Cheez |

Hi all,

We have a party of four playing The Elven Entanglement. We just encountered the Demonic Horde barrier.

The card says: "each character randomly chooses a character to summon and encounter this adventure's servitor demon. To defeat this barrier, all of the summoned demons must be defeated,"

Our question is: is it possible that a single character in the group will have to defeat multiple summoned demons? Example: player 1 randomly chose player 3 and then player 2 randomly chose player 3. Or, do you think each character only fights one?

We're rather certain that poor player 2 has to fight two demons but thought it wouldn't hurt to double check.

Thanks,

Ben and lil' Abby

Sandslice |

1 person marked this as a favorite. |

Take heart, though. Here are some fancy probabilities.

1. The chance that each character fights one demon: n!/(n^n)

2p: 1/2

3p: 2/9

4p: 3/32

5p: 24/625

6p: 5/324 (1.5%)

2. The chance that someone fights all of them: 1/(n^(n-1))

2p: 1/2

3p: 1/9

4p: 1/64

5p: 1/625

6p: 1/7776

3. The chance that you personally don't fight any: ((n-1)/n)^n

2p: 1/4

3p: 8/27 (just under a third)

4p: 81/256 (just over 5/16, or 31.25%)

5p: 1024/3125

6p: 15625/46656 (~33.5%)

So it's likely that someone will fight more than one; but each such is one that someone else (hopefully, HOPEFULLY someone squishy and not packing weapons) won't face. :)

Andrew L Klein |

Nice Sandslice, though the 2p ones (maybe others) are off. For each fighting one, it's 1/3 (A fights both, B fights both, A and B each fight one). That also makes it 2/3 for one person fighting all in 2p, and 1/3 for you not fighting any.

I do hope your "one fights all" odds for 6p are accurate though, I don't even want to imagine someone getting so screwed that that happens lol.

Sandslice |

Nice Sandslice, though the 2p ones (maybe others) are off. For each fighting one, it's 1/3 (A fights both, B fights both, A and B each fight one). That also makes it 2/3 for one person fighting all in 2p, and 1/3 for you not fighting any.

I do hope your "one fights all" odds for 6p are accurate though, I don't even want to imagine someone getting so screwed that that happens lol.

There are two "A and B each fight one" results: *AA*, **AB, BA**, BB. That's why it's a 50% chance for each of two characters. :)

As for the 6p number: it doesn't matter where the first demon lands. The other five have to land on the same character, a 1/6 chance each: (1/6)^5 = 1/7776.

iMonkey |

I'm glad someone else on here thinks of things in math terms. This is certainly a game that makes you assess risk and probabilities. There is (usually) a 1 in 10 chance to encounter the henchman as the top card of a location deck. Doing that twice in a scenario? 1 in 100. Ever done it three times in one scenario? I have once. 1 in 1000.

(I've actually done that twice, but the other time it happened, we realized after the third henchman that no one actually shuffled the decks after setup. Oops.)

pH unbalanced |

I'm glad someone else on here thinks of things in math terms. This is certainly a game that makes you assess risk and probabilities. There is (usually) a 1 in 10 chance to encounter the henchman as the top card of a location deck. Doing that twice in a scenario? 1 in 100. Ever done it three times in one scenario? I have once. 1 in 1000.

(I've actually done that twice, but the other time it happened, we realized after the third henchman that no one actually shuffled the decks after setup. Oops.)

When we started S&S (party = Seltiyel, Feiya, Lirianne, and Lini), we always had Seltiyel go first, and he had a streak of six straight scenarios (starting with the first one in the game) where he successfully closed a location of his first turn. After the one that broke his streak, he then added two more.

I think the henchman was only on top once or twice during that streak.

Riff Conner |

I'm glad someone else on here thinks of things in math terms. This is certainly a game that makes you assess risk and probabilities.

The other night I was actually researching multi-dice probability curves, to learn how to better estimate my chances of hitting a target.

(What I think I sussed out: add together the max values of your dice, cut that in half, and add .5 per die. You have a little better than a 50% chance of rolling that or higher.)

useful site: http://anydice.com/

iMonkey |

iMonkey wrote:I'm glad someone else on here thinks of things in math terms. This is certainly a game that makes you assess risk and probabilities.The other night I was actually researching multi-dice probability curves, to learn how to better estimate my chances of hitting a target.

