9

(NOTE : Final revision of my original answer) The odds of the first tileset not having a single valid word is exactly 91,595,416 / 16,007,560,800 or .5722%, with it occuring once every 174.76378 games. This value is calculated by using the dictionary found in this answer, but can be adapted for any other dictionary. This was brute-forced via python. Code ...


9

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 ...


6

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 ...


3

I used the beginning of the program in @DenisS's answer to build the Scrabble dictionary, then I used it to write a small monte-carlo program to estimate the probability that no word can be formed with seven random tiles. The result is a 0.58% +- 0.27% probability that no word can be formed. Output $ python3 get_proba.py 1000 50 loading dictionary total ...


3

This is actually deceptively simple. A full set includes every card of that type in the game. If you possess a card in a specific set, then nobody else can complete that set. If you possess cards from every set, then no set can be completed without trading with you. The flip side is that if there is any set that you lack cards from, then that set can ...


3

Let's go through this: The first Terror (the one you cast) comes out, the March triggers, but not the Terror. The March resolves, and you flip coins, getting N heads, meaning N Terrors enter the battlefield. Including the one already there, they all see all the others enter the battlefield, see rule 603.6a (emphasis mine): 603.6a Enters-the-battlefield ...


3

This is not math.stackexchange, so I don't have access to any formula formatting. I'll try to make it readable regardless. The basic mathematical construct used here are the so-called binomial coefficients: (a choose b) means a!/(b!*(a-b)!), it says how many different possible hands of b cards there are when you have a unique cards to choose from. I will ...


3

Some slightly watered down math (that still gives the right answer). You have a 100% chance of your first card being a "good" card. There are then 51 cards left, of which 12 will be a card that matches suit, so you have a 12/51 chance that the second card will be "good". 11/50 chance that the third card will be "good". .... So, you have 1x(12/51)x(11/50)x(...


2

91592097 in 16007560800 which is approximately 0.572% (or 1 in 175). Some of what follows is already covered in @DenisS's answer and I have used the same dictionary of words (Collins Scrabble Words (2019)) for easy comparison. Note in particular in that answer the argument for discounting blanks when looking for valid combinations without words (i.e. that ...


2

L. Scott Johnson is mostly correct except for 2 things: You specified it wasn't a straight flush (there are 7 8-card straight flushes per suit he should have removed.) You asked about the 7 card flush, which you were dealt before the draw, not the 8 card flush you drew into. This gives us: 13C7 as the nCr we care about, giving us 1716 possible flushes of ...


2

There's a lot of ways to weaken the effects of the dice. Only the best X dice count - this means the more dice you have the more likely you will get a better overall result, but you are still stuck in the same x to 6x range of values - kind of like how D&D uses rolled ability scores (roll 4 keep 3) or how texas holdem works with hands (best 5 out of 7 ...


1

You don't have to actually use all of the dice that you roll. For example you could be looking at the results of the best 3 dice of the 5 that you rolled. Example for needing a roll of 12 on 3 dice: Roll 5 dice get 1, 2, 3, 6, 6 Chose the 3, 6, 6 for a total of 15 Meet the goal Example for Rolling 10 dice Roll 10 dice get 1, 1, 2, 2, 2, 3, 5, 5, 6, 6 ...


1

How about "number of dice that show a 6", as is used in games like Neanderthal by Phil Eklund? With 5 Dice, you can expect to roll at least one 6 about 60% of the time With 10 Dice, you can expect to roll at least one 6 about 84% of the time So you are more effective with more dice, but it's diminishing returns. You can tweak the probabilities by ...


1

I modified that program by Andrew Inwood to count them (including all the three-of-a-kinds, still), and it finds 5,379,372 11-card hands that contain a 10-card gin hand. Out of C(52,11) hands, that makes the odds about 1 in 11,228 or 0.00891%


1

I'm going to make an estimate from the following assumption: Any hand that contains at least one vowel, y, or a blank allows a valid move. Any hand that contains entirely consonants does not. Obviously there are exceptions, but they should be rare enough to have a negligible effect (and the false positives and false negatives work to cancel each other out). ...


1

Suppose you're playing a three player game with wheat, barley, and corn. For one of the trading players to win with wheat, the two trading players must have nine wheat between them, which is the same as the shut out player not getting any wheat. To get a hand, we choose 9 cards out of 27. There are C(27,9) ways of doing that, where C is the binomial ...


1

(I'm mostly ignoring Bull and Bear rules here, but I think they can be added in with a little fiddling.) If you just want to know whether it's possible, then it's not a question of probability but of showing that there is at least one arrangement of cards that would allow a player to win without trading. And it's pretty easy to see that this must exist, ...


1

According to a Ted-Ed article entitled Here's how to win at Monopoly, according to math experts ... the average game of Monopoly takes about 30 turns per competitor... Reference https://blog.ed.ted.com/2017/12/01/heres-how-to-win-at-monopoly-according-to-math-experts/ So the answer to your question is 120 turns (4 players times 30 turns/player) By ...


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