Math help needed with "expensive" cards

When I pretend, that the second card had a value of 1. There are only 8 cards of these remaining.

Chances are now all multiplied by 9/44 (20,5%):
Drawing a 1 yields a total of 2: 8/43 = 72/1892
Total of 3: 8/43 = 72/1892
4: 7/43 = 63/1892
5: 6/43 = 54/1892
6: 5/43 = 45/1892
7: 4/43 = 36/1892
8: 3/43 = 27/1892
9: 2/43 = 18/1892

I have to do the same for when the second card is a value of 2, 3, 4 etc. all up to 8.

The biggest problem is that I can't put all the results together for the next draw of a card. Since each pile is going to be different. So, no cutting branches.

Thank you KiltedNinja.

By writing out the problem, it is indeed a simulation worthy problem.

There are 2 simulations to write in total.

But I think it is best to focus on the one that looks at "using" the value 9 card. Before even to look at the value 7 card.

The results should look like this:

Hand of 2:
0% chance of playing the value 9 card.
Hand of 3:
##% chance of playing the value 9 card.

By logical reasoning. Having a hand of 9 actually has almost 100%. And a hand of 10 or more will have 100%.
In a sense. It is possible to discard the low value cards. So it would sound reasonable if the hand of 8 and 9 are calculated as well. Of course, by discarding the lowest value or 2 value's for those hands.

I am going to get to work on the hand of 3. That one is still doable on paper.

Now that he (my cousin) got the '7' card right (the simulation rules). I am starting to have doubts about this card.

Still awaiting the same simulator for this card as if it is '6'.

It is very important for me to know when this card can be played.

With it being '7'. Chances are that it can be played in round 3 or 4 already. But the chance for it to fail completely is 5/6th.

In round 5 and 6. The chance is multiplied again by 5/6th. But knowing that the sacrifice is doubled for this. I don't know.

Chances to fail are multiplied every time by 5/6th. And the highest chance to get to this next level seems to be growing longer and longer.

I am starting to believe that it is best to use the card as sacrifice instead. No choice.

So, I await the results on it being value 6 instead. I have good hopes. And the print is already adjusted. But next time, it isn't the value that is going to be adjusted. But the chance on having an effect instead.

But I don't know yet what to do with that.

And change it into:

Or change it into:

Thanks. I will look into it, somewhere this evening.

In the meantime. I need to do something about my "lotery" card. It is still that card that I described in my last post.

I realized that players could count a bit to much. So, when one deck ends. The entire graveyard deck is going to be shuffled. Before it is used again.

But that doesn't take away the luck that some players can have with the lotery card.

So I need to take away this luck. When a player gets it a bit to much.

Some partial solutions:

A higher chance on the bonus. But limited to one. I see no other way around it any more. So basicly choice 3 from the last post.

Increased cost when you have played the card before. But some games allow for more AP or less AP. It also depends on the campaign. The value should remain 6 for sacrifice.

But the cost for using the card? I am thinking about several stats.
Starting with 6? And add 6 when a player is +1 AP ahead on the rest?
Or just adding 2 or 3 for the costs?
1 seems a bit to low.

I'll keep pondering.

All players have 7 AP per round.
There are cards which change this number for 1 round.
But the problem child is the one which does it permanent.
But using the card with a die on it is also a good idea. This way, it isn't permanent either. Making the AP optional to spend, makes it better.

No stacking over time. Maybe only a tempory stack.

Things where a bit more complicated then the link you provided. But it can have its use for future efforts.

Also a big thanks to KiltedNinja for the simulator that he send me. It helps a lot for the first 7 rounds with 6 players.

Clearly, the dreadful lotery card that I needed to look at, needs to be worked on. Lewisher doesn't even know that one of his video's helped me as well. So here it is. Thumbs up.

I am not going to make a new topic for that card. Since it is decided to simplify that bastard to a 1 time dice roll per each usage. 50% chance the card works. And a cumulative activation cost. WiTh cumulative sacrifice for the big picture. But maybe the last part can be left out. In theory, every player will pick the card once every 45 rounds. And with 50 percent chance of it working. It translates to activation every 90 rounds.
In game time, that translates to 15 hours for 2 players. So yeah, no big picture :).

Anydice might have the last word in this. Instead of cumulative costs. Which is clunky. Maybe using multiple dice for one roll. And equal these to the number of ap that the player can spend.

The card gets introduced when players can spend 3 ap per round in the single player campaign. So the minimum dice is going to be 3. So the fail part could be if 3 dice roll a... 6?

This evening I will find out.

Yes... you are a tiny bit late. But that doesn't matter. All feedback is welcome. And you have showed some numbers as well that I have seen as well. 1776, I remember that one. And looking back in my papers, there is only one thing that I don't understand.

How come, I have 31,3%? What I forgot to do was to multiply this with 3/45. Which is the chance to have the 9. Which is indeed 2,1%.
Thank you for pointing that out.

Same goes for the 5%. In my case, I had 70,8%. The chance to get the 9 card is 4/45. Wait, I get 6,3%. Where did I go wrong?

I think that your suggestion is the road I need to walk to see the playability of the 9 card. INCLUDING drawing it by random.
If the card is going to be played by any player. For that, KiltedNinja has supplied me with a nice simulator, that includes some "ai" as well.

***

Feedback from my gamegroup. 2 players suggested to have 1 card per team after the game. And that players decide together, in how to spend that card. But I dislike that. Why? I think that the decision of a card or piece should stay with one player. co-op is ok. But if multiple players decide for one piece or card. This brings a higher chance to "unrest" between "co-operative" players.

Am I right? Or is this a matter of opinion?

***

Now, to change my lotery card. I had trouble finding time to even think about the details. The way how I look at it now is to have the card spend only one time. It costs 6 AP. How the card could work?

You roll an equal ammount of dice as the number of AP that you may spend in total by default, each round.

This means that you may roll 7 dice for most games. There are games that start with only 1 AP per round, but the card is introduced at 3 AP. Then all cards are introduced actually. Main reason is the Assault/Retreat card. Normally this tactic costs 3 AP. But with a certain card, it only costs 1 AP. Huray!

So, 3 dice is the minimum. And I want the player to have a fail roll when he/she rolls at least 3 dice with a, 6?
5 and 6?
4, 5 and 6?

Kinda weird to do this in any dice. I need to invert the dice rolls. Telling it that a fail will be worth 1. And simply look at the chances to roll at most 2. In short words. I am fooling the calculator to do a calculation in a simple way.

It is a shame that I can't activate "at most" and "transposed" at the same time.

With 6 being the fail.
Number of dice: chance in % to succeed
1 and 2: 100
3: 99,54
4: 98,38
5: 96,45
6: 93,77
7: 90,42 and let me stop right here. This chance is to big. My goal is around 50% for 7 AP increased to 8 AP

With 5 and 6 being the fail.
Number of dice: chance in % to succeed
1 and 2: 100
3: 96,30
4: 88,89
5: 79,01
6: 68,04
7: 57,06
8: 46,82
9: 37,72 still rather big on 7 but 9 seems the opposite already. On the other hand, I need to consider the other AP cards and the chance to actually get this card in the team. Further, to get to 9, you need to pass the 8. Thus getting the card AND getting the upgrade.

With 4, 5 and 6 being the fail.
Number of dice: chance in % to succeed
1 and 2: 100
3: 87,50
4: 68,75
5: 50,00
6: 34,38
7: 22,66 waaaaay to low.

It is a decent result imho. Playtest imminent.

How to write the card??? Will this do?