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19

The calculation you are looking for is called a Hypergeometric Distribution. This calculated your chances of drawing a particular number of "successes" from a population, without replacement. Population Size: 60 cards Successes in Population: 4 Birds of Paradise Sample Size: 7 cards Successes in Sample: exactly 1 Results: 33% The online calculator will ...


15

All of the detailed probability calculations and Markov analysis posted by Eric P. and ire_and_curses can be distilled into a simple set of Risk attack heuristics: Large battles favor the attacker but only very slightly. For small battles, attack if you have more armies, stop if you don't. The rationale for these guidelines is outlined below. A large ...


15

The odds of drawing a particular card in a 60-card deck are obviously 1/60. If there are four such cards, the odds are 4/60. The odds of NOT drawing one of those cards in the first draw is 1 - 4/60 = 56/60. To calculate the odds of the entire first hand, we can do it backwards: The odds of not having any of the four cards in the first card is 56/60 (as I ...


13

Since the edit, there are two parts to the remaining question: "When have you definitely lost?" and "How can you tell when you've probably lost?". The first is easier to answer than the second, so I'll start with that. The minimum you can roll on a go is a two and a one (note that a double one would give you four moves). The maximum is double six. So ...


12

Looking at the work that's been done so far, it seemed to me that this is basically just a Markov chain, and in looking for ways to calculate Markov chains, I found that someone has been there and done that. The data from his work matches the simulations here, which should be no surprise given that the rules are very easy to code: Minimum game length: 3 ...


9

The mathematics to take into account the fact it cuts off at 0 and 10 (i.e. spinning -2 when you are on 1 will leave you on 0, not -1) and the "loose all" result are quite complex. Instead I've opted for a simulation. This was run with 10,000,000 simulations, which should be more than enough to get a good result. Here is a histogram of the result. The ...


9

Are you up for a DIY Monte Carlo simulation? I don't know of any specific applications, but if you're comfortable with coding I suggest you gin up a little Monte Carlo simulation. In case you're not familiar with how that would work I'll explain (and can elaborate later if necessary): Populate some sort of database or data structure with your cards. ...


9

A good paper by Jason Osborne can be found here. (It's a correction to an earlier paper by Tan.) He uses Markov chain calculations to get the exact probabilities. You'll especially want to look at Table 3 on page 6, which has these probabilities rounded to three decimals for up to 10 armies per side. I've reproduced it below: As to ease of use: just print ...


8

I cannot imagine why you ask this. But anyways, with the rules you just put it's easy to calculate. (Though I have not played with the last card discarded rule.) You can win from the start with either Three+Straight or 7 Straight or Three+Poker. Now we just add the probabilities. Assuming you play with 4 jokers Using the Hypergeometric distribution you ...


8

Check out the Axis & Allies Combat Simulator/Calculator. You can choose which variant of the game you're playing (Original, Revised Edition, etc.), how many simulations to run, and a number of other options. (There's even a iPhone and Android versions!) The combat simulator then tells you the average IPC loss for both sides and the average winning ...


8

Short Answer: No. Long Answer: The odds of getting a 5:2 split over the more common 4:3 split is unchanged, even if the first card is intentionally chosen to be a copper. Mathematically, this can be proven using hypergeometric distribution. This can be calculated using the following function, where N is the size of the deck, n is the number of cards ...


8

As stated in the Risk FAQ the expected losses per attack for standard Risk rules is about 6 to 7. This means the attacker is expected to lose 6 armies for every 7 defender armies destroyed. Since we're talking about expected values, this represents the mean (average), which is most akin to a 50th percentile or 50% chance that that is what will happen. ...


7

I ran a script to calculate probabilities of attacker losing two troops (D wins), defender losing two troops (A wins) and each losing one troop (Tie). The following table shows these probabilities, along with the difference from standard play in brackets. The final column shows the troop loss ratio per die. Number of | Win Probabilities (Difference due ...


7

Although the full calculation to discover whether you will win a sequence of battles is difficult to make, it is easy to calculate the chances of winning any particular combination of attacker and defender dice. I reproduce here the table of expected losses described in this paper. Defender Dice 1 2 ...


7

Since no one has provided the correct answer, at least not they current way you have it worded, I will take a stab at it. What are my odds of drawing at least 1 of that card in my initial 7 card hand? Answer: 1-Hypergeometric Distribution(population=60,successes in population=4,sample size=7,successes in sample=0) = 1 - .600500 = 39.95% Everyone got this ...


