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Hello everyone!

So today I was desperately trying to figure out the math-hammer for the following problem, and I am normally pretty good at it (or at least passable) but there are just too many re-rolls and "pick the highest of multiple dice" problems for me to really grasp a good chance.

The weapon is fired at Ballistic Skill 4, and is 2 shots at Str. 10 AP1. Fair enough. The hard parts are considering the other rules it has: Ordnance (2d6 and pick the highest for armour penetration) and Tank Hunter (re-roll failed armour penetration dice, can choose to re-roll glances.)

The weapon is targeting Armor 15 and is seeking a Pen, so I -will- re-roll any glances.

Help!

*Edit - first doesn't mean best, my calculations below are incorrect*

Well, here's what I got. Hopefully my math is right, the internet is watching!

So, easy part. 2 shots at BS4 = 2 * 2/3 = 4/3 to hit, or 133%.

Slightly harder part. You're str10, shooting at 15, and it's a pen or nothing. So, you need a 6. So, you have a 1/6 chance of succeeding.

But you can reroll failures due to tank hunter. So, you'll be rerolling 5/6 of the time. And those rerolls still have a 1/6 chance of succeeding.

So, your odds of penetrating are (1/6 + 5/6 * 1/6). Again, it's a 1/6 chance of succeeding the first time, and you try again 5/6 of the time with the same odds.

You have 2 dice from ordnance, so your odds of penetrating is twice the above, or 2 (1/6 + 5/6*1/6).

To put it together, multiply your chance to hit with your chance to pen and you're done.

4/3 * 2 * (1/6 + 5/6*1\6) ~ 81%.

That percentage is higher than I would have initially expected, honestly, so I might be wrong. If anyone sees an error, please let me know.

That's.... a salty peanut, as a friend would say, if it's really 81%.

For what it's worth, what I am doing is trying to see what chance a 30k Valdor Tank Hunter with a Solar Auxilia Tank Commander has of penetrating the armour of a Warlord Titan and therefore making all its guns snapfire, provided its void-shields are removed some other way. (If I'm fighting a Warlord with two Valdors I have ~2200 other points to spend on gubbins as well).

I'll try to explain as I go.

BS4, 2 shot --> Number of shots * chance to hit each shot = amount of hits --> 2 * 0.66666. = 1.3333333....
I asume not further explanation to be required here. :p

You roll 2 dice, looking for a 6 to penetrate the AV 15.
The first dice has a 1 in 6 chance (or 16.66% chance) of penetrating. the formula for this is: Number of hits * chance to roll a 6 = Chance to penetrate. So 1.33333*0.16666=0.213333333.

The second dice has a 1 in 6 chance of penetrating as well, but it only matters in case the first doesn't penetrate (remaining 83.3333% chance).
The formula for this is: Number of hits * chance the first shot misses * chance to roll a 6 = Chance to penetrate on the second shot.
So that's 1.33333333*0.833333*0.166666666 = 0.1851.

We add both to get the chance to pen without re-rolls. This is 0.39843333333 (or 39.84% chance to pen).

The re-rolls only come into play when both dice do not penetrate, this is the remaining 60.15% chance.

The first re-roll has a 1 in 6 chance of penetrating aswell, but only comes into play after the first two fail.
Formula this time is: Amount of hits * chance both shots miss *chance to roll a 6= chance to penetrate when you re-roll the first dice.
In numbers: 1.3333333*0.6015*0.166666= 0.1336666 (or 13,36%)

We add this to our previous hit chance of 39.84 so it equals 53.2% chance to penetrate.

The re-roll on the second dice has a 1 in 6 chance of penetrating, but only comes into play after the first two dice rolls AND the first re-roll fails to penetrate. Thus only the remaining 46.8% applies to it.
Formula for this one is: Amount of hits * chance both shots miss and the first re-roll * chance to roll a 6 = chance to penetrate when you re-roll the second dice.
This equals 1.333333333*0.46.8*0.1666666=0.10399999 (or 10.39%)

We add this to our previous hit chance aswell. 10.39+53.2= 63.59% which is the final result.

2 shots, fired at BS 4 at Str 10 have a 63.59% of penetrating a AV15 target if they have Tank hunter and ordinance.

