Comparing odds
Now here we are really at the essence of why understanding all those odds is necessary.
You could calculate the odds in decimal numbers but they are no good for you that way because you need a calculator to do it or at least time. But you don’t have that in a game, you don’t want to have all those odds in your head either, so the easiest way is really to be able to calculate them quickly to be able to compare them.
Just because you know the likeliness of some actions doesn’t help you anything. It only helps you if you know wether or not it is more likely than another alternative action.
A common example is the question what is better: To be able to reroll a roll or to get +1 on the result. (Accurate or Pass dilemma). The truth is: The better odds give you a better chance on the reroll too but they are less likely to come in effect. So lets have a look.
A 2+ roll is 5/6 =30/36. A 3+ reroll roll is the counterchance of 4/36 -> 32/36. Now that’s a pretty neat extra 2 out of 6 36ses there.
3+ -> 4/6 -> 24/36 compared to 4+ rr -> CC 9/36 ->27/36.. We see again the reroll chance is a little better, shows us just how much a reroll really is worth.
4+ -> 3/6 ->18/36 compared to 5+ rr -> CC 16/36 -> 20/36
5+ -> 2/6 -> 12/36 compared to 6+ rr -> CC 25/36 -> 11/36. Here you see a breaking point from what you have seen before, because the 6+ roll with only half the base chance of the 5+ roll is just too unlikely to even the odds even with a reroll.

Now looking at this you might be tempted to always take the reroll skill over the +1 skill but that would often be a mistake. The problem is that, while most of these odds are pretty decent, in Bloodbowl we have to rely on our rolls as good as possible. A reroll skill is better if you don’t have team rerolls at your disposal to do the same thing. But almost all rolls in bloodbowl are of critical importance. The advantage of a +1 skill is that it works in addition to a possible team reroll, giving you combined a much better effect. For instance the 35/36 chance of a 2+ reroll roll rather than a unreliable 32/36 chance 3+ reroll roll (Of course because you can only use it once per turn that isn’t entirely true but that would go out of hand to explain that. Lets just say you don’t fail as much as you think. But if you do you should better have a plan.)
The reroll skills are good but dead ends, you get the maximum out, you can get.. It means you plan to burn through your rerolls and will be short handed anyways. (Wich is fine, just keep the dilemma in mind.)

Block Dice
As mentioned in the beginning block dice are just some normal dice with a different set of symbols. There could be numbers on them it wouldn’t make a difference. Because the regular case is that you can choose among 2 block dice it requires a moment of thought how this can be translated into odds. The easiest way is to do it the same way you would look at a roll you can reroll. In other words you look at the first dice result and you have two scenarios: Either it gives you the desired outcome or it doesn’t. Then you look at the second d6. Assuming you aren’t greedy and don’t target the ballcarrier your main goal on a block will be not to cause a turnover for yourself. A pow is nice but seriously there is lots you can do later if you just don’t get knocked down yourself. You get knocked down means it’s over.
There are generally two results that knock you down wich are the skull and the both down result, meaning an unskilled player fails 2/6 of his 1 dice rolls, or every 3rd roll if you prefer that. With 2 dice that means we have an equal fail chance to a 3+ reroll with 4/36 scenarios that will go south.
In comparison, if you get the Block/Wrestle skill, a Both Down will be harmless to you. Instead of a 2/6 failchance per dice it’s now a 1/6 chance. That means just by using Block you have doubled your chances of success. If you translate this to two dice your failchance is 1/36 now instead of 4/36. You now have quadrupled your chances. You see how that equates to a 2+ roll instead of a 3+ roll.
An interesting implication we can learn from this is this:

If you block with 1 dice you fail 1/6. That means such a block has an equal chance to an ag4 dodge wich is also a 1/6 fail.
If you have no block, your failchance is 2/6 with a single dice. If you have ag3 your failchance is also 2/6. That means having ag4 on the dodge roll instead of ag3 is mirroring exactly the behavior of having or not having the Block skill on a block.
+ST on the other hand works in a similar way on blocks as the Dodge skill works on dodges. Having a ST4 player stand against a ST3 player will give him 2 dice. Assuming he has no block his fail chance would be now 4/36. If he was to attempt a dodge with the Dodge skill on ag3 his fail chance would also be 4/36.
If you have the block skill and the higher strengh your fail chance is 1/36. If you have ag4 and the dodge skill your fail chance is 1/36.
You can see how having the Block skill for blocks equates to having ag4 for dodges here while ST+ for block equates to having a reroll on a dodge. That's no coincidence. It's because those things have exactly the same number of sides on a d6.
There is one difference tough. As long as you have team rerolls in addition to everything you usually rather want to do the blocks, because they are rerollable, even if you already used 2 dice. Thats why they are preferable over everything. You can get much better odds than 5/6 and still use a reroll on it afterwards. That's nowhere else in the game allowed.


