In the card game Spades, what are the odds for a trick making it around the table without being trumped (or someone playing a card of a different suit) for the first time the suit is played, second, third? In other words what are the odds that everyone will have a card of a particular suit (not counting trump suit) once, twice, three times around the table
In my experience:
First trick: ~90%
Second: ~75%
Third: ~25%
I have no idea if there is a way to calculate this.
-USCDiver
I don’t know the rules of spades, you may want to make your question explicity statistical in order to get an explicit answer. I believe you are asking (is this true):
Given a random distribution of the deck into 4 groups of 13, and that I have at least one card of suit x, what are the odds that each of the other three groups also has a card of suit x. Then same question, once 4 cards of suit x have been taken out of the population.
Is that correct? Also, do you want to know the answer when you only have one card of suit x, or n cards of suit x? Obviously it makes a difference, if you had 13 hearts, the odds are zero everyone else has it, whereas if you have only one heart, it seems pretty likely everyone else has at least one.
I have no idea how to calculate the odds. Even if you knew the odds, they probably wouldn’t be helpful in actual playing because all it takes is one funky deal to throw the odds out the window.
Experience has told me never to count Kings if I’ve got four of that suit, and never count the Ace if you’ve got five of that suit. It’s better to discard it (or take the bag if you can’t) than get set.
It’s more effective just to go with your gut feeling based on what the card distribution looks like. If you’ve got
Hearts Clubs Diamonds Spades
2 10 A 2 3 5 7 10 J Q K A J ---
An odds calculation would certainly say that the AH is a winner, but I’d be very reluctant to bid on it because of the funky distribution.
No offense, Att, but that seems like a… uh… non-productive way to play cards. Odds are exactly what they say the are, they don’t pretend there is never a funky deal. In your above example, you have no way to know what the other players have, and the point of odds it to give you helpful guidance about the probabilities.
You might as well live each day as if it’s your last, because one funky asteroid or alien invasion makes any long term planning pointless.
The odds may work out over the long haul (a thousand hands of playing), but a game of spades can be lost in 4 or 5 hands. Relying on odds in the first or 2nd hand of spades is like relying on odds in the first or second deal of blackjack. Knowing the odds is sometimes just not too helpful. Better to duck the odds by playing defensively, IMHO.
Ignoring clues to distribution in the bidding and prior play, the probabilities are based upon the number of cards the leader has in the suit led.
I am reading from a probablity of card distribution for bridge, which should be the same as four-handed spades with a standard deck.
Two cards - hitting a singleton in at least one hand about 12%, a void 1%.
Three cards - 4-3-3 27.5%, singleton about 18%, void 2%.
If the lead is from a seven card suit, a singleton or void will be hit about 86%. More than seven, always.
Like I said, I don’t know the exact rules of spades, but I still have a hard time following you. Perhaps you are thinking of different things by “the odds” than I am. To use your blackjack analogy, if you don’t rely on odds extensively in the first or second deal of blackjack, you are playing very badly. If I am dealt a 20, I do not take a hit, because the odds are I will lose. If I am dealt an 11 against a dealer 6, I double down, because the odds are I will win. If I am dealt an 8, I hit, because the odds are I will improve my hand at no cost. True, card counting is not reccomended at this point (I am guessing that’s what you referring to), but that is not because the odds are to be ignored, it is because the odds say that there’s no point to it.
Yes, my terms are unclear, and blackjack was a bad example (I was thinking of card counting).
In the example hand I gave, there is a pretty good chance that one of the other three players has an oddball card distribution and may even be void in hearts, making the Ace worthless. Further complications come into play during the bidding. If I hold the above hand and three bids before me are 4, 2 and 6 (12 total), I am going to err on the side of caution and not count my AH. BUT- if the bids are 4, 6 and 2 (still 12, but now my partner has the strong hand), I’ll risk bidding the Ace since my team shows overall strength. If the bid totals 9 or 10 by the time it gets to me, then there are so many bags up for grabs that I’ll count the AH, and probably the AC too.
I guess what I’m trying to say is that, although there may very well be a way to calculate the odds in the OP, there are too many variables involved to make relying on those odds worth while.
OK then… now I’m with you! 
You said, “there is a pretty good chance that one of the other three players has an oddball card distribution and may even be void in hearts, making the Ace worthless.” I think that “pretty good chance” is exactly what the OP wanted answered, and I think what aahala answered while we were arguing semantics.
I just remembered that we have this wonderful symbol font and I didn’t use it 
So here:
[symbol]©[/symbol] 2 10 A
[symbol]§[/symbol] 2 3 5 7 10 J Q K A
[symbol]¨[/symbol] J
[symbol]ª[/symbol] —
Cheers for Attrayant!
One very useful strategy in this game is to pass on cards so that you leave yourself short on one suit, this then allows you to dump dangerous cards neutrally and perhaps even clear yourself of another suit.
Its this passing on of cards to the next player that makes probability calculations go out the window since this is based upon strategy and gut feeling.