I just tried 10 times myself, and got a pen 80% of the time. Try it yourself and see, I'm curious to see how you do.

But tank hunter and ordnance is a solid combo.

Still, I am planning on taking two so-equipped Valdors, which hopefully will make an enemy Warlord snapfire pretty reliably (even if 63% is right, that is pretty good, per Valdor). There are other problems with the strategy, such as the short range of the Neutron Laser or the comparatively thin armour of the Valdors, but they are superheavies (so surprisingly durable) and -very- powerful.

 Kap'n Krump wrote:
Well, here's what I got. Hopefully my math is right, the internet is watching!

So, easy part. 2 shots at BS4 = 2 * 2/3 = 4/3 to hit, or 133%.

Slightly harder part. You're str10, shooting at 15, and it's a pen or nothing. So, you need a 6. So, you have a 1/6 chance of succeeding.

But you can reroll failures due to tank hunter. So, you'll be rerolling 5/6 of the time. And those rerolls still have a 1/6 chance of succeeding.

So, your odds of penetrating are (1/6 + 5/6 * 1/6). Again, it's a 1/6 chance of succeeding the first time, and you try again 5/6 of the time with the same odds.

You have 2 dice from ordnance, so your odds of penetrating is twice the above, or 2 (1/6 + 5/6*1/6).

To put it together, multiply your chance to hit with your chance to pen and you're done.

4/3 * 2 * (1/6 + 5/6*1\6) ~ 81%.

That percentage is higher than I would have initially expected, honestly, so I might be wrong. If anyone sees an error, please let me know.

Your chances do not double when you roll 2 dice, this is where you make a mistake. 1 dice has 50% chance to roll a 4+, that does not mean that if you roll 2 dice you have a 100% chance to roll a 4+.

 Unit1126PLL wrote:
That's.... a salty peanut, as a friend would say, if it's really 81%.

For what it's worth, what I am doing is trying to see what chance a 30k Valdor Tank Hunter with a Solar Auxilia Tank Commander has of penetrating the armour of a Warlord Titan and therefore making all its guns snapfire, provided its void-shields are removed some other way. (If I'm fighting a Warlord with two Valdors I have ~2200 other points to spend on gubbins as well).

Warlord Titans are immune to crew shaken and stunned and do not fire snapshots if they get a penetrating hit. Unless your tank has a special rule, the warlord is firing as regular.

DaPino wrote:

 Kap'n Krump wrote:
Well, here's what I got. Hopefully my math is right, the internet is watching!

So, easy part. 2 shots at BS4 = 2 * 2/3 = 4/3 to hit, or 133%.

Slightly harder part. You're str10, shooting at 15, and it's a pen or nothing. So, you need a 6. So, you have a 1/6 chance of succeeding.

But you can reroll failures due to tank hunter. So, you'll be rerolling 5/6 of the time. And those rerolls still have a 1/6 chance of succeeding.

So, your odds of penetrating are (1/6 + 5/6 * 1/6). Again, it's a 1/6 chance of succeeding the first time, and you try again 5/6 of the time with the same odds.

You have 2 dice from ordnance, so your odds of penetrating is twice the above, or 2 (1/6 + 5/6*1/6).

To put it together, multiply your chance to hit with your chance to pen and you're done.

4/3 * 2 * (1/6 + 5/6*1\6) ~ 81%.

That percentage is higher than I would have initially expected, honestly, so I might be wrong. If anyone sees an error, please let me know.

Your chances do not double when you roll 2 dice, this is where you make a mistake. 1 dice has 50% chance to roll a 4+, that does not mean that if you roll 2 dice you have a 100% chance to roll a 4+.

Cool! 63% per tank isn't bad though.

DaPino wrote:

 Unit1126PLL wrote:
That's.... a salty peanut, as a friend would say, if it's really 81%.

For what it's worth, what I am doing is trying to see what chance a 30k Valdor Tank Hunter with a Solar Auxilia Tank Commander has of penetrating the armour of a Warlord Titan and therefore making all its guns snapfire, provided its void-shields are removed some other way. (If I'm fighting a Warlord with two Valdors I have ~2200 other points to spend on gubbins as well).

