Recommend
3 
 Thumb up
 Hide
14 Posts

Chaos in the Old World: The Horned Rat Expansion» Forums » Strategy

Subject: Vengeance - Improved Probabilities rss

Your Tags: Add tags
Popular Tags: [View All]
Joel Schuster
Germany
Bretten
Baden-Württemberg
flag msg tools
badge
Avatar
mbmbmbmbmb
I am not a math geek at all. Vengeance seems like a powerful upgrade, especially when playing for the dial as Khorne. But how powerful is it really ? I am trying to give it some numbers to fathom its power. I am sure there are some math geeks around who can verify or proof wrong my assumptions.

Lets look at a regular Khorne warrior, rolling 2 dice. Your chance of scoring at least 1 hit (not considering multiple hits) is 75% which leaves 25% chance of 0 hits.

With Vengeance, you roll 3 dice, which increases your chances for at least 1 hit to 87.5% and only leaves 12.5% for no hit. Effectively halves the bad luck.

Because of explosions rerolls, each die you roll produces an average of 0.6 hits. So, from 2 dice you may expect 1.2 hits on average, from 3 that is 1.8 already. So close to expecting 2 hits ?

There have been a number of cards and upgrades which raise the cultists defense, like RainOfPus or the Slaanesh cultist upgrade. With the expansion, we got PleasureShield and StrengthInNumbers. Of course you may also be faced with the situation that the only applicable target is a figure of 2 defense.

With just 2 dice, your chance of getting at least 2 hits is about 31.9%, including an explosion reroll. Math experts, correct me if I am wrong.

Now, the chance of getting 2 hits from 3 dice is considerably higher, its 56% including an explosion reroll, according to my calculations.

I'd be glad if someone could verify or proof me wrong. I am quite sure I got some error in all that

Edit: See slightly corrected values below.
2 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Joel Schuster
Germany
Bretten
Baden-Württemberg
flag msg tools
badge
Avatar
mbmbmbmbmb
There is one thing that is quite likely I got wrong. Its the same for 2 or 3 dice so lets deal with it for 2 dice as that is easier.

For 2 dice there are 36 possible results. 8 of these directly provide the desired result, which is 2 hits. 4-4, 4-5, 4-6, 5-4, 5-5, 5-6, 6-5 and 6-6. Results including one 6 but failed to get 2 hits otherwise, I treated as half successes. Because one 6 grants one hit and the reroll has a 50:50 chance of producing the 2nd desired hit.

So, without considering explosion rerolls, the chance of getting 2 hits right away is 8/36 = 22.2% When you add the explosion reroll as half a success, you get to the 31.9% Not sure if that is as easy as that though.
1 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Joel Schuster
Germany
Bretten
Baden-Württemberg
flag msg tools
badge
Avatar
mbmbmbmbmb
So, assuming my assumptions are correct, Vengeance does 3 things.

a) Increase the chance of getting at least 1 hit from a warrior from 75 to 87.5%

b) Increase the chance to get at least 2 hits from a warrior from 33 to 56%.

c) Increase damage output of a single warrior from 1.2 to 1.8 hits on average.


Quite considerable.

Edit for accurate values.
1 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Rob Corn
United States
Columbus
Ohio
flag msg tools
badge
How appropriate. You fight like a cow.
Avatar
mbmbmbmbmb
For two dice with explosions, treat the "6" results (that only result in one hit, like a 3-6) as having two possible states: a 3-6-1 and a 3-6-0; basically a coin flip to determine the explosive hit.

That adds to the 36 possible values by 6:

1-6 2-6 3-6
6-1 6-2 6-3

Of which 15 are double hits (the 9 non-explosive double hits plus the 6 from above).

That gets you 15/42, or a 35.7% chance for a double hit on two dice (ignoring explosions gets you 9/36, or 25%). Edit: The percentage of the hits should be added, not the overall ratio -- 25% plus 50% (the hit rate) times 6/36 = 12/36 and not 15/42. See Jeff's explanation below.