7

The final answer will be obtained using P(winner) = P(winner on brain storm) + ( P(cantrip but no winners on brainstorm) * P(winner on cantrip) ) There 23 wins to be found in 50, so there are 50-23 = 27 non-win to be found in 50. P(no winners on brain storm) = (27/50)(26/49)(25/48) = 15% P(winner on brain storm) = 1 - P(no winners on brain storm) = 1 - ...


5

This is actually a pretty interesting question. I also decided to write a simulation. The results are quite interesting, and tend to show a bit about how careful you have to be to get a good simulation. For those who care, here's a bit of code: [Edit: replacing code due to stupid mistake] [Edit2: added code to compute average gain per spin] #include ...


5

I wrote something to enumerate the relevant states and then print out the probabilities it found that way. Somehow a simulation felt like a cop-out to me :) I believe this game is for more than one player? And it ends when ANY player gets 10 cherries, right? The code and results below assume that. For the two-player case, I got a mean end turn of around ...


5

Magic Workstation besides many other tools for collection management, deck building, and online play has a very powerful probability calculator. It will go beyond opening hand and will let you see by what turn are you likely to have drawn the combo that you need.


5

The reason H2,H3,H4,H5 is more likely than H6,S6,C6,D6 is simply the rules of the game. If a heart is led, it is mandatory to play a heart if possible, so most tricks contain four of the same suit, and a trick with one of each suit is extremely rare. When you add in the requirement for all four to be of the same rank, your second example is vanishingly ...


4

I believe a probabilistically equivalent question is "If a single card revealed from an opening hand is copper, does it tell you anything about the starting deck?" Order doesn't matter (even in your example situation, as every hand is guaranteed to have copper, that the first card is copper isn't significant). Bayes' Rule for conditional probability says ...


4

I've worked out an approximation based on a seven card hand with a standard deck. I leave it as an exercise to do the same calculations for a five card hand :-) I think there are four ways of winning the hand: 4 of a kind and 3 of a kind 4 of a kind and 3 straight 3 of a kind and 4 straight 4 straight and 3 straight (this includes a 7 straight hand) In ...


3

If you are online, the easiest way to determine whether you have a good chance of winning a battle is to use this calculator: http://armsrace.co/probabilities It emphasizes a non-trivial conclusion: if you have the choice, always attack the big guys first in your sequence! For instance, if you have 6 on a territory, and want to attack a 2 and a 1 (and you ...


3

Surprisingly, the odds for Scenarios 1 and 3 are the same! The chances for each: Scenario 1: 0.0047% Scenario 2: 0.0535% Scenario 3: 0.0047% Here are the calculations: Note: The order you consider cards in these calculations doesn't matter, so I'll use the most convenient ordering possible. Scenario 1: First, there are 24 cards in a Euchre deck. To ...


3

Well there are only 52 C 8 ≈ 750 million different 8-card hands (eight because of the one you pick up at the start of your turn). That shouldn't take too long to brute force. I wrote a quick C# program that does just that. It runs in under 10 minutes on my PC (faster, by the way, than the equivalent non-parallel program in C). According to it, there are ...


3

The chance of drawing the card will be 1 - (chance of not drawing). I know that sounds silly, but bear with me. You get to draw 7 cards. The chance of avoiding it on your first draw is 56 / 60 because there's 56 cards that are not that card. The next card is a chance 55 / 59. This carries on for 7 cards. 56 * 55 * 54 * 53 * 52 * 51 * 50 ...


2

So my precious calculations were simple, just to get a rough idea since you only wanted to know what was needed to get above 60% Now more advanced calculations show (this is not exact, but close) 1 sub: less than 50% (around 48%) 2 subs: about 87.5% 3 subs: 98-100% You can calculate it more precisely, but it won't make much sense though.


2

Two good rules of thumb are, never resign in a position with contact, and resign if the result is "obvious". If you need to stop and think to work it out then it's probably quicker to just play on to a position of certainty anyway - certainly so online where forced moves (and optionally, greedy bearoffs) can be played automatically. It's generally not ...


2

I have personally found that Enclave of the Bear's The defender subtracts 1 from his lower defense die in the first territory you attack during your turn ability to be the most beneficial starting power, both at the start and throughout the campaign.


2

Die Mechaniker 1) Your starting HQ is always treated as FORTIFIED (+1 to both dice) when you defend it. 2) If your defense roll is two natural 6s, that territory cannot be attacked again for the rest of the turn. Reason: The odds of getting two natural 6s on a single roll is 1/36. Therefore, the odds that we have not rolled a pair of 6s by the nth attack ...



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