Warlord Titans are immune to crew shaken and stunned and do not fire snapshots if they get a penetrating hit. Unless your tank has a special rule, the warlord is firing as regular.

The tank has a special rule:

Solar Auxilia Army List, HH4 pg. 282 wrote: Shock Pulse: Any vehicle (including Super-heavy vehicles) that suffers a Penetrating hit from a weapon with this rule may only fire Snap Shots the following turn.

Comment redacted. Huh, I wouldn't have thought they'd nerf such a rare and expensive model.

The Valdor Tank Hunter existed long before the Warlord Titan... *goes to look up year*

I have trouble with quotes, but I'm confused by this:


"Your chances do not double when you roll 2 dice, this is where you make a mistake. 1 dice has 50% chance to roll a 4+, that does not mean that if you roll 2 dice you have a 100% chance to roll a 4+."

I thought this was correct. Statistically, you do have a 100% of getting a 4+ on two dice. This doesn't always happen in reality, but statistically, I believe that is an accurate assumption, because 2 x 50% odds = 100%.

If that is incorrect, than what is the odds of rolling 1 4+ on 2 dice?

Furthermore, if your odds don't double when you use two dice, how do you get 133% to hit? That's 2 shots at BS4, or 2*2/3, which is 4/3, or 133%. it's the same operation, and we agree on the to-hit percentage. How is applying that logic to two dice for penetrating hits any different?

The Valdor has rules dating back at least to 2012.

And that sort of math problem is why I asked the experts >.>

So some of the math above is incorrect.

Rolling 2 shots hitting 66% of the time don't allow you to hit 133% of the time (in the first place you cannot hit more than 100&% of the time. The easiest way to avoid this trap is to look at how often you will miss both shots.

So each shot misses 1/3rd of the time. That means missing both shots has a probability of 1/3 * 1/3 = 1/9 or 11.11111%. Which means you hit 88.9% of the time.

To penetrate you need at least one 6 on your 2D6 roll. There are 11/36 possible combinations that contain at least one 6 (so about 30.6%). So you have a 25/36 chance of not rolling a 6 on 2D6, or 69.4%. So with a re-roll the odds of failing on both attempts is 25/36*25/36 = 625/1296=48.2%. Which means you penetrate 51.7% of the time.

Which mean your chance of penetrating is 0.88889*0.517747 = 0.460219 or about 46% of the time.




Automatically Appended Next Post:
The odds of rolling at least one 4+ on 2 dice is 1-(1/2*1/2) = 1-(1/4) = 3/4 or 75%. Essentially it is 1 minus the odds of it not happening. You would only add if the events are mututally exclusive as in there is no case when both events can be true. So rolling either above or below 4+ on a single dice would add the 2.

Breng77 wrote:
So some of the math above is incorrect.

Rolling 2 shots hitting 66% of the time don't allow you to hit 133% of the time (in the first place you cannot hit more than 100&% of the time. The easiest way to avoid this trap is to look at how often you will miss both shots.

So each shot misses 1/3rd of the time. That means missing both shots has a probability of 1/3 * 1/3 = 1/9 or 11.11111%. Which means you hit 88.9% of the time.

To penetrate you need at least one 6 on your 2D6 roll. There are 11/36 possible combinations that contain at least one 6 (so about 30.6%). So you have a 25/36 chance of not rolling a 6 on 2D6, or 69.4%. So with a re-roll the odds of failing on both attempts is 25/36*25/36 = 625/1296=48.2%. Which means you penetrate 51.7% of the time.

Which mean your chance of penetrating is 0.88889*0.517747 = 0.460219 or about 46% of the time.

That's less reliable. I may have to take three vehicles, though one of them won't have BS4 or Tank Hunters. If we break 3k points (so if he brings really anything else), I'll ask him to play apocalypse and then I can upgrade them all again.

Breng77 wrote:
So some of the math above is incorrect.

Rolling 2 shots hitting 66% of the time don't allow you to hit 133% of the time (in the first place you cannot hit more than 100&% of the time. The easiest way to avoid this trap is to look at how often you will miss both shots.

