Using a simple card game as an encounter

[Fugu]

My villain loves games and challenges of wit. In tonight's session, I want her to propose a quick card game to my players in lieu of battle. I looked online for a good strategic game that would be quick to teach and play, but for lack of a game that satisfied me, I designed one. I'm posting it here because I want to spot problems in advance, if possible. Let me know if you have a way to improve the game, or if you have experience with this unusual type of encounter. Even if you read this after tonight, go ahead and comment. In a few days, I'll share how it went down.

I designed the game for ease of learning, pace of play, and the appearance of strategy. It doesn't have to be all that deep of a game since we're only going to play it once or twice (if the players even accept the villain's challenge).

The game is for 2 players. (The characters will play as a team. The villain will deal.) The object is to collect the highest total value in each suit.

After shuffling, each player receives 4 cards face-up and sorts them by suit. Then the dealer reveals the top four cards of the deck and asks, "1st or 2nd?" If the other player chooses 1st, he selects 1 card and places it with the others of its suit, then the dealer chooses 2 of the remaining 3 cards and places them with their suits. The 4th card is discarded. If the other player chooses 2nd, the dealer picks 1 card, the non-dealer picks 2 cards, and the 4th card is discarded. Play continues until the deck is empty. Score by determining which player has the highest face-value total within each suit. If a player controls 3 or 4 suits, that player wins. If there is a tie, discard all Aces and score again. Repeat this process with Kings, Queens, etc., until the game is no longer a tie.

[erikun]

I would find it very odd if the players lose by simply choosing "second" all the time. We're talking about a card advantage of 28 to 16; it will be remarkably easy to simply allow the villian the one suit they want, while going after two other suits (and allowing the third to languish). The only thing that I could see different would be if face cards could become unusually more valuable.

Other than that, pretty interesting! It's nice to see a good non-combat encounter. I expect you are planning on handing out XP if the players win? (They'd probably be disappointed if they missed an XP reward by choosing the interesting route.)

[Surrealistik]

http://www.vgcats.com/comics/?strip_id=12

[Fugu]

I think the main thing that prevents 2nd from always being correct is that as the game develops, each player will get far enough ahead in certain suits that they aren't really important anymore. The "swing suits" are the important ones, and choosing 2nd could give the opponent a high card in a swing suit. I played two test games (once with a real opponent), and 1st was usually the more appealing play. But I could have been playing badly!

I think if you choose 2nd every time, your opponent will, on average, get a card in the upper quartile of the best suit available, and you'll get two cards in the center quartiles of the second and third best suits. It's misleading to call that an advantage of 28 to 16. Sometimes a Queen to swing Hearts will be more important than 8 and a 5 to pad Clubs and Spades. There is probably a mid-game point when switching from 2nd to 1st becomes correct.

(If this were an actual card game, choice of 1st or 2nd would alternate between players, but I want to weight it in favor of the group since I have the natural advantage of having designed the game.)

[erikun]

Mathematically speaking, it would be very hard for the 1st choice to win, at least against an opponent uninterested in losing.

Consider the following cards in each suit: 23456789JQKA. If we give P1 the top 1/4th, give P2 the middle 2/4ths, and discard the lowest 1/4th, we get an approximation of you game. This gives us the following numbers:
234 = discarded
56789J = P2 hand = 45 points
QKA = P1 hand = 36 points

As you can see, Player 1 (the one with the 1st choice) clearly loses if trying for the highest card in each suit. It seems doubtful that Player 1 could even win three suits, with the card advantage that Player 2 has. Player 1 could clearly dominate a single suit, but doing so would lose the other 3 suits to Player 2, thus losing the game. As best, we're looking at Player 1 trying to bring the game to a tie, winning 2 suits, but never fully succeeding.

Not that that is a problem in your case. For one, the players may not use the optimal strategy. For another, it means you have a fairly good idea of how the game shall proceed. I'm just pointing out that, unless I missed some big rules somewhere, then whoever is making the choices can easily control the winnings.

[Hidden Sanity]

The biggest peoblem is that whoevers deciding '1st or 2nd' will always be able to pick the best option; leaving the other player to table-scraps. I.e. if three cards of similar vaule come up, you clearly pick 'second' and if there's only one decent card, you pick 1st, ect... and no matter how much schemeing or playing the dealer does, the dealer will always fall behind against a skilled foe... then again, it might take a game or two for the players to figure this out.

[Blazen]

The easiest solution I can come up with is the choice of 1st or 2nd alternates each turn.
You may also want to put in some form of bluff or stealth check for the players if they decide to go a dishonest route.

[Fugu]

We didn't get to the card game tonight; it will happen next week. But I can change the game before then if I want. Now, numbers! I'm going to do a lot of probability type stuff and it's, well, probable that I got some of it wrong. Please correct me if you can!


234 = discarded
56789J = P2 hand = 45 points
QKA = P1 hand = 36 pointsAre we using the same point values for each card? It looks like you've skipped 10 and reduced the face card values by 1 each. But your model is on the right track: 1st chooses the highest every time, 2nd gets the middle two, and the lowest is discarded. Also, there are only eleven rounds, because each player received 4 random cards at the beginning. We have to account for the absence of the 12th and 13th cards from the chosen rounds, which means expressing each player's point total as a range of scores given the possibility of any two cards being removed. If, for simplicity, we examine a suit in which P1 chooses three cards (which will be the case for three out of four suits), we get the following:

P1 = three of 10JQKA = 33-39 points
P2 = six of 45678910J = 39-51 points

Each player also adds that random card to their score, which could be anything from 2 to A. That means:

P1 = 35-53
P2 = 41-65

So if P1 always chooses first and always takes the highest card, he'll win suits some of the time thanks to lucky pulls at the start. We can't say how often, since we haven't measured how many games result in, say, a P1 score of 50. It's more complicated a measurement than I care to make.

Details: There's only one way each to achieve the extremes of each range, and only one way to achieve the penultimate on either side (change the value of the random card added by 1). But for P1's extreme +/- 3, for example, there are at least three ways, because you can adjust the random added value or the value of either of the two missing cards 2. The further one goes from the extreme, the more dizzying becomes the array of possible adjustments to the random value added and the randomly missing cards.
But we also have to account for the fact that a rational player choosing first won't go for QKA in all four suits. He will ignore one suit and seek JQKA in each of the other three. That means P1's range is:

P1 = four of 910JQKA + a random card = 44-64

Assuming that the suit P1 is ignoring is obvious, P2 can eschew cards of that suit as well, but only in exchange for the low-value cards that would otherwise be discarded. P2's range becomes:

P2 = eight of 2345678910J + a random card = 46-74

TL; DR
So if Player 1 always chooses first, Player 2 always chooses second, and both players make rational selections, it looks like Player 2 will win more often than Player 1. The outcome depends on the average score more than on the ranges, but the average is too difficult to obtain. In any case, it is not clear that Player 2 will win more games always choosing second than he would have by occasionally choosing first. Therefore, the optimal strategy is still hidden to me. My guess is that choosing second is best early game when you have poor information, and choosing first becomes increasingly better as the game progresses.