While I've been playing I was wondering something. What are the odds, when you attack, that the defender will draw into something that could beat you? Attacking with something like 3 Aces is obviously pretty safe, but in situations with marginal cards, I was wondering if it made sense to attack or not.

The odds are actually quite a bit different than normal poker odds. Here, you and the opponent are each drawing from a separate deck, so your having drawn a particular set of cards doesn't prevent your opponent from drawing them too. More significantly, attacking in Aces & Adventures is quite different than poker. Rather than making the best hand you can using five cards, you sort of play "hands" directly, and attack with just a high card, pair, three- or four-of-a-kind and so on, out of the cards in your hand. And then the opponent draws cards (equal to their defense minus your offense), but in matching your attack they have to match the number of cards you used. So, if I'm attacking with a pair of 10s, and the defender draws eight cards, the question is "How likely it is for them to draw a higher pair than me?", not a how likely it is for them to draw into a better hand overall.

I decided to try and calculate the actual probabilities, and it was more complex than I thought at first since the combinatorics of it ended up being pretty involved. It's easy to calculate the probability of drawing into no pairs, but it's much more difficult to calculate "What's the probability of drawing two, or more, Jacks, or better, within eight cards?". It took enough work to do that I'd figure I would share the results in case anyone else was curious.

• All the probabilities below are only considering the pure case of comparing your attack hand to the defender's. I'm completely ignoring:
  
  • Abilities
    
  • Safe Attack
    
  • Any and all other special cases due to conditions, traits, equipment, everything else in the game, etc.
• I'm also ignoring Jokers (except where noted) and Wild Range/Suit.
  
  • They don't matter much for the defense since the enemies don't get their own wild cards.
    
  • The player does get wild cards, but I'm not really talking about the odds of getting a hand in the first place, just the odds of winning when attacking with it. In that case, two Kings and a Joker function exactly the same as a natural Kings three-of-a-kind.
• I might have gotten some calculation or another wrong, or otherwise made a mistake somewhere.
  
• And, most of all: None of this really matters very much! This is just something I thought was interesting, but the impact on how you'd play the game is pretty small.
  
  • If you just care about playing the game and having fun you can safely ignore all of this.
    
  • If you want to play better, read any of the other guides, they'll be a bigger help.
    
  • If you care about playing the game very well, then this still doesn't matter much because Abilities matter a lot more than the raw card probabilities.

This is all mainly just something that's interesting to consider, with a few things that have strategic impact in a couple of corner cases. So with that in mind, let's take a look.

To attack and win with a single card means the defender needs to not draw any cards that are equal or higher than the one you're attacking with. The more cards they can draw, the more likely it is they will do so.

So if you attack with a single 8, and the defender draws eight cards, the only way to win would be if they only draw 2s, 3s, 4s, 5s, 6s and 7s in those eight cards, which isn't impossible but is very unlikely (0.1% chance). However, it doesn't matter if they actually draw into any pairs or better though! With 8 cards of ranks 2-7, the defender is guaranteed to have at least a pair or better, and maybe even a straight, but it doesn't matter for them, since they defend with just a single card.

The charts below are showing the percentage of the time you will win, when attacking with a single card of a given rank (each row), versus a defender who is drawing some number of cards (each column).
So for example, the top left cell shows that if you attack with a single [2], and the defender is drawing a single card, you have a 0% chance of winning (because anything they draw will beat or draw to you). The bottom right cell shows that if you attack with a single Ace, and the defender draws 8 cards, you have a 50.1% chance of beating them.

All the tables below shows the odds of winning only. To save space, I am not showing separate tables for losing and for drawing, because they are mostly quite similar. Although there is some nuance, like how Aces can draw but never lose, and 2s can draw but never win. However for the most part, just looking at straight win percentage is illustrative enough. Think of it as "the chance you will damage the opponent", and take it as slightly underestimating "the chance you won't take counterattack damage".

• You can see the odds of attacking with a single card are generally not that great, and fall off rapidly if the opponent can draw more than a few cards.
  
• The dropoff as the opponent draws more cards is steeper than I would have guessed. Attacking with a Jack versus them drawing a single card, you'll win more than 2/3rds of the time. However if the opponent can draw just two cards, attacking with a Jack is worse than a coin toss.
  
• Aces perform well however! Attacking with a single Ace means you're guaranteed not to take counterattack damage.
  