For three dice, there are 216 base combinations (6 cubed), of which 108 have at least two hits, a flat 50% before explosions.

Again, only the single hit "6"s are interesting for this case, which is 27 sets:

1-1-6 1-2-6 1-3-6 | 2-1-6 2-2-6 2-3-6 | 3-1-6 3-2-6 3-3-6
1-6-1 1-6-2 1-6-3 | 2-6-1 2-6-2 2-6-3 | 3-6-1 3-6-2 3-6-3
6-1-1 6-1-2 6-1-3 | 6-2-1 6-2-2 6-2-3 | 6-3-1 6-3-2 6-3-3

So adding 27 additional to the 216 possible results and adding 27 to the possible successes gets you 135/243, or a 55.6% chance for a double hit using three dice. Edit: Similarly, this should be 50% of 27/216, or 27/432 added to 216/432 = 243/432 and not 135/243.

Additional explosions don't matter since you're only interested in one additional hit. Calculating the odds of, say, four hits on six dice is a lot more complicated!

Edit: Nevermind! See below for the right answers.
2 
 Thumb up
0.25
 tip
 Hide
  • [+] Dice rolls
Joel Schuster
Germany
Bretten
Baden-Württemberg
flag msg tools
badge
Avatar
mbmbmbmbmb
Ok, I did what Rob said, just had an error in the 2 dice calculation, so I was off at 31.9 instead of the 35.7%. The 2hits from 3dice was quite accurate with 56 to 55.6%.

I left out the >2 hits indeed. Think you can cover that by the estimation of 0.6 hits per die. What I wanted to look at was how savely you get your DAC from a single bloodletter and how Vengeance improves that.

Thanks for your input guys, much appreciated
1 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Jeff Stafford
United States
Phoenix
Arizona
flag msg tools
mbmbmbmb
I don't know if I'm thinking about this correctly, but I don't think those probabilities are correct, but they are very close and could serve as an estimate.

The problem is that you have two independent probabilities when you consider the exploding 6. Let me give an example that will make sense. Let's say the Nurgle card is played in a space that reduces the dice by one, and so the bloodletter can only roll one dice. What are the chances of getting two hits?

Here are the possible outcomes:
1
2
3
4
5
61
62
63
64
65
66 (etc).

If you simply add the 2 hit outcomes divided by the total number of outcomes, you get 3/11 = 27%. That is obviously way to high, as the probability of even rolling a 6 is 1/6 or 16%.

So, you have to compute first - the probability of rolling a 6, which is 16% Then you have to multiply that by the chance of rolling a 4,5, or 6 with the second roll, which is .50%. So your actually probability of rolling two hits with one dice is actually around 8.33%.

What is your chance of getting two hits with two dice?

Here are the possible outcomes:

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

So, first, what is the probability of getting a straight up two hits - no exploding 6s? 9/36 or 25%.

Now, what is the probability of getting an exploding 6 to bring you up to 2 hits?

These are 1-6, 2-6, 3-6, 6-1, 6-2, and 6-3 or 6/36 or 16%. Now, what is the chance that that second roll will be a six? It's 50% again, so 50% x 16% = 8.333%.

Then you have to add the two probabilities together: 25% + 8.333% = 33.33%.

Nitpicky, right?






3 
 Thumb up
0.25
 tip
 Hide
  • [+] Dice rolls
Joel Schuster
Germany
Bretten
Baden-Württemberg
flag msg tools
badge
Avatar
mbmbmbmbmb
hotcoppersky wrote:
Now, what is the chance that that second roll will be a six? It's 50% again, so 50% x 16% = 8.333%.


I think you want to say "that second roll be another hit" actually, not six. That aside. You may be right

The 25% is to get 2 hits straight from 2 dice. If you dont, there is the chance to get 2 hits out of a single die by explosion, which is a 16% chance on a six and then a 50% for the reroll to get another hit.

