Wayne Walker
United States Chuluota Florida

So I am starting my second ever try of Solo Conquest after failing pretty miserably the first time with 0 conquered cities.
Like last time, I am playing with Goldyx. I select Great Start for my tactic.
5 Movement to go through forest and onto ruins. Gained a white crystal through a card play (can't remember the card name). Used two white mana dice from the source (using Mana Storm).
Put three white mana on the altar for 7 fame!
I have only played three games, but that is by far my best start.
What is your best start?

Jonathan
United States Minneapolis Minnesota

Wow, that's a pretty slick turn you had there! 7 fame right from the get go. Nice job. I can't recall my best start, but usually it involves getting to the closest village, recruit, pillage.

howl hollow howl
United States Oregon

Yup, if you get one of those ruins on the opening map, a combination of Mana Draw, Crystallize, and/or Mana Draw makes that a fairly easy grab. Always a nice gift, especially as it's rare in a Conquest scenario that you'll make your way back for a more difficult ruin.



Perhaps you used Concentration (or Goldyx's version in this case) to get 1 white mana token?

howl hollow howl
United States Oregon

Yes, I meant a combination of Concentration, Crystallize, and/or Mana Draw.



Dave wrote: Yes, I meant a combination of Concentration, Crystallize, and/or Mana Draw. One of the tactics cards can be quite helpful too. (#3: Mana Steal?)

Robin Reeve
Switzerland StLégier Vaud
Looking for a game session in Switzerland? Send me a pm!

Of course, one needs to be lucky with the Mana in the Source.

Carlos Pérez Cantalapiedra
Spain Zaragoza Zaragoza

1/6 + 1/6 + 1/6 = 1/2 to have a white dice, activating the advanced effect of mana draw?
If you use the basic effect, is that because you would prefer not to have another white dice in the next turn?
maybe I am wrong, it is late now and thinking a bit fuzzy

Magnesi
Australia Melbourne Victoria

urkary wrote: 1/6 + 1/6 + 1/6 = 1/2 to have a white dice, activating the advanced effect of mana draw? Gold dices also work, and you can't start the round with any dice configuration (at most 1 dice shows nonbasic colors). I'm too lazy to compute it, but I think the probability of having the right dices to steal one and activating the mana draw is around 0.5.
Is much more difficult to have the right cards, the right tile and the right ruin.
Edit: Anyway, you can't add probabilities in this way. The probability of having at least one white in three dices (if all combinations are possible) is 1(5/6)^3, which is around 42%

Trevin Beattie
United States Eugene Oregon

syzygia wrote: urkary wrote: 1/6 + 1/6 + 1/6 = 1/2 to have a white dice, activating the advanced effect of mana draw? Gold dices also work, and you can't start the round with any dice configuration (at most 1 dice shows nonbasic colors). I'm too lazy to compute it, but I think the probability of having the right dices to steal one and activating the mana draw is around 0.5. Is much more difficult to have the right cards, the right tile and the right ruin. Edit: Anyway, you can't add probabilities in this way. The probability of having at least one white in three dices (if all combinations are possible) is 1(5/6)^3, which is around 42%
That may be, but all combinations are not possible are they? Given that if two dice are nonbasic colors they must be rerolled, and also allowing for a gold die to substitute for any basic color, the probability of having at least one white or gold die is closer to 66¼%.

Magnesi
Australia Melbourne Victoria

Yeah, sure. My edit was talking about the 1/6+1/6+1/6 way of computing the probability of getting 1 white in three unconstrained dices. The "around 50%" I talked about (which is something I didn't compute, so it may be wrong) was about having the available dices for the altar, which (let's assume is the blue altar for simplicity) are 3 blue dices, 1 blue and 1 gold, 1 white and 1 gold or 1 white and 1 blue.
Ok, let's compute that....
Possibilities: 4^3 with a 0 nonbasic dice + 3*2*4^2 with 1 non basic = 160 BBB: 1 BBG: 3 BWG: 6 BGX: (where X is neither B nor W; so green or R): 12 WGX: (X=green,R): 12 BBW: 3 BWW: 3 BWX: (X=R,green, black): 18
Total: 58
Probability: 58/160 = 0.3625
Is it right? Notice that the WWW altar has a lower probability without crystallize at hand; and the same with crystallize. GGG/RRR altarhave the same probability if you don't have crystallize, and slightly higher if you do have it.

Magnesi
Australia Melbourne Victoria

syzygia wrote: Notice that the WWW altar has the same probability with crystallize at hand. This sentence was wrong, I don't know what I was thinking...
Now I realize that you can also have concentration at hand, making the things easier. I was simply thinking in mana steal+mana draw, so my computations are with this on mind. If you have other cards, the odds are different.