• A single ace has a better than coin-flip chance of winning, even if the opponent is drawing 8 cards(!)

• You shouldn't attack with a single card if you could attack with a pair+, it isn't worth splitting an KK attack into two single-K attacks, even if you have the attacks to spare (of course with the very big exception of unless you have cards that care about single-card attacks, like the Hunter, but we're just talking about hands in isolation).
  
• If the defender is only drawing a single card, odds are good attacking with a 9 or higher
  
• Only attack with a single King if they're drawing 3 or less, a Queen if they're drawing 2 or fewer
  
• But attacking with even a single Ace is always safe, and has better odds than even some low-rank pairs.

It's quite easy to make a pair. Drawing 5 cards, and excluding jokers, you'll have at least a pair about half the time (49.3%), and if you can draw more cards or have wild cards the probability is even higher.

Pairs automatically beat a defender drawing only a single card, if your offense is higher enough than the enemy's defense. But even outside of that, the odds of winning are much more in the player's favor with pairs than single card attacks. Why?

Let's say you're attacking with a single Jack and the opponent is drawing two cards. They have a lot of 'outs' that will draw or beat you: any J, Q, K or A, each of which they have four copies of, so they have sixteen opportunities to prevent a win (four to force a draw, twelve to beat you). So they just need to draw any of those sixteen cards in their two draws, which they have a 52.5% chance of doing.

But if you attack with two Jacks, the math is quite different, because now they need to draw a pair, they need JJ, QQ, KK or AA. They need to draw the right card twice, and for it to be the same one (so King + Ace won't do). There are 1,326 distinct two-card hands (52c2, see the last section). Of those, just 24 will prevent a win. The four ranks that can beat/draw us, in any two of the four suits. That's a 24/1326 = 1.8% chance, dramatically lower than the almost even odds of defending against a single Jack.

• Attacking with any kind of pair has good odds if they aren't drawing too much in defense. Even a lowly pair of 2s will reliably win against four or fewer drawn cards.
  
• But the odds for low rank pairs do drop off if the opponent can draw a lot though
  
• Aces are still great, but they're not that much better than Kings or Queens (unlike with single card attacks).
  
• There could be instances where it'd be worth breaking up a high rank 4-card attack into two pairs, or triple Aces into a single and pair. I don't think it'd come up often, but if you were in a situation where multiple attacks would be better, the odds increase going from a high pair to three- or four-of-a-kind is so marginal that you could split them up without hurting your chances.

• Any pair is a good bet against a <5 card opponent
  
• 10 through Ace pairs are generally good even if they're drawing a lot of cards

Three-of-a-kind hands are much rarer than pairs. Whereas almost half of five card hands will have a pair, only 2.1% will have a three-of-a-kind. But that rarity goes both ways- if you are able to attack with trips, the odds the defender will also draw into it is very low. As a result, three-card attacks will almost always win, as you can see in the chart below.

In fact, the odds of winning with three-of-a-kind is so much higher for low-ranked hands, especially for low-ranked pairs, you will actually get more marginal utility out of using Jokers on low-rank pairs over high-rank ones.

Here's what I mean: Say you have two attacks, and the defender is going to draw into eight cards, and in your hand, you have a Joker, two 4s and two Queens. If you start to attack, the game will try to make the best hand possible, which in this instance will suggest the two Queens and Joker to make three-of-a-kind Queens. If you attack with that, the QQQ will have a 97.2% of getting through, leaving you with the 44 on the second attack, which has only a 18.7% of succeeding. So the combined probability is a 97.2*18.7 = 18.2% of landing both attacks.
But if instead, you use the Joker with the 4s, the 444 will have an 89.8% chance of winning, followed by the QQ with a 71.1% chance. This combined probability is much higher, 89.8*71.1% = 63.8%! We've increased our chance of winning both attacks by over x2.5 fold as a result of upgrading the low-rank pair instead. I find that pretty surprising, even if the situation when this applies isn't that common.

• Trips will automatically beat a defender drawing only one or two cards.
  
• And will reliably win even if the defender is drawing a lot of cards.
  
• Even low ranked triples! Triple 2s has better odds of winning against an opponent drawing seven cards (94.9%), than a single Ace has against an opponent drawing a single card (92.3%)

• Attacks with three-of-a-kind will probably succeed most of the time, especially if it's a mid rank or higher.
  
