Features of this Coin Tosser
- Toss a coin
- Uses American dime for the toss
- Also allows side landing according to the probability of 1 in 6000
- Start and Stop the coin at your own will
Statistics of this Coin Flipper
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Flip 82 Coins
Total Possible Combinations 83
Number of combinations are calculated using the formula
[ (2+82-1) choose (82) ] or [ 83 choose 82 ]
where 2 is the number of sides of the coin and 82 is the number of coins.
You can try generating all the combinations of Heads and Tails using the following combination generator
All possible combinations of 82 coins -
Flip 82 Coins
Total Possible Permutations 4.8357032784585E+24
(around 5 septillion)
Number of permutations are calculated using the formula [ 2^82 ]
where 2 is the number of sides of the coin and 82 is the number of coins.
You can try generating all the permutations of Heads and Tails using the following permutations generator
All possible permutations of 82 coins
Probabilities of this 82 Coin Tosser
Assuming that all the coins are fair coins and probability of getting either Heads or Tails is 1/2 or 50%Probability of getting a Heads when flipping a coin 82 times
In technical terms, this is equivalent of getting atleast one Heads.
Probability of getting atleast one Heads is close to 1., about 100% percent.
The probability of getting atleast one Heads is calculated by multiplying all the probabilities of not getting a Heads for each coin and then subtracting the answer from 1.
P(Heads>=1)
= 1 - [ P(H=0)Coin1 x P(H=0)Coin2 x P(H=0)Coin3 x P(H=0)Coin4 x P(H=0)Coin5 x P(H=0)Coin6 x P(H=0)Coin7 x P(H=0)Coin8 x P(H=0)Coin9 x P(H=0)Coin10 x P(H=0)Coin11 x P(H=0)Coin12 x P(H=0)Coin13 x P(H=0)Coin14 x P(H=0)Coin15 x P(H=0)Coin16 x P(H=0)Coin17 x P(H=0)Coin18 x P(H=0)Coin19 x P(H=0)Coin20 x ... ]
= 1 - [ 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x ... ]
You can generate all possible permutations and combinations of 82 Coins containing atleast one Heads using the following permutations and combinations generator
All possible combinations of 82 coins with atleast one Heads
All possible permutations of 82 coins with atleast one Heads
To verify the answer, you can divide the number of permutations containing atleast 1 Heads by number of total possible permutations of 82 coins.
Probability of not getting a Heads when flipping 82 Coins
Probability of not getting any Heads is close to 0.000000000000000000000000207, about 0.000000000000000000000021% percent.
Probability of getting zero Heads is calculated by multiplying all the probabilities of not getting a Heads for each coin.
P(Heads=0)
= P(H=0)Coin1 x P(H=0)Coin2 x P(H=0)Coin3 x P(H=0)Coin4 x P(H=0)Coin5 x P(H=0)Coin6 x P(H=0)Coin7 x P(H=0)Coin8 x P(H=0)Coin9 x P(H=0)Coin10 x P(H=0)Coin11 x P(H=0)Coin12 x P(H=0)Coin13 x P(H=0)Coin14 x P(H=0)Coin15 x P(H=0)Coin16 x P(H=0)Coin17 x P(H=0)Coin18 x P(H=0)Coin19 x P(H=0)Coin20 x ...
= 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x ...
Probability of getting exactly 1 Heads or 81 Tails when flipping 82 Coins
Probability of getting exactly 1 Heads is close to 0.000000000000000000000017, about 0.0000000000000000000017% percent.
You can calculate this using counting or by using the binomial distribution
Using Counting
You can think of 82 coins as spots to be filled. There are 4.8357032784585E+24 ways to fill these spots with either a Heads or Tails (all possible permutations). Now, we need to find how many of these permutations contain exactly 1 Heads. In other words, in how many different ways can we place 1 Heads in 82 spots? Or, in how many ways can we pick 1 out of 82 spots? The answer can be caclulated using the formula [ 82 choose 1 ]. So, there are 82 ways to choose 1 out of 82 spots. The probability is calculated by dividing the number of ways to pick 1 spots by the number of total permuations.
P (H=1)
= Ways to pick 1 out of 82 spots / Total permutations of 82 coins
= (82 choose 1) / (2^82)
= 82 / 4.8357032784585E+24
Using Binomial Distribution
In Binomial Distribution, we look for k number of successful events in n number of trials where p is the probability of each successful event. Here, a successful event means getting a Heads and an event means a coin flip.
To get exactly 1 Heads, we need k=1 successful events out of total n=82 events. Each event has a probability p=1/2 of occuring.