So each shot misses 1/3rd of the time. That means missing both shots has a probability of 1/3 * 1/3 = 1/9 or 11.11111%. Which means you hit 88.9% of the time.

To penetrate you need at least one 6 on your 2D6 roll. There are 11/36 possible combinations that contain at least one 6 (so about 30.6%). So you have a 25/36 chance of not rolling a 6 on 2D6, or 69.4%. So with a re-roll the odds of failing on both attempts is 25/36*25/36 = 625/1296=48.2%. Which means you penetrate 51.7% of the time.

Which mean your chance of penetrating is 0.88889*0.517747 = 0.460219 or about 46% of the time.

You have two shots, so statistically, yes, it is possible to hit more than 100% of the time. You may be thinking it's impossible to hit over 100% with just one shot, but he gets two.

You can look it the other way, like you said. You'll miss 1/3 of the time, with two shots, that's 2/3. So, out of 2 tries, you'll miss 2/3 of the time in total, so 2-2/3 = 4/3 is your hits, which is still 133%.

edit. wat.

 Kap'n Krump wrote:
I have trouble with quotes, but I'm confused by this:


"Your chances do not double when you roll 2 dice, this is where you make a mistake. 1 dice has 50% chance to roll a 4+, that does not mean that if you roll 2 dice you have a 100% chance to roll a 4+."

I thought this was correct. Statistically, you do have a 100% of getting a 4+ on two dice. This doesn't always happen in reality, but statistically, I believe that is an accurate assumption, because 2 x 50% odds = 100%.

If that is incorrect, than what is the odds of rolling 1 4+ on 2 dice?

No. Your assumption is fundamentally flawed (and I'm not trying to be mean or condescending). Statistics are based in reality. if it happens statistically, it also happens in reality. Now that doesn't mean that if you roll 6 dice that you will automatically have a 6. That's because the sample size is to small. If you'd throw a ridiculous amount of dice (for instance all dice thrown by all 40K players combined since the first game of 40K), you'd observe that about 16,6666% of those dice come up as a 1, 16,666666% of those dice would show a 2, ....., and 16,666666% of those dice would show a 6. Slight deviations are possible, but I asure you that in 9.99 out of 10 cases you won't see a result dip under 15.5% or go over 17.5% if fair dice were us.

Roll a single dice you have 50% chance of rolling a 4+. The second dice is only required if you miss your first shot. You have a 50% chance to miss and when that happens your second dice has a 50% chance to roll a 4. 50% * 50% = 25%. This means your second dice adds 25% chance to roll a 4 to your first dice. 50% + 25% = 75% to roll at least one 4 when rolling two dice

But you don't miss 2/3rds of the time with 2 shots that makes no sense. That would mean statistically you miss more with more shots. Which is only true if I care about hitting every shot. Your theory doesn't hold up because you can never be assured of hitting 100% of the time. Each shot is a seperate miss, so you would not hit 100% of the time you would hit 2/3rds of the time on each shot, which as show above means you hit ~89% of the time, because 1/9th of the time both miss. You don't add the probability as it is possible for both to hit and both to miss.

 Kap'n Krump wrote:

Breng77 wrote:
So some of the math above is incorrect.

Rolling 2 shots hitting 66% of the time don't allow you to hit 133% of the time (in the first place you cannot hit more than 100&% of the time. The easiest way to avoid this trap is to look at how often you will miss both shots.

So each shot misses 1/3rd of the time. That means missing both shots has a probability of 1/3 * 1/3 = 1/9 or 11.11111%. Which means you hit 88.9% of the time.

To penetrate you need at least one 6 on your 2D6 roll. There are 11/36 possible combinations that contain at least one 6 (so about 30.6%). So you have a 25/36 chance of not rolling a 6 on 2D6, or 69.4%. So with a re-roll the odds of failing on both attempts is 25/36*25/36 = 625/1296=48.2%. Which means you penetrate 51.7% of the time.

Which mean your chance of penetrating is 0.88889*0.517747 = 0.460219 or about 46% of the time.

You have two shots, so statistically, yes, it is possible to hit more than 100% of the time. You may be thinking it's impossible to hit over 100% with just one shot, but he gets two.