• The odds increase from two-of-a-kind to three-of-a-kind is very significant for low rank cards. As a result, you'll get more marginal utility from using Joker with a low-rank pair than a high-rank pair.

Four-of-a-kind hands are quite rare, just 0.024% of five card hands, which means you can expect to see one once every 4,164 deals on average. Having Jokers makes them significantly more common, though still quite rare- if you have two Jokers, four-of-a-kinds would be over 1000% more common, but still well less than 1% of the time.

Consequently, they're extremely likely to win. Quad 2s is 99.7% likely to win even against eight cards, and everything else is better from there.

But because three-of-a-kind hands already were so likely to win, the same thing applies about using Jokers. It generally isn't worth it to use a Joker to turn a three-of-a-kind into a four-of-a-kind, especially if they're high ranked cards. You're likely to win anyway, and so you're better off saving the Joker for something else.

• It's hard to beat a four-of-a-kind, especially since the enemies can't use straight flushes.

• Attacks with four-of-a-kind will almost always succeed, even if low ranked.
  
• But because three-of-a-kinds already have such high odds to win, it may not be worth spending Jokers to make one into a four-of-a-kind.

Normally the only kind of four-card hand you can use is a four-of-a-kind. But a two pair hand would also be four cards, except that ordinarily you can't use two pair as a valid attack hand. However, there is a daily challenge condition that permits using two-pairs (and maybe it pops up elsewhere in-game somewhere, I'm not sure).

I haven't played with it very much so I'm not sure exactly how they work here- can two-pair attacks only be defended against by another two-pair? And is it high pair wins? (eg, JJ + QQ loses to KK + 44). If so, then two-pair attacks have higher odds of winning than single pairs, but are worse than three-of-a-kinds. I won't show the table for them since I'm not sure how the game logic works exactly, and they're not part of the normal gameplay anyway.

5 card hands include, in order of their rarity: flushes, straights, full houses, straight flushes, and 5-of-a-kind hands, which are possible using Jokers. The probabilities vary, but all them are fairly rare (straights are the most common, and they're only 0.39% of five card hands). But in game terms, there's no difference between them. Because enemies can't use 5-card hands, these all automatically win regardless of how many cards they can draw in defense!

We don't need a chart for this one. It's 100% under any circumstance!

• It is kind of funny the enemies can't use five card hands, I wonder why the designers settled on that. Since there so rare, maybe it just wasn't fun for any enemy who draws 5+ cards to have an extremely rare chance of beating you randomly? Or maybe it makes the hand optimization calculation too finicky.

• Straights and flushes are the only way to attack with otherwise unmatched cards, so they're usually worth using when they come up, if only to cycle the five cards.
  
• Five-of-a-kinds are almost never worth spending a Joker on (well, besides doing it once for the achievement), because a four-of-a-kind already has extremely high odds, and so does a high-ranked three-of-a-kind
  
• If you care about making multiple attacks, it might be worth breaking a fullhouse into a separate pair and triple. The odds will be lower, but if the pair is high ranked, it may not be that much lower.

I'll come back to this and lay out how I calculated the numbers above. It's a little hard to do math formatting in Steam guides though. But I'll fill this in later for anyone who might be interested.

Okay I was dragging my feet about this since I didn’t think anyone would really be interested in seeing the math, but somebody was!, so I figured I would oblige.

Probability mathematics can get quite complex and involve a lot of calculus. But thankfully, calculating the probability for things with fixed outcomes like dice and cards can be fairly straightforward, and you can get pretty far with just some basic algebra. There are many good resources for learning more about this. For starting at the basics, you could try something like Khan Academy’s series on probability[www.khanacademy.org]. But if you'd like a more in-depth treatment of this, one of the best resources I’ve come across is the excellent textbook the The Probability Lifesaver (2017) by Steven J. Miller[web.williams.edu], which I highly recommend.

In general, there are two approaches to calculating hand probabilities in situations like this. You can solve it analytically (deriving the precise formula that calculates the probability you want), or you could solve it via numerical approximation (generating a statistically significant number of observations and then sampling the results). Approximation using Monte Carlo integration is generally the easiest solution. In an environment like MATLAB, you can write up a quick program to draw a million five card hands, and then count how many are full houses (or whatever). But that can also be a little disappointing since you never learn the underlying formula. It can also be tedious when you have parameters like the number of cards the opponent draws, since you'd need to generate a new million hands with them drawing one card, another million for two cards, and so on. So I went the analytical route.

