# Tag Info

## Hot answers tagged probability

29

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 ...

24

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 ...

18

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 ...

15

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 ...

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

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. ...

12

You can do this calculation using the multivariate hypergeometric distribution. The setup is as follows: The deck of 60 cards consists of: 4 cards of type A, 4 cards of type B, 4 cards of type C, 4 cards of type D, and 44 cards of type E (other). Your criteria are that a hand of 7 cards contains at least 1 card of type A, at least 1 card of type B, at least ...

11

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 ...

11

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 x-...

11

The above picture reflects the probability of drawing the full set of Exodia, dependent on the number of cards drawn overall. For example, there is a 0.000152% chance the player will draw Exodia in their opening hand, and a 87.5% chance they will draw it after seeing 39 cards out of their 40 card deck. If all 40 cards are drawn, there is a 100% chance the 5 ...

9

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. (...

9

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

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 ...

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 get....

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 ...

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

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 ...

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 - ...

7

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 ...

6

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

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.

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 ...

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

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 ...

4

The probability of getting a Complete Destruction in 1 turn is roughly 20%. So, the expected number of turns is 1/0.2 or 5 turns on average. The 20% number comes from two sources. First, I found a lengthy discussion at board game geek on this topic, which gives the final result of approximately 20%. http://boardgamegeek.com/thread/1155539/king-tokyo-odds/...

4

Deal out the four cards in a line. How likely is it that the one on the far right is the traitor card? 1 in 4 right? Now have the players pick up everything except the far right card. What are the chances that they didn't pick up the traitor card? It's the same question. So your answer is still 1 in 4.

3

I think your calculations look roughly correct (my calculations disagree at 4 decimal places, but I doubt you're using that many in an actual game!) Here's my calculated probabilities for a single defence die (plus bonus die): Defender wins 4109/7776: 0.528421 Attacker wins 3667/7776: 0.471579 Here's my calculated probabilites for two defence dice (plus ...

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 ...

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