Carlos Pérez Cantalapiedra
Spain Zaragoza Zaragoza

I am not an expert on probability, but the third axiom of probability is not about adding probabilities for independent events? http://en.wikipedia.org/wiki/Probability_axioms In my simplification for getting at least a white directly I dont see the problem of adding probabilities. Of course if you consider the full rules, they are not completely independent and gold can be used as white...
I edit my self It is the substracting the probability of not getting 1, you are right!

Trevin Beattie
United States Eugene Oregon

Math is hard, so I cheated. I wrote a lisp function to generate all 216 combinations of 3d6 in emacs, then used a regular expression to remove the lines containing at least two nonbasic colors, then simply used egrep  wc to count the number of lines containing either white or gold.
(I’m a programming geek as well as a board game geek. )



urkary wrote: I am not an expert on probability, but the third axiom of probability is not about adding probabilities for independent events? http://en.wikipedia.org/wiki/Probability_axioms In my simplification for getting at least a white directly I dont see the problem of adding probabilities. Of course if you consider the full rules, they are not completely independent and gold can be used as white...
That's some heavy stuff you're quoting for something as simple as a d6 roll ; )
First, to see that simply adding probabilities can't work, just consider what results you'd get for six or more dice ...
I'll assume by "independent" events you mean disjoint events, as in the article. "Independent" is to my knowledge never used for events, as it refers to random variables where the outcome of one does not influence the other. Usually, three (ideal) d6 rolls are independent variables. No matter what the first two dice do, the third still has a 1/6 chance for each possible result. In Mage Knight solo, if for some reason you roll the initial three dice one after another, they are NOT independent: if the first is showing gold or black, the other two can not (you'd have to reroll these, removing them from the list of possible results and raising the probabilities for each basic colour to 1/4).
If you are talking about "events" for the three initial dice, then your random variable would be "initial state of the three mana dice". An "event" would for example be "whitegreenblue", or "at least one die is usable as white". Mutually disjoint events, for which you can add probabilties, are such that they are mutually exclusive, i.e. they can never hold at the same time. For example, "at least one die is usable as white" and "no die is usable as white". As one of these two will always be true, their probabilities add up to one, which is how the computations above me for "at least one die is usable as white" were done. Alternatively, you could have computed the probabilities of the three events "exactly one die is usable as white", "exactly two dice are usable as white" and "exactly three dice are usable as white", which are mutually exclusive, and added them up.
To make a probably unnecessary post even longer: If you were talking about the events "The first die is showing white", "the second die is showing white" and "the third die is showing white", then these are not mutually disjoint events, as can be readily seen from the fact that "all three dice are white" is part of all three events  so you can't simply add their probabilities. Furthermore, since there is the rule about rerolling dice, neither is the probability of all three events 1/6 ...

Magnesi
Australia Melbourne Victoria

Jan's answer is impeccable, but I'll try to explain the same in other words, using plain language instead of technical.
In general, when you want to add probabilities, you should use the proposition
P(A union B) = P(A) + P(B)  P(A intersection B)
If A and B are mutually exclusive, P(A intersection B) = 0 and we have a particular case of the third axiom.
Notice that setting B = not A, we have the inclusionexclusion principle, namely
P(A) + P(not A) = 1
That is all what I applied for computing the probability of getting at least one white in unconstrained dices (if 2 or 3 nonbasic colours are allowed):
P(at least one white) = 1  P(0 whites) = 1  (5/6)^3
As Jan explained, the problem of omitting the intersection in the proposition is clear when you take 6 dices. A simpler case: Take for instance heads or tails when you're flipping two coins.
If you add probabilities, then we have that
P(at least one head) = 1/2 + 1/2 = 1 P(at least one tail) = 1/2 + 1/2 = 1
so, you always have 1 tail and 1 head when flipping two coins. The correct computation is:
P(at least one head) = P(head in the first coin) + P(head in the second coin)  P(head in both coins) = 1/2 + 1/2  1/4
Trevin wrote: Math is hard, so I cheated. I wrote a lisp function to generate all 216 combinations of 3d6 in emacs, then used a regular expression to remove the lines containing at least two nonbasic colors, then simply used egrep  wc to count the number of lines containing either white or gold. (I’m a programming geek as well as a board game geek. ) Well, you counted the number of favorable cases and the number of possible cases, using classical probability. I did the same thing when I computed the probability of being able to active the blue altar in the first turn. I just did it manually and you used emacs, but we both used the same principle
In fact, the greatest thing of the third Kolmogorov axiom is that is for arbitrarily large sets (even uncontable). For the examples here classical probability is all we need.
Edit: Sorry, for the offtopic of he offtopic we are already having, but I had to add that:
Trevin wrote: then used a regular expression https://xkcd.com/208/

Richard Sharpe
United States Georgia

I'll bet playing games with you number crunchers is a real hoot.