(What I think I sussed out: add together the max values of your dice, cut that in half, and add .5 per die. You have a little better than a 50% chance of rolling that or higher.)

useful site: http://anydice.com/

Oooh that website is nice. My nerd brain is tingling. I will definitely be using it next time I play.

I use a slightly different process for my mental math but I get the same outcome as you. I add each dice individually by: (max value / 2) + .5 + modifiers, and if the total of all the dice isn't more than what I need to roll to succeed at a check, I ask around for blessings.

Fayries |

Each character randomly chooses a character to summon and encounter this adventure's servitor demon. To defeat this barrier, all of the summoned demons must be defeated.

Is it acceptable to have 1) each and every character randomly choosing a character to summon and encounter the servitor demon, and 2) then do the summoning and encountering? (You would know, for example, that you have three servitor demons to beat and would spend resources accordingly.)

Or do you have to resolve the summoning and encountering for the first encounter before randomly choosing who will do the second encounter, and so on?

Theryon Stormrune |

Demonic Horde wrote:Each character randomly chooses a character to summon and encounter this adventure's servitor demon. To defeat this barrier, all of the summoned demons must be defeated.Is it acceptable to have 1) each and every character randomly choosing a character to summon and encounter the servitor demon, and 2) then do the summoning and encountering? (You would know, for example, that you have three servitor demons to beat and would spend resources accordingly.)

Or do you have to resolve the summoning and encountering for the first encounter before randomly choosing who will do the second encounter, and so on?

The first thing that happens is that every character randomly chooses another. Reread the first sentence. It says "to summon and encounter". That "to" is letting you know that the encounters come after. So the choosing comes first. Then the encounters happen. Those encounters, though simultaneous, happen in any order.

Kumarei |

There is (usually) a 1 in 10 chance to encounter the henchman as the top card of a location deck. Doing that twice in a scenario? 1 in 100. Ever done it three times in one scenario? I have once. 1 in 1000.

This isn't quite right. Playing a one player game, there's a 1/1000 chance to have the villain and both henchmen on top of the deck. More locations increases the chances substantially, though. In a four player game, there's about a 1.5% chance to get 3 henchmen/villain on top of the decks, and an 11% (!) chance to get two. In a six person game the chances are even better of getting three henchmen/villain on top, with a 3.8% chance (so you should expect it a little less than one out of every 25 games).

Theryon Stormrune |

We only play two character games, and we always roll the dice at the same time and sort out the encounters later. Though I guess by the letter of the law you're probably right, that you roll dice and deal with encounters one by one.

There are tons of questions about sequencing on the boards. A lot of them go unanswered by Mike, Vic and other *official* sources or without a FAQ entry. Personally I think it's because they want us to enjoy the game without being tied down by rules. Not everything needs a rule or a FAQ. But when there are exceptions needed to sort out wording that doesn't work … Valeros's Tactician power that let's him move when another player encounters a villain. But this sequencing about Demonic Hordes is silly.

Letter of the Law, it doesn't say roll and encounter. It says randomly chose a character **to** summon and encounter. Why would the designers make it so that you don't know what resources you'd need as a party for the encounters? We've always played it where I establish how we decide who gets an encounter (You're 1&2, you're 3&4, you're 5&6, I'm 7&8. Everyone roll a d8.) and we roll. Then we go around the table doing the encounters.

I guess that if you want to make the game harder, you could play it that everyone chooses a character that summons and encounters the servitor demon right then and then move to the next. (But that's not what the card says.)

Theryon Stormrune |

We play with 4 so everyone just rolls a d4 to see who fights the demons. You could imagine how thrilled our Alain player was to be stuck fighting 3 Blood Demons.

We usually play 3-4. One time Enora faced 3 of them. Oops!

Theryon Stormrune |

Our Amaryllis got blasted with 3 of them one time as well, however you can only take so much damage. Now if these were carrion golems she was fighting, would be a different story.

On a side note, over the weekend (at a con), we never encountered the golem in-game but we definitely found them lurking when we closed locations. I always like to show casters what they avoided ... hehe

Frencois |

The stat that keeps irritating us is the chance in a 6p game to actually encounter 3 or more demons.

... Especially vulture ones if you have a crapy constitution.

Chance that at least one character in the party will meet

- 2+ demons: 45936/46656=98%

- 3+ demons: 17136/46656=37% (that's more than 2 to 1 odds)

- 4+ demons: 2436/46656=5%

- 5+ demons: 186/46656=0,4%

- 6 demons: 6/46656=0,01%