You can look it the other way, like you said. You'll miss 1/3 of the time, with two shots, that's 2/3. So, out of 2 tries, you'll miss 2/3 of the time in total, so 2-2/3 = 4/3 is your hits, which is still 133%.

No, this is just wrong. See my above comment.

Yeah if you have the chance to miss, your hit chance literally can not be 100%. I had to edit my previous post because I didn't really know how to phrase what I meant to say, and then misread something so responded to it incorrectly. But both shots have a chance to miss, ergo your hit chance is not ever going to be 100%, let alone 133% (which isn't possible)

DaPino wrote:

 Kap'n Krump wrote:
I have trouble with quotes, but I'm confused by this:


"Your chances do not double when you roll 2 dice, this is where you make a mistake. 1 dice has 50% chance to roll a 4+, that does not mean that if you roll 2 dice you have a 100% chance to roll a 4+."

I thought this was correct. Statistically, you do have a 100% of getting a 4+ on two dice. This doesn't always happen in reality, but statistically, I believe that is an accurate assumption, because 2 x 50% odds = 100%.

If that is incorrect, than what is the odds of rolling 1 4+ on 2 dice?

No. Your assumption is fundamentally flawed (and I'm not trying to be mean or condescending). Statistics are based in reality. if it happens statistically, it also happens in reality. Now that doesn't mean that if you roll 6 dice that you will automatically have a 6. That's because the sample size is to small. If you'd throw a ridiculous amount of dice (for instance all dice thrown by all 40K players combined since the first game of 40K), you'd observe that about 16,6666% of those dice come up as a 1, 16,666666% of those dice would show a 2, ....., and 16,666666% of those dice would show a 6. Slight deviations are possible, but I asure you that in 9.99 out of 10 cases you won't see a result dip under 15.5% or go over 17.5% if fair dice were us.

Roll a single dice you have 50% chance of rolling a 4+. The second dice is only required if you miss your first shot. You have a 50% chance to miss and when that happens your second dice has a 50% chance to roll a 4. 50% * 50% = 25%. This means your second dice adds 25% chance to roll a 4 to your first dice. 50% + 25% = 75% to roll at least one 4 when rolling two dice

No apology required. I am incorrect, as is my calculation. I page on this topic was helpful for me: http://math.stackexchange.com/questions/729920/probability-of-head-in-coin-flip-when-coin-is-flipped-two-times

What made the most sense to me was the same question asked differently - odds of getting one heads of two coin flips. Only four possibilities exist - two heads, one head and one tail, one tail and one head, and two tails. in 3/4 of those eventualities, you get one heads, so the odds of success is 75%, like you said.

Anyways, I'm clearly incorrect. I'll be quiet and reread your post to see how to do it correctly.

*panics* I'll come back later when this is solved hopefully.

Da Pino is correct. No matter how many dice you roll there is never a 100% chance that you will get the number you want on any of them because the events are random. While it is possible to get more successes than you are looking for, it is also possible to never hit at all. For instance even if I had 100 dice to throw to hit at a 3+, I would still have a 0.0000000000000000000000000000000000000000000000019% chance of missing all 100 shots. So while it is functionally a 100% chance it is not in actuallity any such thing.

 Unit1126PLL wrote:
*panics* I'll come back later when this is solved hopefully.

All I know is don't listen to me, I didn't do it right. Sorry! It's fair enough, I thought my result of 81% was a bit optimistic.

Unit126PLL someone can check my math but I'm pretty sure I got it right.

I love mathammer!

its usually easiest to not make any mistakes if you find the chance for success for a single shot first, then work to multiple shots like you normally would.

The chance to hit is simple here, and although the chance to penetrate looks complicated its the same as a single reroll but with the caveat of picking the highest dice. Since you need a 6 to penetrate its not so hard, out the 36 possible combinations for 2D6, 11 of them have a 6 so the chance to pen is 11/36 before the reroll. The chance of failing is then (1-11/36) and the chance at succeeding in the reroll is also 11/36 but is obviously only done if you fail the first time so the total chance to pen is the chance that you succeed with the first roll, plus the chance you succeed at a reroll. the total expression with chance to hit is then:

(2/3) * [11/36 + 11/36(1-11/36)] = 34.52% chance to pen for a single shot.

