


I wrote a simple python script to simulate the dummy player drawing cards in a solo game. It shows that the average 6round game gives you about 31 turns assuming random tactics. Of course you can get more turns per game by choosing lowernumbered tactic cards or by choosing tactics that give you an extra turn.
Here's the script: http://pastebin.com/BbcDrxQQ
Note: the script assumes that the solo player always moves first.
Some interesting stats: The dummy starts with 2 reds and a blue. If you let the dummy take a red crystal and red card the first day, that'll cost you about 0.4 turns assuming random picks thereafter. If you do it again that night, it'll cost 0.5 turns. 3rd time costs 0.6 turns. 4th time costs 0.4 turns, and the 5th time costs 0.15 turns. The difference between bestcase and worstcase scenarios is 3 turns.
So when deciding which action/spell to pick, I'm probably never going to worry about what the dummy player is getting. Half a turn usually won't make a difference compared to picking the best card.



I figured half the time the dummy player would go first. So that's 3 turns you'd miss.

Alex Brown
Australia Sydney NSW

More than half? I'm never picking 1 2 or 6 firstturn.



If the dummy starts with 2xred and 1xblue, is the difference of expected amount of turns really only 3 between the following two cases? Assume the player always goes first on each of the 6 rounds.
Worst case: Every round, dummy gets a red card and red crystal. Best case: Every round, dummy gets nonred, nonblue card and red crystal.
I only assumed these are the extreme cases. If there are worse or better cases possible, feel free to correct me.



During first round the dummy has 16 cards.
1/4 of his cards are red > +2 cards 1/4 of his cards are blue > +1 card 1/2 of hi cards are green/white > + 0 cards. So each turn he reveals 3+ (1/4 * 2 + 1/4 *1 + 1/2*0) = 3.75 cards. With 16 cards he gets he will reveal the last card during turn 5 resulting in 6 turns. (7 for the player if he goes first)
worst case scenario: During round 6 the dummy has 21 cards. 9/21 of his cards are red > +7 cards 4/21 of his cards are blue > +1 card 8/21 of his cards are green/white > + 0 cards. So each turn he reveals 3+ (9/21 * 7 + 4/21 *1 + 8/21*0) = 6.19 cards. With 21 cards he gets he will reveal the last card during turn 4 resulting in 5 turns. (6 for the player if he goes first)
best case scenario: During round 6 the dummy has 21 cards. 4/21 of his cards are red > +7 cards 4/21 of his cards are blue > +1 card 13/21 of his cards are green/white > + 0 cards. So each turn he reveals 3+ (4/21 * 7 + 4/21 *1 + 13/21*0) = 4.52 cards. With 21 cards he gets he will reveal the last card during turn 5 resulting in 6 turns. (7 for the player if he goes first)
So if the cards revealed are by chance you will have on average only one turn more in the best case.
If in the worst case scenario you always reveal red cards the dummy will have only 4 turns (player 5). If in the best case scenario you always reveal green/white cards the dummy will have 8 turns (player 9). So the difference can be up to 4 turns in round 6  if you are really lucky/unlucky.
edit: sorry, I just realize that this doesn't answer the question. As I only look at round 6. But if it is only 1 turn in round 6 it is very likely that there are only 3 turns more during the entire game.

Klaude Thomas
New Zealand

jaafit wrote: Some interesting stats: The dummy starts with 2 reds and a blue. If you let the dummy take a red crystal and red card the first day, that'll cost you about 0.4 turns assuming random picks thereafter. If you do it again that night, it'll cost 0.5 turns. 3rd time costs 0.6 turns. 4th time costs 0.4 turns, and the 5th time costs 0.15 turns. The difference between bestcase and worstcase scenarios is 3 turns.
So when deciding which action/spell to pick, I'm probably never going to worry about what the dummy player is getting. Half a turn usually won't make a difference compared to picking the best card. Good post. I've started feeling that for all the hassle of setting up and managing the dummy it would be better to use a simpler system, such as using a dice roll to decide if you get a 4, 5, or 6 turns each round.
For example at end of player's turn 4 the round ends on 6 at end of player's turn 5 the round ends on 56 at end of player's turn 6 the round ends on 46 at end of player's turn 7 the round ends
So your odds are ~28% you'll go to turn 7, and of course ~17% the round will end on turn 4. The odds of a game being only 24 turns long are small: it's possible.
an alternative is to give player a turn after the round end is called at end of player's turn 3 round end is called on 6 at end of player's turn 4 round end is called 56 at end of player's turn 5 round end is called 46 at end of player's turn 6 the round ends
Your odds are the same, but you are forewarned that the end of round is coming. In either case player can call the end of round early if they run out of cards. As for skills, simply do as in the Volkare missions, i.e. pick an MK and stack their skills, flipping one each time you choose skills.
Obv you could map the numbers to manacolours and use manadice if preferred. Let's say 6 is black, 5 is red, and 4 is white.



