Making a long and bording board game statistically winable

I'm currently designing a board game (purely for novelty sake. Only one will be made and sold for a charity auction) that's supposed to be long and boring.
While I don't think anyone would actually play it (not for the suggested 8+ hours per player, anyway) I do want to make sure I get the math right.
I'll rephrase the rules for simplicity sake:

A player starts from space 0 and must roll a 1d6 die in succession until he reaches to space 10080.
He starts with a score of 3 and must not get a total score of 0 or 6. Getting to either a 0 or a 6 means game over.
A player starts the game by picking 2 cards from a pile 500 and with each roll of a die must pick up as many cards as his roll.
The cards are +1, -1, 0 and "skip".
The player must discard as many cards as his roll in each turn.
Each card affects the player score, with 0 and skip not adding or detracting from it, however the player can't use three 0s in row. Using three zeros in a row (in a single turn) means game over (skips can be used with no limitation).
At the end of the turn, after discarding the cards, the player detracts a point from his score (again, 0 means a loss)
Cards are shuffled once the large majority have been discarded, and once the player reached space 10080, he wins.

The question is, what distribution of cards would make it a fair game (say, a 50% winstate)?

Questions:

1. What does "skip" do?

2. When a player discards cards, is each one played "separately", or all at once?

If it's separately, then obviously the chance of winning is zero, since the score can change only by 1 point at a time, which means eventually you will get to 6 (or 0) before you get to 10080.

Could you explain the rules in detail, so we can have a better idea of how the game works?

Skip does absolutely nothing. Think of it as a free discard.
Example turn:
Player has a +1 card and a -1 card with the score of 4. He rolls a three, advances three spaces and picks up +1 card, 0 card and a skip card. He plays -1>3>+1>4>0>and ends his turn with the automatic -1 for a 3.

Quote: noy2222

Skip does absolutely nothing. Think of it as a free discard.
Example turn:
Player has a +1 card and a -1 card with the score of 4. He rolls a three, advances three spaces and picks up +1 card, 0 card and a skip card. He plays -1>3>+1>4>0>and ends his turn with the automatic -1 for a 3.

You left out the part where the player adds the die roll to the score.

Question: does the game end with a loss if the player has 0 or 6 at any time, or just at the end of the turn (after the -1 end-of-turn adjustment)?

Also, the way you describe it, once a player gets past, say, 10, it's almost impossible to lose, unless there are a lot of -1 cards, in which case it's going to be hard to get past 6 in the first place. Are 0 and 6 the only losing numbers, or is it all multiples of 6, such as 12, 18, 24, and so on (except for 10,080, of course)?

If I'm understanding this correctly (would which be a miracle since you aren't being clear)...

You have a score which is adjusted by the card values, and a position adjusted by the roll of the die. If the score is ever 0 or 6 (the number adjusted by the cards), player loses his turn, goes back to starting position, etc.

If that's the case, and your score can only change by -1/+1 at a time, then the player will end up losing his turn (hitting 0 or 6) if he gets THREE more -1 cards than +1 cards (or if he gets three more +1 cards than -1 cards).

If that is the case, and the other rules given, off the top of my head, I'd say there is no possible way for a card distribution to give a 50% winstate, unless something crazy like 75% of the cards are skips.

ThatDonGuy: I see where I've confused you. There are two separate entities - The space the player is, indicating how close he is to the finish line, and the score the player must balance throughout the game.
The space is purely an indication of how much in total the player has rolled since the start of the game (to use Monopoly terms, space 10 would be Jail, 17 would be community chest, 39 would be Boardwalk, etc).
What I've called in this explanation "score" is the way the player has balanced the point cards, not used too many +1s or -1s to take him past the extreme points (0 and 6).
I hope that makes more sense.

RS: The score carries over from turn to turn and you automatically get a -1 at the end of each turn, which makes the intuitive play to stay on 3 if possible, and if not either 2 or 4.

So let me see if I understand:
The object is to roll a total of 10,800 on the dice.
The player starts with 2 cards, and every time the die is thrown, the player draws that many cards and then plays that many cards (so the player will always have 2 cards in his hand when he rolls the dice); points are added or subtracted based on the card values, and at the end of the turn, 1 is subtracted from the score.
The player wins if his dice total is 10,800 or more before his score ever gets to 0 or 6.

One question: what is the difference between a 0 card and a "skip" card? It seems to me that they both do the same thing - you play a card without changing the score.

You got it.
Skip can be used as many times as a free discard card as the player wants with no penalty.
0 is also a free discard, but can only be used a max of 2 times in a row per turn. If a player is unlucky to have rolled a 3-6 and after picking up additional cards has a hand full of 0s, he instantly lose the game.

Sorry to have made my explanation confusing. It's a difficult game to understand by-design.

Maybe the strategy I am using isn't optimal, but I get about a 25% win rate with 60% +1s, 10% 0s, 10% -1s, and 20% Skips.

It goes up to 36% with 80% +1s, 5% 0s, 5% -1s, and 10% Skips.

Here's my current strategy: ("Level" is the score, to differentiate it from points)
1. Play a +1 if the level is 1-3
2. Play a -1 if the level is 4-5
3. Play a Skip if the previous card was a zero
4. Play a Zero if you can (i.e. if the game would not immediately end in a loss)
5. Play a +1 if you can
6. Play a -1 if you can
7. Play a Skip if you have any
If you can't do any of those, then any card you play is a loss

That sounds pretty good. Can you show your work?

Quote: noy2222

That sounds pretty good. Can you show your work?

Not really - it's based on Monte Carlo simulation, and trial and error for the card percentages.