For the 1 card hands, the math is simple enough we can use a straight combinatoric formula. We want to know, if we have a card of a given rank, what is the probability the opponent will draw an equal or higher one in 1-8 draws? For simple questions about successes in a sample drawn without replacement, we can use a formula based on the hypergeometric distribution.

Let’s say we’re trying to determine the possibility of winning with a 6, when the defender draws one card. There are 36 cards that will tie or beat a 6 (6 through 10 and all the face cards = 9 ranks, multiplied by 4 suits = 36). So we’re asking, if the opponent draws 1 card from a deck of a 52, what are the odds they fail to find any of those 36 cards? This is simple enough, we know this is (52-36)/52 = 30.8%.
But it gets trickier with multiple draws, like the probability of winning with a King, when the defender draws 8 cards. The defender has 8 ways to tie or beat our King (the four Kings and four Aces). But with 8 draws, it isn’t as simple as (44/52)^8. Because as we draw cards from the deck the probability changes. If we’ve drawn 7 cards without finding a King or Ace, the deck is now 45 cards, not 52, and so on. And also, we don’t care about multiples- drawing two or more Aces is fine. This is why we need a formula.
In spreadsheet software, we can use =HYPGEOM.DIST for this. It works like this:

So for the probability of winning with a 6, when the defender draws 1 card, it’s
For the probability of winning with a King, when the defender draws 8 cards, its:

And that lets us calculate the whole table of possibilities very quickly.

The basic way you calculate any probability is by counting the number of things you want to happen, and then dividing by the number of possible outcomes.

The number of possible outcomes, the divisor, is actually pretty easy to calculate. We have a 52 card deck, and the opponent is drawing N cards. How many ways can you draw 1 card from 52? 52 ways. How many ways can you draw 5? It's 52 choose 5, which is =COMBIN(52,5) in Spreadsheet software, and is written out as a tall bracket. So to figure out the probability of any given situation, we need to find out how many ways the opponent can make a hand, and then divide it by 52 choose N, where N is the number of cards they drew.

Figuring out the number of ways to make a hand can be challenging, especially when there are more cards than we need in the hand. You can see what I mean by looking at the Wikipedia page for Poker probability[en.wikipedia.org]. As an example, here's the number of ways you can make a full house in a hand of five cards:
To translate: This says, we take 13 ranks and choose 1, and we multiply that by taking the 4 suits and choosing 3. That gives us the number of ways to get three-of-a-kind. Then, we take the 12 remaining ranks and choose 1 and we multiply that by taking the 4 suits and choose 2, to get the number of ways to draw into two-of-a-kinds. The end result tells us there are 3,744 ways of drawing a full house in five cards. Intuitively this seems pretty easy, the combinations line up nice and neatly.

But here's the formula for the number of ways to make a full house in seven cards. Uh oh:
What happened? Why is this so much more complex? Because with five cards, there's basically only one way to make a full house (3 of one kind, 2 of another). But with seven cards, we actually have three cases we need to consider. First, we could draw a three-of-a-kind, a two-of-a-kind, and then two unrelated cards- that's what the third row shows. But we could also draw a three-of-a-kind, a two-of-a-kind, and another different two-of-a-kind, which is what the second row shows. Lastly, we could also draw a three-of-a-kind and a second three-of-a-kind, and an unrelated card, which is what the top row shows. All of those would be full houses from seven cards, so now we need to add all those up separately.

So this means calculating something like, "How many can you draw a three-of-a-kind with seven cards", means you need to anticipate the permutations like drawing two different three-of-a-kinds, or a three-of-a-kind and a four-of-a-kind, and so on. As far as I know, there's no universal way of generating these besides figuring it out by hand. So that's what I did. To generate the numbers in the tables above, I had to figure out each case and calculate the combinations. It was... pretty messy and tedious. This is what part of my spreadsheet for this looks like:


Yikes. I was hoping to find a general method, like, a single formula to output "What's the probability of drawing an X-of-a-kind, or better, in an N card hand", and then I could check the output against the work I did. But I don't think there is a general method, so I sort of gave up on it, and just cleaned up the output enough to finish the tables referenced above. Long story short, I should've just brute forced it by generating millions of hands! But I hope something here is useful or interesting to you.