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The probability to win all 5 coin flips with Krark's Thumb is 0.75^5, or 23.7%, up from 3.1% without Krark's Thumb. Normally, you would have a 50% chance to win a single coin flip. Since multiple coin flips are independent of each other, you can just multiply their probabilities, so you get a (0.5)^5 = 0.031 (3.1%) probability to have 5 coin flips go your ...


3

P(i successes) = ((# of ways to get i 6s and no 5s) + (# of ways to get i-1 6s and some 5s)) / 6^n (# of ways to get i 6s and no 5s) = (n choose i) * 1^i * 4^(n-i) [pick which dice must be 6s, 1 way for each of those to be a 6, 4 ways for each of the others to be neither 6 nor 5] (# of ways to get i-1 6s and some 5s) = (n choose i-i) * 1^(i-1) * (5^(n-(i-1)) ...


3

Assuming a Euchre deck with 24 cards (9, 10, J, Q, K, A of each suit, no joker) and perfect shuffling, the chance that the 21st card in the deck (the card that gets turned up) is a particular card is 1/24. Thus, if you play 4 deals, there is a (1/24)^4 = 1/331776 = 0.0003% that all four deals will turn up the Jack of Spades. There's much higher chance of all ...


3

I used to be a member, and for a time the President, of the MSU Mah Jong club. The custom used there built the wall as a square that overlapped at the ends, not as a spiral as shown in the linked video. From a game mechanics point of view, building the wall is comparable to shuffling cards and leaving a scuffled deck to draw from. in terms of randomizing ...


3

In theory yes, you can play with a single pile. The reason people setup the wall this way (aside from historical reasons) is because the layout suits the nature of the game. You typically play on a square table and tiles are fairly large so stacking them in a single tower would be pretty dangerous. Also, after setting up the wall, players draw their hands ...


2

I used [Any Dice] website to work this out. The code below is probably not the cleanest but I only learned this site today. The function below will count 1 success if at least one 5 and add to that how many sixes rolled. This is getting results of to 5d6. function: contains VALUES:s in SEQUENCE:s { FIVES: 0 SIXES: 0 loop P over {1..#VALUES} { if (5 = ...


2

You want to go through each of your dice from lowest to highest and, if your chance of success is at least as good if you re-roll that die than if you do not re-roll that die, add that die to your list of re-rolls. As soon as you find a die that worsens your current chance of succeeding, stop, and re-roll all the previous dice in this procedure. If you ...


1

Expected Value Expected value is the statistically 'average' result of a random event. That is to say, if you did the randomization action a million times, what would be the average value of your results. It's calculated as P(R1)*V(R1) + P(R2)*V(R2) + .. + P(Rn)*V(Rn), where P(Ri) is the Probability of Result i, and V(Ri) is the Value of Result i. For a ...


1

The odds of turning up a given card that occurs once in the deck are 1 in N, where N is the size of the deck. Since you say the shuffles were all good, this means the trials are all independent. So the odds of it happening four times are (1/N)^4 If N is 24, this comes out to 1 in 331776. If N is 32, this comes out to 1 in 1048576. If you don't restrict it to ...


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