In one effort to juggle with the values I got this result as well. Thats why I ask, being no expert on this field. You may convince me of this or that result. Adding and multiplying like this makes sense. I already signed the former result as well though Another opinion to verify either conclusion ?
1 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Jeff Stafford
United States
Phoenix
Arizona
flag msg tools
mbmbmbmb
Exactly, I meant hit, not six. Trying to figure out these probabilities makes my brain hurt!
 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Joel Schuster
Germany
Bretten
Baden-Württemberg
flag msg tools
badge
Avatar
mbmbmbmbmb
Well, I've had ~32% first, then Rob led us to 35 and this is 33, I think we can agree its not far off from 1/3 chance to get 2 hits out of 2 dice.

Is the change for 2 hits out of 3 dice much different then ? Straight chance is about 50% and then the explosion rerolling effect kicks in to increase it by a few digits. Should be slightly more than for 2 dice.
 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Joel Schuster
Germany
Bretten
Baden-Württemberg
flag msg tools
badge
Avatar
mbmbmbmbmb
hotcoppersky wrote:
Exactly, I meant hit, not six. Trying to figure out these probabilities makes my brain hurt!


Ah, its not even close to rocket science And the geek in you wants to know, otherwise you wouldnt have started thinking about it
1 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Jeff Stafford
United States
Phoenix
Arizona
flag msg tools
mbmbmbmb
So..if we did three dice for at least two hits?

Well, 108/216 is two hits out of the box - 50%.

Then there's the 27 one hit with a '6' rolls. 27/216 = 12.5% And again, the chances of getting a hit on the second roll is 50%. So, 12.5% * 50% = 6.25%.

50% + 6.25% = 56.25%.

So, the chances of getting two hits from Vengeance is from 33.33% to 56.25%.


2 
 Thumb up
0.25
 tip
 Hide
  • [+] Dice rolls
Rob Corn
United States
Columbus
Ohio
flag msg tools
badge
How appropriate. You fight like a cow.
Avatar
mbmbmbmbmb
hotcoppersky wrote:
Here are the possible outcomes:
1
2
3
4
5
61
62
63
64
65
66 (etc).

If you simply add the 2 hit outcomes divided by the total number of outcomes, you get 3/11 = 27%. That is obviously way to high, as the probability of even rolling a 6 is 1/6 or 16%.

Aha, you got it -- I've been using a spreadsheet for calculating Space Hulk odds, and tried to map it to the exploding 6s and didn't think through the problem enough. Great explanation.
1 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Bryan Myers
United States
Midlothian
Virginia
flag msg tools
mbmb
This was made by MrBlarney at Penny-Arcade; it shows the probabilities for various hits given the number of dice:



If you want to calculate "at least x hits", you can either add up everything from that spot down or subtract the ones above it from 100%.
1 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Rob Corn
United States
Columbus
Ohio
flag msg tools
badge
How appropriate. You fight like a cow.
Avatar
mbmbmbmbmb
Here's that same table reworked to show "at least" hits:

Battle Dice
Hits
(at least) 1 2 3 4 5 6
0 0.5 0.25 0.125 0.062 0.031 0.016
1 0.5 0.75 0.873 0.936 0.967 0.983
2 0.083 0.333 0.561 0.728 0.837 0.905
3 0.014 0.09 0.249 0.433 0.598 0.729
4 0.002 0.021 0.081 0.196 0.341 0.492
5 0.005 0.022 0.071 0.157 0.275
6 0.001 0.005 0.022 0.06 0.129
7 0.001 0.006 0.019 0.052
8 0.001 0.005 0.018
9 0.001 0.005
10 0.001
Expected 0.6 1.2 1.8 2.4 3 3.6
1 
 Thumb up
1.00
 tip
 Hide
  • [+] Dice rolls
Front Page | Welcome | Contact | Privacy Policy | Terms of Service | Advertise | Support BGG | Feeds RSS
Geekdo, BoardGameGeek, the Geekdo logo, and the BoardGameGeek logo are trademarks of BoardGameGeek, LLC.