End Phase:
1. Hero Tokens are resolved - The Hero in the Empire kills a seductress
2. Resolve Old World cards - Reavers look, but find no peasants to kill
3. Score Ruined Regions
The Empire: Nurgle scores 12 VP Slaanesh scores 6 VP
The Badlands: Slaanesh scores 8 VP Tzeentch scores 4 VP
5. Check for Game end
The game end conditions are checked one at a time, in this order:
Dial Victory - no one has achieved DIAL victory
Victory Point Victory - Yes; at least one player is over 50 points Slaanesh is the victor, as the lands fall into an orgy of celebration.
All 5 ruination cards drawn (most VPs wins)
No cards in the Old World deck (All players lose)
Congrats to Slaanesh; good game all. I'll be back with a final board, an animated gif, and assorted commentary in a few hours; I'm late for my game night here. Also, game three starting soon - should we go with a new thread, or keep using this one?
Congrats to Slaanesh; good game all. I'll be back with a final board, an animated gif, and assorted commentary in a few hours; I'm late for my game night here. Also, game three starting soon - should we go with a new thread, or keep using this one?
An animated gif?! Impressive.
And yeah, I'm gonna second locking this thread and starting a new one.
Whew, great game! This is one of those board games I absolutely love and it still makes me sad I don't get to play it as much. In fact, I was supposed to play tonight, but people got lazy :P
Thanks for hosting this Darian, it was nice to finally play Nurgle
Setting up to move the warpstone back out of there in a future turn, if necessary
Fun facts:
* If both warpstones had been left in The Border Princes, it would have been ruined automatically, (first place to Khorne, second tie between Tzeentch and Slaanesh), giving Nurgle the game.
* If Khorne manages one more hit in the Empire, the game isn't over yet. I have to sit down and think about the probability of that for a bit; certainly less than 50%, since you only expect a bit more than one hit per two dice, but I don't know a formula off-hand to deal with exploding hits.
New game thread is up, but feel free to continue to use this thread to talk about strategy and offer constructive criticism to the players and the host.
Setting up to move the warpstone back out of there in a future turn, if necessary
Fun facts:
* If both warpstones had been left in The Border Princes, it would have been ruined automatically, (first place to Khorne, second tie between Tzeentch and Slaanesh), giving Nurgle the game.
* If Khorne manages one more hit in the Empire, the game isn't over yet. I have to sit down and think about the probability of that for a bit; certainly less than 50%, since you only expect a bit more than one hit per two dice, but I don't know a formula off-hand to deal with exploding hits.
Hopefully my math is correct...
Let p = prob(hit), q = prob(exploding hit); r = prob(nonexploding hit) = p - q.
p(x hits) = {(1-p), x = 0; (q)^(x-1)*(r+q(1-p)), x >= 1}
E(hits) = ((r+q(1-p))/((1-q)^2)
For Chaos in The Old World, p = 1/2, q = 1/6; r = 1/3.
For 6 dice, the expected number of hits is 3.6. Calculating the probability of making 6 or more hits on those dice is a bit trickier and requires some multinomial calculations (and probably some shortcuts that I haven't bothered to look up), but it boils down to this:
p(0 hits over 6 dice) = 1.6%
p(1 hit over 6 dice) = 7.8%
p(2 hits over 6 dice) = 17.6%
p(3 hits over 6 dice) = 23.7%
p(4 hits over 6 dice) = 21.7%
p(5 hits over 6 dice) = 11.9%
p(6 or more hits over 6 dice) = 15.7%
So... there was a reasonable chance in fact that the game would have continued.
Hmm... maybe I should make a table to post into the next thread...
For 6 dice, the expected number of hits is 3.6. Calculating the probability of making 6 or more hits on those dice is a bit trickier and requires some multinomial calculations (and probably some shortcuts that I haven't bothered to look up), but it boils down to this:
p(0 hits over 6 dice) = 1.6%
p(1 hit over 6 dice) = 7.8%
p(2 hits over 6 dice) = 17.6%
p(3 hits over 6 dice) = 23.7%
p(4 hits over 6 dice) = 21.7%
p(5 hits over 6 dice) = 11.9%
p(6 or more hits over 6 dice) = 15.7%
So... there was a reasonable chance in fact that the game would have continued.
Hmm... maybe I should make a table to post into the next thread...
So the sweet spot is 3 hits over 6 dice, but 6 hits has a higher chance of happening than 5 hits?
For 6 dice, the expected number of hits is 3.6. Calculating the probability of making 6 or more hits on those dice is a bit trickier and requires some multinomial calculations (and probably some shortcuts that I haven't bothered to look up), but it boils down to this:
p(0 hits over 6 dice) = 1.6%
p(1 hit over 6 dice) = 7.8%
p(2 hits over 6 dice) = 17.6%
p(3 hits over 6 dice) = 23.7%
p(4 hits over 6 dice) = 21.7%
p(5 hits over 6 dice) = 11.9%
p(6 or more hits over 6 dice) = 15.7%
So... there was a reasonable chance in fact that the game would have continued.
Hmm... maybe I should make a table to post into the next thread...