To get the chance to get at one or more pens is exactly the same as the chance for both shots to not fail. The chance to not fail to pen is 1 - 0.3452 = 0.6548. Then the chance for both to fail is 0.6548^2 = 0.4288. This means that 42.88% of the time you will miss both shots, so 1 - 0.4288 is the chance you will NOT miss both shots. Thus you will pen atleast once 57.12% of the time, on average.

TL;DR = 57.12%

Quick & Dirty Explanation: You have to find the probability of a penetrating hit for a single shot that hits, and for both shots hitting. You then use a weighted average to give you the odds of at least one hit resulting in at least one penetrating hit.

---------------

Here's my math on it. I like to run it backwards from where we have to end up at. Makes it mentally easier.

First, we need the probability of a penetrating hit, P(Pen) on 2d6 picking the highest. Since you're S10 against AV15, you need a 6, or one sixth of the possible results. Thankfully, you get two rolls to make! That means you need either or both to be sixes. To get the probability that "at least one" will result in a 6, we take 100%, and subtract the probability that NEITHER die will roll a 6.

P(Pen)=1-(5/6)*(5/6)
P(Pen)=1-(25/36)
P(Pen)=11/36
P(Pen)=30.56%

Now we also get to RE-ROLL this if neither are a pen. Thankfully, since you said you'll always reroll a glance, we can just ignore that as junk data. We do the same thing here, finding out the possibility that neither will be a six, multiplying them together, and subtracting that result from 1.

P(Pen)=1-(25/36)*(25/36)
P(Pen)=1-(625/1296)
P(Pen)=671/1296
P(Pen)= 51.77%

So right now we're at the probability, given a hit, that we will end up with a penetrating hit. In order to know our overall probability though of getting a penetrating hit, we need to find average probability that each throw will result in a Pen. First we need to find out how likely each possible combination of hits there are:

P(Both Hit) = 2/3*2/3
P(Both Hit) = 4/9
P(Both Hit) = 44.44%

P(Neither Hit) = 1/3*1/3
P(Neither Hit) = 1/9
P(Neither Hit) = 11.11%

P(Just One Hit) = 1-P(Both Hit)-P(Neither Hit)
P(Just One Hit) = 1-4/9-1/9
P(Just One Hit) = 4/9
P(Just One Hit)= 44.44%

We need to now find how frequent a penetrating hit is as each other these possible outcomes. Obviously, if neither hit, the probability is 0%. Easy. If one hits, well we already got that, 51.77% If Both Hit, we just take probability that neither hits will pen, and we subtract that from 100%!

P(At Least One Pen) = 1-(P'Pen)*(P'Pen)
P(At Least One Pen) = 1-(625/1296)*(625/1296)
P(At Least One Pen) = 1-(390625/1679616)
P(At Least One Pen) = 1288991/1679616
P(At Least One Pen) = 76.74%

Great! So now, we just take the average of each possibility, and total that up to get our actual chances!

P(Neither Hit)*P(Pen on Neither Hit) = 11.11% * 0% = 0%
+
P(Just One Hit)*P(Pen on Just One Hit) = 44.44% * 51.77% = 23.01%
+
P(Both Hit)*P(At Least One Pen) = 44.44% * 76.74% = 34.11%
=
TOTAL: 57.12%


So there you go! Class dismissed!

EDIT: Added colour codings to make easier to follow where the numbers are coming from.

Can confirm Yarium's result

You'll need 3 Valdors for an 82% 4 Valdors for a 90% chance to pen, should be a good value.

Breng77 wrote:
Unit126PLL someone can check my math but I'm pretty sure I got it right.

You were right all the way up until the last line. You are correct about the chance for at least one hit (0.8889) and the chance to pen (0.5177) but multiplying them together doesn't correspond to anything meaningful because you can only multiple a single hit times the chance to penetrate not "at least one hit" which is what 0.8889 is.

I am only able to use 2 HQ choices for Auxilia Tank Commanders, so only 2 Valdors get the special rule.

Is that reliable enough power (with two shots each, which they have) to shake the Warlord Titan after its shields are down almost every turn? Or should I ask to play apocalypse and bring more tanks with more Commanders?