Need math help with card hand odds.

So,

You can only play EITHER the light sides of the cards OR the dark sides of the cards BUT NOT BOTH,

AND

EACH SUIT can appear on EITHER OR BOTH of a light side or dark side of any card?

and you are trying to find the probability, say, of having a certain hand given the fact that the hand has 2 possible ways of being played, even though the "objective" may appear only on the light side, and not on the dark side (example - you have a 4-of-a-kind with the dark side of your cards, but the corresponding light sides do not form a 4-of-a-kind)?

and you have 4 copies of each combination of each suit's light-dark combination (4 suits light * 4 suits dark * 4 copies of each card = 4x4x4 = 64 cards)?

If so, I can work this out for you (and explain it), but I would need to know that this is the set of parameters you are working in...

If you want to see the rest like 3 of a kind or pair, you can use the same formula just adjust it.
16/70 = 0.228571429 (first light 1)
15/69 = 0.217391304 (second light 1)
14/68 = 0.205882353 (third light 1)
54/67 = 0.805970149 [b](anything but a light 1 card)[/b]
53/66 = 0.803030303 (anything but a light 1 card)

Which gives you 0.006621161. Add them together = 0.013242322 (for both colors)

You can use this program or similar ones to get the fraction value, but its not in simplified.
http://www.webmath.com/dec2fract.html

Short answer, yes.

hoywolf had it right in his explanation. And, since what I am gathering from your other explanations, the "special" cards will most likely help equally well across the board. If this is true, you should use the raw 64-card deck as your baseline to figure relative value inherent in each card combination. Trying to scale it up to 70 cards using probabilities including special cards that carry a wide variety of "ifs" will negligibly affect the probability outcomes, and waste a lot of time.

This is a "relative" probability, however, so it is actually MUCH more like blackjack than it is like poker, which has an "absolute" probability.

If I gather this explanation, you are turning in (or discarding) cards in order to claim some objective card. The moment you draw any cards from the deck, the probability is now tainted by what you no longer have the ability to play. This means that if you happen to discard 4 light 1 cards, and they have 1-2-3-4 dark sides, the ability for you to get a 1-2-3-4 is now cut by 1/8, since there are 4 combinations for light and 4 for dark, but one of those is DEFINITELY gone with playing the 1-1-1-1 light combination. This becomes further complicated by the fact that your discarded light and dark combinations won't always line up so cleanly.

The probabilities of poker are based upon the fact that eah player gets one set of cards to make a hand out of a deck that is made complete before more probabilities are figured from it (i.e.- new hands are dealt from a full deck each time).

A possible fix to this is to have the gain from completing the objective cards relatively the same. It will shift the strain from valuating objective completion against probability toward the players' strategy toward not being limited in their endgame (ending up with cards that complete no objectives).

Sure, the benefit for putting together a certain more difficult combination should outweigh that of an easier combination, but if completing difficult combinations keeps you from ever making the easy ones at the end of the game, your opponent will likely win by making several easy combinations while you are trying to line up the more difficult ones. This is also compounded by the fact that repeated play by the same players (and you should hope that you get repeat players) will reveal this, leaving both players to avoid the more difficult combinations in favor of not being stuck with no options at the end. This will change the game to one of luck, in which the player who happens to get the right difficult card combination ends up being the victor more often than not if they can score the additional payout for combinations both players are trying to avoid.

If you valuate all card combinations relatively equally (or make multiples of each objective with heavy payout and light payout versions of each one), you should find that planning the USE OF THE CARDS IN YOUR HAND will make the game more challenging than trying to hope for a high-value, difficult-to-obtain objective.

This is just a suggestion, but additionally, you will find that with such even distributions (the number of card types exactly matches the number of times they appear in each combination) the probability of 4 of a kind is so close to 1-2-3-4 that valuing the two objectives differently is relatively moot.

The reason the probabilities vary so greatly in rummy and poker with a standard deck of cards is that the number of cards in each suit varies in a 3-to-1 ratio to the number of suits.

No, I would stick with 3,4,5 or even 6-card combos (depending on hand size), but make the benefit for completing them relatively even (only slightly better for getting 4-of-a-kind than 3-of-a-kind). If you do not, your slippery-slope mechanic would unbalance the game too often.

For example - let us say that the chance of getting 6-of-a-kind (or a 3-3-3-2-2-1 combo, or whatever combo you chose) is 1,520,386 times LESS LIKELY than getting a 3-of-a-kind. Would you make your player get over one million more of the benefit just because he got the luck of the draw? It seems ridiculous, but when dealing with probability - it happens.

Look at the back of a "Quick Draw" lottery ticket (or review the odds for KENO - it's the same game). You will notice that getting 10 numbers right pays out $500,000 in some cases, but getting 8 numbers right pays $50,000. That's a $450,000 difference for getting 2 more numbers right. The difference between getting 1 number right and getting 3 numbers right (still only a 2-number difference)? $2.

I would make a set of payouts that are relatively balanced FIRST, and assign them to the cards in a RELATIVE way (i.e. - 4-of-a-kind is harder to get than 1-2-3, so assign it a "better" payout), not an ABSOLUTE SCALED way (as stated above - so that someone on their first turn gets $1 for playing a 3-of-a-kind and the opponent, on their first turn, gets over $1,000,000 based on their lucky hand - the game would not necessitate a second turn at that point). This will keep the game balanced and make runaways less likely.

That said, and regarding the "quest":

If there is another overall objective inherent in the game (some arc-story-driven quest), you might want to add a choice when completing smaller objectives -

help yourself (collect some payout)

or hinder the opponent (make them discard cards or even a completed objective).

You could then allow the player to do BOTH if they complete a difficult combination, as rated by the previously discussed probability valuation.

This would allow complexity and strategy for gameplay without limiting turn options, and still allow for use of the probability valuation.

Yes, or even, if you think that's still unbalanced, try

3 cards = 1 pt
4 cards = 3 pts
5 cards = 6 pts

because, in addition to completing harder combinations, you're limiting further options (for each 3 five-card combos, you COULD complete 5 three-card combos).

The idea is to use the probability to create a scale, not to create a gauge. A scale is a relative rating which can be "scaled" up or down to fit a set of parameters. A gauge is an absolute reference, which tells you exactly where something is relative to something else.

Also, I like your idea of having the light/dark combos have different meanings or playability in the game. I look forward to seeing how it turns out!