So the sweet spot is 3 hits over 6 dice, but 6 hits has a higher chance of happening than 5 hits?
6 or more has a higher chance of happening than 5 exactly
Setting up to move the warpstone back out of there in a future turn, if necessary
Fun facts:
* If both warpstones had been left in The Border Princes, it would have been ruined automatically, (first place to Khorne, second tie between Tzeentch and Slaanesh), giving Nurgle the game.
And that's why it was moved. Unfortunately, I had to rely on Khorne to kill things in order to save the game.
Hmm, so I messed up the probabilities somewhat in yesterday's calculations. It would have been a 13.0% chance for Khorne to have gotten 6 or more hits, not 15.7%. Still, not that far out of a chance to have occurred. I made a table for expected results up to 6 battle dice. Cells without values have probabilities below 0.1%. There is some rounding, so the values won't be exact, but they'll be close enough for most purposes.
Posts
1. Hero Tokens are resolved - The Hero in the Empire kills a seductress
2. Resolve Old World cards - Reavers look, but find no peasants to kill
3. Score Ruined Regions
The Empire:
Nurgle scores 12 VP
Slaanesh scores 6 VP
The Badlands:
Slaanesh scores 8 VP
Tzeentch scores 4 VP
4. Advance Threat Dials
Khorne advances 2 clicks: 23 - Upgrade; 28 - Upgrade
Nurgle advances 1 click: 22 - Upgrade
Tzeentch advances 1 click: 21 - 2 Warpstones
Slaanesh advances 1 click: 24 - Upgrade
VP after dial ticks:
Khorne: 28
Nurgle: 41 + 12 = 53
Tzeentch: 42 + 4 = 46
Slaanesh: 45 + 14 = 59
5. Check for Game end
The game end conditions are checked one at a time, in this order:
Dial Victory - no one has achieved DIAL victory
Victory Point Victory - Yes; at least one player is over 50 points
Slaanesh is the victor, as the lands fall into an orgy of celebration.
All 5 ruination cards drawn (most VPs wins)
No cards in the Old World deck (All players lose)
I think a new thread for the third game? And maybe ask them to lock this one? I don't know what the rules are for that sort of thing.
Inquisitor77: Rius, you are Sisyphus and melee Wizard is your boulder
Tube: This must be what it felt like to be an Iraqi when Saddam was killed
Bookish Stickers - Mrs. Rius' Etsy shop with bumper stickers and vinyl decals.
An animated gif?! Impressive.
And yeah, I'm gonna second locking this thread and starting a new one.
Thanks for hosting this Darian, it was nice to finally play Nurgle
Inquisitor77: Rius, you are Sisyphus and melee Wizard is your boulder
Tube: This must be what it felt like to be an Iraqi when Saddam was killed
Bookish Stickers - Mrs. Rius' Etsy shop with bumper stickers and vinyl decals.
Fun facts:
* If both warpstones had been left in The Border Princes, it would have been ruined automatically, (first place to Khorne, second tie between Tzeentch and Slaanesh), giving Nurgle the game.
* If Khorne manages one more hit in the Empire, the game isn't over yet. I have to sit down and think about the probability of that for a bit; certainly less than 50%, since you only expect a bit more than one hit per two dice, but I don't know a formula off-hand to deal with exploding hits.
Hopefully my math is correct...
Let p = prob(hit), q = prob(exploding hit); r = prob(nonexploding hit) = p - q.
p(x hits) = {(1-p), x = 0; (q)^(x-1)*(r+q(1-p)), x >= 1}
E(hits) = ((r+q(1-p))/((1-q)^2)
For Chaos in The Old World, p = 1/2, q = 1/6; r = 1/3.
p(0 hits) = 1/2 = 50.0%
p(1 hit) = 5/12 = 41.7%
p(2 hits) = 5/72 = 6.9%
p(3 hits) = 5/432 = 1.2%
E(hits) = 3/5 = 0.600
For 6 dice, the expected number of hits is 3.6. Calculating the probability of making 6 or more hits on those dice is a bit trickier and requires some multinomial calculations (and probably some shortcuts that I haven't bothered to look up), but it boils down to this:
p(0 hits over 6 dice) = 1.6%
p(1 hit over 6 dice) = 7.8%
p(2 hits over 6 dice) = 17.6%
p(3 hits over 6 dice) = 23.7%
p(4 hits over 6 dice) = 21.7%
p(5 hits over 6 dice) = 11.9%
p(6 or more hits over 6 dice) = 15.7%
So... there was a reasonable chance in fact that the game would have continued.
Hmm... maybe I should make a table to post into the next thread...
So the sweet spot is 3 hits over 6 dice, but 6 hits has a higher chance of happening than 5 hits?
Inquisitor77: Rius, you are Sisyphus and melee Wizard is your boulder
Tube: This must be what it felt like to be an Iraqi when Saddam was killed
Bookish Stickers - Mrs. Rius' Etsy shop with bumper stickers and vinyl decals.
6 or more has a higher chance of happening than 5 exactly
Nintendo Network ID: AzraelRose
DropBox invite link - get 500MB extra free.