This is some very hard math. That I can't do this time. (I am to stupid for this, honestly).

So... I need some help.

So sorry if the questions sound like a math test.

***

For a game that I am working on. Players are allowed to have a maximum of only 7 cards in their hand.

When an 8th is about to be drawn. Players are forced to play or discard some cards from their hands.

Only 1 card, is drawn per round, per player. This happens at the end of a round.

A card can only be played, if other cards are sacrificed to gather the costs of that one card. So the real first time that an Event Card can be played, is at the start of round 3.

A card with value 5, needs for example a card of value 4 and 1 to be sacrificed. A card with value 6 may also be sacrificed for this, but any excess sacrifice is lost. There is no mana burn.

The value's:

1x9, 2x8, 3x7, 4x6, 5x5, 6x4, 7x3, 8x2 and 9x1.

A total value of 165.

There are 45 cards in total.

The average value is 3.667 per card.

The game has a maximum of 6 players. If all 6 players have 7 cards. The remaining deck is only 3.

With 4 players, the remaining deck is 17 and the maximum hand can be 11 if we choose to leave 1 card as deck.

To keep things simple, I want 7 cards to remain the maximum.

I am worried that 7 cards per player is not enough for certain cards.

If this is the case. I need to double print the deck for 6 players. Or think of an exception rule.

***

The most expensive card has a value of 9.

Assuming that this card has been drawn and the player plans on using it.

How big is the chance that this card can be played with a hand of 2?

A hand of 3?, 4?, 5?, 6? and 7?

The same for the 2 cards with value 8. From which, one is drawn.

And one of the 3 cards with value 7?

There is no need to look at the value's 6 or less.

***

There is a card with value 7 that is allowed to be played multiple times before returning it to the deck. This automatically means, paying multiple times the value.

How many times can this card be played, given that the player has reached 7 cards?

What if the hand is 8 cards?

9? 10?

***

Whoever answers these questions.

I will be very grateful.

Cheers,

X3M

Stormyknight1976,

Sorry. That doesn't help me at all. Since it is an obvious fact actually.

What I asked for was the situation:

A deck of 45 cards. A player draws the card with value 9 and wants to use it.

How big is the chance that this player will be able to use this card each round. While drawing 1 card each round?

Knowing that 7 is the maximum hand.

If the range is going to be somewhere at 10% or so at 7 cards. I must reconsider adding more cards for the hand. And thus print the deck twice.

If the range is going to be somewhere at 60% or so. It is ok.

This is how far I got:

Other players draw cards as well. But we don't know which ones. So, with 6 players, we can still pretend, there is only one player.

Round 1:

Player draws the value of 9 card. It cannot be spend.

Round 2:

Player draws any value of 1 to 8 by on of the other cards. No matter what it draws. The value of 9 cannot be spend.

Chances to have the following value's ready for sacrifice:

1: 9/44

2: 8/44

3: 7/44

4: 6/44

5: 5/44

6: 4/44

7: 3/44

8: 2/44

In round 3, the chances differ. Since in round 2, any card could have been drawn. And is removed from the deck. Not only that. But the value's ready for sacrifice will range from 2 to 16. And the chance in order to spend the value 9 card after round 3. It needs a sacrifice of value 9 or higher.

I don't know how to put this in the right math for a fast calculation. I guess I need to take the slow road. If Value 9 is satisfying. I don't even have to consider value 8 or less. :)

Except for that value 7 card that can be used multiple times.