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Hello dudes, I am terrible at maths (or math for my American hombres). Maybe you can help me out.

If I use Typhus who gets a 2+ save, a 5+ FNP and a 6+ roll on top if he takes the warlord trait, does that mean there is a 4% chance of a wound getting through?

I used this maths:

5/6 * 1/3 * 1/6 = .04 = 4%

Assuming no AP on the hit, yes.

ItsPug wrote:
Assuming no AP on the hit, yes.

Great thanks, this is handy so I can work out different AP levels.

Chance to wound after hit and wound rolls were successful should be: 1×1/6×2/3×5/6=10/108=9,25%.

Your approach is flawed because in the case of 2-5 on the first roll the other two rolls aren't made (as is respectively the case for the second roll). Unless i'm utterly confused of course... you would have to add steps up:
5/6+(1/6×1/3)+(1/6×2/3×1/6)=
90/108+6/108+2/108=98/108=90,74%
Which would add up with the above chance to take a wound to a sum of 100%

 Rippy wrote:
Hello dudes, I am terrible at maths (or math for my American hombres). Maybe you can help me out.

If I use Typhus who gets a 2+ save, a 5+ FNP and a 6+ roll on top if he takes the warlord trait, does that mean there is a 4% chance of a wound getting through?

I used this maths:

5/6 * 1/3 * 1/6 = .04 = 4%

Your individual probs are all reversed.
The probability of a wound going through a 2+ armor save is 1/6 (it only goes through on a roll of 1). Then 4/6=2/3 on the FNP (the wound remains on a roll of 1-4), and 5/6 on the warlord trait (only the 6 cancels it, so the other 5 rolls go through).
So Cpt. Icanus is right, it's 9.26%.

Okay cool, big thanks here guys.

I made (or am making this)

http://www.meltatotheface.com/wp-content/mathhammer.php

bit bare bones atm, give it a go it has basic functionality for standard stats. Will improve it for d3 damage, other ignore wound ruless, mortal wounds etc

if i'v messed it up let me know!

I still don't understand the maths behind it


Automatically Appended Next Post:

 Latro_ wrote:
I made (or am making this)

http://www.meltatotheface.com/wp-content/mathhammer.php

bit bare bones atm, give it a go it has basic functionality for standard stats. Will improve it for d3 damage, other ignore wound ruless, mortal wounds etc

if i'v messed it up let me know!

Thanks Latro, I will check it out.


Automatically Appended Next Post:
Looks like it needs feel no pain added.

aye its pretty basic i just whipped it up together this morning, if i get a chance i'll develop it a bit more

 Latro_ wrote:
aye its pretty basic i just whipped it up together this morning, if i get a chance i'll develop it a bit more

It's very handy though, good job

 Rippy wrote:
Hello dudes, I am terrible at maths (or math for my American hombres). Maybe you can help me out.

If I use Typhus who gets a 2+ save, a 5+ FNP and a 6+ roll on top if he takes the warlord trait, does that mean there is a 4% chance of a wound getting through?

I used this maths:

5/6 * 1/3 * 1/6 = .04 = 4%

The best formula is this...

(to hit) * (to wound) * (failed save) * (failed FNP) * (wounds from shot) = damage done per shot.

So a bolter from a marine looks like this..

2/3 * 1/3 * 1/6 * 2/3 = 2/81 or 2.46% chance of getting wounded from a marine

 Latro_ wrote:
I made (or am making this)

http://www.meltatotheface.com/wp-content/mathhammer.php

bit bare bones atm, give it a go it has basic functionality for standard stats. Will improve it for d3 damage, other ignore wound ruless, mortal wounds etc

if i'v messed it up let me know!

I LOVE IT . Do you know many trees are the victims because of my doing calculations on paper? Not many because I use a calc but still I love it. Good Job.

TBH, I am kind of surprised that no one has made an app for things like this. The calculations (and the code) isn't complicated but if you don't know how to do the math, it seems pretty insurmountable. I feel like there is definitely money to be made there.

Great, thanks Labmouse!

This actually seems like the right thread for this as I am also terrible at Mathhammer.

What's the probability of making a 9" charge with a re-roll (assuming a re-roll failed charge ability).

I know the initial roll is 5/18 or 27.78% (rounded), but for the life of me I don't know how to figure out what it becomes with a re-roll.

Audustum wrote:
This actually seems like the right thread for this as I am also terrible at Mathhammer.

What's the probability of making a 9" charge with a re-roll (assuming a re-roll failed charge ability).

I know the initial roll is 5/18 or 27.78% (rounded), but for the life of me I don't know how to figure out what it becomes with a re-roll.

To find a statistic for "does this happen at least once out of X number of chances," the best way is to find the chance of it not happening at all, then reverse it. For example using your rounded number, there's a 72.22% of not making that charge. So if you rolled twice, you'd multiply by itself for a result of 52.16 (also rounded.) That is the chance of not making the charge at all, and since a chance is always out of 100, 100-52.16 = 47.84 is your chance of at least one charge roll working out.

 niv-mizzet wrote:

Audustum wrote:
This actually seems like the right thread for this as I am also terrible at Mathhammer.

What's the probability of making a 9" charge with a re-roll (assuming a re-roll failed charge ability).

I know the initial roll is 5/18 or 27.78% (rounded), but for the life of me I don't know how to figure out what it becomes with a re-roll.