Calculating Rows of Actions
Being able to look at a single roll and tell what is more likely is a good first step but don’t neglect your intuition especially if more throws are involved. There are often cases where you have to make only one hard roll to get something. Alternativly you could make a medicorely difficult roll but you have to make a lot of high success chance rolls to get there. If you calculate odds here to schematically neglecting the low risk actions you might fail to see that there lies a much better shot with the one risky action.
To tell the truth you are still mostly better off taking the many rolls because if you have a turnover half through your turn that’s still better than if you go on all or nothing and fail your first action.
The stuff we are doing now you mostly be not able to calculate in your head. But seeing how it works might help you to estimate your chances roughly or allow you to calculate certain odds in advance f you wish so. Anyhow here is how you calculate Rows of Actions.
You already possess all the tools to do that. The AND OR principle is the key here (In case of AND you multiply, in case of OR you add.).
Let’s say you want to calculate a dodge (3+), two GFI and a Blitz (2d with Block skill).
We would go on about this like that:
You have to succeed:
A dodge AND a gfi AND a gfi AND a blockthrow OR a blockthrow.
2/3 * 5/6 * 5/6 *( 5/6+ (1/6*5/6) )
Now a couple of things need to be said to this. The 2/3 chance is the 3+ dodge. I would have described this before as 4/6 but if you actually want to multiply probabilities you are still better off finding the least common multiple (2/3). The two 5/6 are the GFI, easy to grasp. In the breaks we have 5/6 for the first block. Now I said we add the second block like stated within the AND OR principle. But you see that that isn’t exactly right, because I write there 1/6*5/6 instead of just 5/6.
I explained this before. If you add probabilities you can only add for the second scenario where the first scenario didn’t take place. The first scenario is 5/6 here for you to not turnover on the block. The counter chance for that is 1/6 that you fail and this is what you multiply to the second block (5/6) again. Hope you can follow me there. In short it means you can look at the two dice rolls as one even with entirely own probability within the events of OR statements.
That would look like this:
2/3 * 5/6 *5/6 * 35/36
Pretty likely that block not to screw you, eh? If we had rerolls that would be the point where we would have to apply them, but more about that in a minute.
Now we won’t be able to make this that much easier for us so lets grab a calculator and we roughly get to a 45% (0.45) chance of success. We can easily translate this into 36ses by just multiplying it with 36 and we get: 16.2/36, an easy number to compare.
Generally it isn’t really important to you how likely a whole row of actions is, it is only important wether or not it is more or less likely than another row of actions. Therefore it is advisable to just ignore any action you would do in either scenario.

Calculating Rows with Team-Rerolls
Now taking rolls out of the equation doesn’t really work when team rerolls are involved.
In fact you are far off what the average person can do in the head in a matter of seconds. Well it’s possible to do but why try so hard, right? Math is supposed to make things easier, not more complicated.
However if you wanted to calculate stuff, here is how you would do it:
Let’s have a look at our example from before:
a dodge (3+), two GFI and a Blitz (2d with Block skill).
2/3 * 5/6 *5/6 * 35/36
You could apply a team reroll to any of those four rolls if they were to fail. So this is how we go on about it. We know already the variant written up there gives us the 45% chance we will definatly succeed even without the reroll. Now we add a bunch of other scenarios.
The entire combined odds would look like this.
(Odds for everything to go right) AND ( only Dodge needs to be rerolled) AND (only first GFI needs to be rerolld) AND (only second GFI needs to be rerolld) AND (Blitz needs to be rerolled)
These are the four extra scenarios that can happen. Now if we look at the numbers it would look like this:
All goes right:: 2/3 * 5/6 *5/6 * 35/36
+
Dodge fails: (1/3*2/3) * 5/6 * 5/6 * 35/36
+
1st GFI fails: 2/3 * (1/6*5/6) * 5/6 *35/36
+
2nd GFI fails: 2/3 * 5/6 * (1/6*5/6) * 35/36
+
Blitz fails: 2/3* 5/6 * 5/6 * (1/36*35/36)
=
Probability to succeed with a team reroll

However we are in luck. Because we have a common base chance(=all goes right above) that is available in all scenarios and the only thing that changes is the failchance within it we get this simple formular.
2/3*25/36*35/36 ( 1 + 1/3 + 1/6 + 1/6 +1/36)
Or to write it abstract:
P(Base Chance) * ( P(All goes right= 1) + P(Dodge fails) + P(1st GFI fails) + P(2nd gfi fails)+ P(Blitz fails) )
And you get:
2/3 * 25/36 * 35/36 * 61/36 = 76.27% = 27.5/36
P in this context means basically: Probability of (action).
Anyways, I tried to explain it easy as I could. If you could grasp the concept to this point you have most of the necessary understanding about chances you need for this game. Anyhow all things I’m going to teach you about it for now.
Now you are ready to learn to the game!