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: 1 (the minimum number we want to draw)
Results: 39% ...
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 ...
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 ...
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 ...
The key change in reversing the direction of movement is the fact that players now leave Jail in the opposite direction. The oranges and reds are particularly profitable because they are within 1-2 rolls of Jail, among other reasons. When moving counterclockwise, the dark purples and dark blues are now in a more often visited area, so I would expect their ...
The first mulligan is free in Brawl[CR 903.11g], so it's 3 mulligans leaves you with 5 cards. I shall reply accordingly.
After removing the Commander, 59 cards remain in the deck. 58 of those aren't the card.
The probability that the 1st card drawn isn't the card is 58/59.
The probability that the 2nd card drawn also isn't the card is 57/58.
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 - ...
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.
Symbolizing the cards that give you a draw a D, and the cards that do not as N, there are 6 possible permutations of draw vs no-draw.
In each permutation, the second N card represents the farthest you can get through your deck on this turn. I have bolded it for easier visibility.
This has been a problem since the days when dice, hand-carved out of sheep's knucklebones, could never be considered mathematically fair. Historically, there have been two principal ways of improving a game:
Make sure that a particular number is not always good or always bad. In Craps, rolling a 7 is good for the shooter on the first roll, but bad ...
The decks here are small enough that we can explicitly list out all the possibilities:
your draw, opponent's draw
2, 2 *
3, 2 *
3, 3 *
4, 2 *
4, 3 *
4, 4 *
I've put an * next to all combinations which result in you escaping. As you can see there are 6 of those out of 16 total. Since each pair of draws in ...
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 ...
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 ...
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.
If you “need” the card in your starting hand and there is no other option—you take as many as you need until you have it, then stop. If you don't have it when you've mulliganed to 1, you concede and hope for better luck next game.
Granted you also need lands for mana, so you probably concede if you don't have it and an island when you mulligan to 2.
The probabilities of a single player getting one of the above hands, from a single deal, are:
Scenario 1: 0.09%
Scenario 2: 1.54%
Scenario 3: 0.05%
Below I have calculated the probabilities and also simulated them to check the calculations.
I have also simulated the scenarios to find the probabilities of any player getting one of the ...
The statistic theory presented by the question is best described Hypergeometric Distribution. During the game set up, 2 Non-betrayal objectives per player, and 1 Betrayal objective, are combined in the opening deck. From that 7 card deck, each player will draw 1 objective, 3 total. Your goal is that 3 non-betrayal objectives are drawn.
N = number of items ...
No player has a greater chance of drawing an even or uneven distribution than any other.
One way of looking at it is to consider permutations of the tokens, where they are laid out in some sequence instead of jumbled in a bag or something. Then, if we shuffle the tokens up, so that they can be in any order, we can distribute the tokens, so that the first ...
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/...
The reason that equity is used instead of winning probability is because it is possible to win a single game, a double game (gammon) or triple game (backgammon).
Let's say that the value of the game, or bet, is $1. (That would occur if the cube is in the middle. If it has been turned, you multiply by 2, 4, or whatever the number is on the cube.)
Answer found through personal research.
Cowry shells have a roughly 30% chance of rolling a 0. Depending on the shell, that can get as low as 18.65% and as high as 39.11%. (At least, with the test subjects I had) It seems that the larger the shell is, the less likely it is to roll 1. Even between shells of similar size, there is significant variation. So, ...
I think that fairness in the game is more dependent upon everyone using the same cowry shells, and less on having a 50/50 probability. The game would not be fair if one player used coins, and another used cowry shells.
The question reminds me of playing Shagai. Shagai is the Mongolian word for an ankle bone (sheep to be specific). Mongolians have come up ...
There is a legit mathematical answer to this question, but it's a bit outside my ability. What I can offer you, however, is an example of how Magic players of various levels of mathematical knowledge have attempted to tackle the question.
This article by Frank Karsten is a good start. Doing it mathematically, you get stuff like this:
The thing is, having none of those cards is not the only way you'd be unable to cast the Geist on turn 3. You'd have to account for a whole bunch of possibilities:
Zero islands, zero plains, and zero Geists
Zero islands, one plains, and zero Geists
Zero islands, two plains, and zero Geists
... (up to 9 plains)
One island, zero plains, and zero Geists
If you really need a card on your starting hand to win, and otherwise you lose, then you have to take mulligans until the card shows up or you run out of mulligans. The chance to have Treasure Hunt in your hand by turn 2 with at 6 mulligans is 87.5%
When the alternative to taking a mulligan is losing the game, then taking another mulligan that draws you at ...
What are the odds of being dealt all 7 cards in the same suit?
The odds of pulling 7 spades in 7 cards the number of ways to pull 7 spades (from among the 13 spades) divided by the number of ways to pull 7 cards (from among the 52 cards).
13c7 = 1716
52c7 = 133784560
1716 / 133784560 = 0.0000128 = 0.00128 %
For general suit (not just spades), multiply ...
If you are online, the easiest way to determine whether you have a good chance of winning a battle is to use this calculator:
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 ...
Let's set a lower bound on the likelihood of winning with doubles, by simply ignoring all cases where it is impossible to win without doubles.
Assumption: All board positions considered are equally likely. This is probably not true, but will approach truth in longer games.
Consider the case of two men only left on the board, both in the home court, and not ...
This questions is substantially different from the first version. Given the mechanics of Exodia, each of the five parts must be collected in hand. To support that, 0 Exodia cards can be drawn prior to the use of Morphing Jar. In this, the situation is similar to the linked question with the emphasized difference.
N = number of items in the ...
27.6% if you go first, or 34.5% if you go second. This does not account for mulligans.
Those numbers were obtained experimentally using the code posted below. There's an error in the math presented below. (David Z's theoretical result matches my experimental result, not my theoretical result which is slightly off.0 I'm leaving it up in the hopes that ...