To find a statistic for "does this happen at least once out of X number of chances," the best way is to find the chance of it not happening at all, then reverse it. For example using your rounded number, there's a 72.22% of not making that charge. So if you rolled twice, you'd multiply by itself for a result of 52.16 (also rounded.) That is the chance of not making the charge at all, and since a chance is always out of 100, 100-52.16 = 47.84 is your chance of at least one charge roll working out.

That is a cute trick! Thanks!

Audustum wrote:
This actually seems like the right thread for this as I am also terrible at Mathhammer.

What's the probability of making a 9" charge with a re-roll (assuming a re-roll failed charge ability).

I know the initial roll is 5/18 or 27.78% (rounded), but for the life of me I don't know how to figure out what it becomes with a re-roll.

The basic rule of probability is two-fold:

Rule 1:

If two events both need to happen, you multiply their individual probabilities. For example, getting two heads in a row on a coin flip would be 1/2 * 1/2, or 1/4. We call this an AND probability, meaning "what is the probability of 1 heads AND 1 heads?"

Rule 2:

If two events can both happen for a single event, we add the probabilities together. For example, if I flip a coin, I can get a heads or a tails, meaning to find the probability of getting one or the other, we would add 1/2 + 1/2, which sums to 1. This is assuming you can't land on the side. We call this an OR probability, meaning "what is the probability of 1 heads OR 1 tails?"

Sometimes you have to use a combination of the two to figure out your whole probability. For example, you correctly identified your initial probability of making the 9" charge. So if you made that charge, you're great! But what if you missed the charge? That would happen 13/18 times. So either you make the charge on the first roll, OR you miss the charge the first time, but make it the second time. So we need to take the initial probability of making the charge (5/18) and add that to the probability that we miss the first charge but make the second.

But that probability is itself a combination of two events, so we'll need to multiply them. The two events are:

1) I miss the first charge (13/18)
2) I make the second charge (5/18)

So 13/18 * 5/18 tells us how likely we are to miss the first charge and then make the second one. This is 65/324, or just about 20% of the time.

Now all we have to do is add that probability to the initial probability that we made the charge in the first place and we have the total chance that we'll make the charge at all with a re-roll. That was 27.7%, so with a straight re-roll, we're looking at 47.7% chances that we make the charge.

If you want to take into account the fact that you could also just choose to re-roll one of the dice with a CP re-roll, then things get a lot more dicey, mathematically speaking (no pun intended) because what if you roll a 6 and a 1 on your original roll? Obviously it's much better to just use a CP re-roll on the 1 because you make that roll 66% of the time, only needing a 3+ to succeed.

Recalling someone doing this math from 7th edition (because this option effectively grants you 7th edition fleet) you have a 57% chance of making the 9" charge with judicious usage of your CP re-roll or your ability to re-roll both dice.

What might judicious usage look like, you ask? Well basically if boils down to this:

If one of your dice is a 4 or higher, re-roll the lowest die to try and make your charge. If neither of your dice are 4 or higher, re-roll them both. Happy charges!

 luke1705 wrote:

Audustum wrote:
This actually seems like the right thread for this as I am also terrible at Mathhammer.

What's the probability of making a 9" charge with a re-roll (assuming a re-roll failed charge ability).

I know the initial roll is 5/18 or 27.78% (rounded), but for the life of me I don't know how to figure out what it becomes with a re-roll.

The basic rule of probability is two-fold:

Rule 1:

If two events both need to happen, you multiply their individual probabilities. For example, getting two heads in a row on a coin flip would be 1/2 * 1/2, or 1/4. We call this an AND probability, meaning "what is the probability of 1 heads AND 1 heads?"

Rule 2:

If two events can both happen for a single event, we add the probabilities together. For example, if I flip a coin, I can get a heads or a tails, meaning to find the probability of getting one or the other, we would add 1/2 + 1/2, which sums to 1. This is assuming you can't land on the side. We call this an OR probability, meaning "what is the probability of 1 heads OR 1 tails?"

Sometimes you have to use a combination of the two to figure out your whole probability. For example, you correctly identified your initial probability of making the 9" charge. So if you made that charge, you're great! But what if you missed the charge? That would happen 13/18 times. So either you make the charge on the first roll, OR you miss the charge the first time, but make it the second time. So we need to take the initial probability of making the charge (5/18) and add that to the probability that we miss the first charge but make the second.

But that probability is itself a combination of two events, so we'll need to multiply them. The two events are:

1) I miss the first charge (13/18)
2) I make the second charge (5/18)

So 13/18 * 5/18 tells us how likely we are to miss the first charge and then make the second one. This is 65/324, or just about 20% of the time.

Now all we have to do is add that probability to the initial probability that we made the charge in the first place and we have the total chance that we'll make the charge at all with a re-roll. That was 27.7%, so with a straight re-roll, we're looking at 47.7% chances that we make the charge.

If you want to take into account the fact that you could also just choose to re-roll one of the dice with a CP re-roll, then things get a lot more dicey, mathematically speaking (no pun intended) because what if you roll a 6 and a 1 on your original roll? Obviously it's much better to just use a CP re-roll on the 1 because you make that roll 66% of the time, only needing a 3+ to succeed.

Recalling someone doing this math from 7th edition (because this option effectively grants you 7th edition fleet) you have a 57% chance of making the 9" charge with judicious usage of your CP re-roll or your ability to re-roll both dice.

What might judicious usage look like, you ask? Well basically if boils down to this:

If one of your dice is a 4 or higher, re-roll the lowest die to try and make your charge. If neither of your dice are 4 or higher, re-roll them both. Happy charges!

And there's the detailed answer! Thank you too!