Probability of k=1 successful events is then calculated as follows.P(k=1)
= (n choose k) pk (1-p)n-k
= (82 choose 1) (1/2)1 (1/2)82-1
= (82 choose 1) (1/2)82
= 82 x (1/4.8357032784585E+24)
You can generate all possible permutations and combinations of 82 Coins containing exactly 1 Heads using the following permutations and combinations generator All possible combinations of 82 coins with exactly 1 Heads All possible permutations of 82 coins with exactly 1 Heads
To verify the answer, you can divide the number of permutations containing exactly 1 Heads by number of total possible permutations.
Probability of getting exactly 41 Heads or 41 Tails when flipping 82 Coins
Probability of getting exactly 41 Heads is close to 0.0878, about 8.78% percent.
You can calculate this using counting or by using the binomial distribution
Using Counting
You can think of 82 coins as spots to be filled. There are 4.8357032784585E+24 ways to fill these spots with either a Heads or Tails (all possible permutations). Now, we need to find how many of these permutations contain exactly 41 Heads. In other words, in how many different ways can we place 41 Heads in 82 spots? Or, in how many ways can we pick 41 out of 82 spots? The answer can be caclulated using the formula [ 82 choose 41 ]. So, there are 4.2478458084879E+23 ways to choose 41 out of 82 spots. The probability is calculated by dividing the number of ways to pick 41 spots by the number of total permuations.
P (H=41)
= Ways to pick 41 out of 82 spots / Total permutations of 82 coins
= (82 choose 41) / (2^82)
= 4.2478458084879E+23 / 4.8357032784585E+24
Using Binomial Distribution
In Binomial Distribution, we look for k number of successful events in n number of trials where p is the probability of each successful event. Here, a successful event means getting a Heads and an event means a coin flip.
To get exactly 41 Heads, we need k=41 successful events out of total n=82 events. Each event has a probability p=1/2 of occuring.
Probability of k=41 successful events is then calculated as follows.P(k=41)
= (n choose k) pk (1-p)n-k
= (82 choose 41) (1/2)41 (1/2)82-41
= (82 choose 41) (1/2)82
= 4.2478458084879E+23 x (1/4.8357032784585E+24)
You can generate all possible permutations and combinations of 82 Coins containing exactly 41 Heads using the following permutations and combinations generator All possible combinations of 82 coins with exactly 41 Heads All possible permutations of 82 coins with exactly 41 Heads
To verify the answer, you can divide the number of permutations containing exactly 41 Heads by number of total possible permutations.
Probability of getting exactly 81 Heads or 1 Tails when flipping 82 Coins
Probability of getting exactly 81 Heads is close to 0.000000000000000000000017, about 0.0000000000000000000017% percent.
You can calculate this using counting or by using the binomial distribution
Using Counting
You can think of 82 coins as spots to be filled. There are 4.8357032784585E+24 ways to fill these spots with either a Heads or Tails (all possible permutations). Now, we need to find how many of these permutations contain exactly 81 Heads. In other words, in how many different ways can we place 81 Heads in 82 spots? Or, in how many ways can we pick 81 out of 82 spots? The answer can be caclulated using the formula [ 82 choose 81 ]. So, there are 82 ways to choose 81 out of 82 spots. The probability is calculated by dividing the number of ways to pick 81 spots by the number of total permuations.
P (H=81)
= Ways to pick 81 out of 82 spots / Total permutations of 82 coins
= (82 choose 81) / (2^82)
= 82 / 4.8357032784585E+24
Using Binomial Distribution
In Binomial Distribution, we look for k number of successful events in n number of trials where p is the probability of each successful event. Here, a successful event means getting a Heads and an event means a coin flip.
To get exactly 81 Heads, we need k=81 successful events out of total n=82 events. Each event has a probability p=1/2 of occuring.
Probability of k=81 successful events is then calculated as follows.P(k=81)
= (n choose k) pk (1-p)n-k
= (82 choose 81) (1/2)81 (1/2)82-81
= (82 choose 81) (1/2)82
= 82 x (1/4.8357032784585E+24)
You can generate all possible permutations and combinations of 82 Coins containing exactly 81 Heads using the following permutations and combinations generator All possible combinations of 82 coins with exactly 81 Heads All possible permutations of 82 coins with exactly 81 Heads
To verify the answer, you can divide the number of permutations containing exactly 81 Heads by number of total possible permutations.
Probability of getting all Heads when flipping 82 Coins
Probability of getting all Heads is close to 0.000000000000000000000000207, about 0.000000000000000000000021% percent.
This is calculated by multiplying together all the probabilities of getting a Heads for each coin.
P(Heads=82)
= P(H=1)Coin1 x P(H=1)Coin2 x P(H=1)Coin3 x P(H=1)Coin4 x P(H=1)Coin5 x P(H=1)Coin6 x P(H=1)Coin7 x P(H=1)Coin8 x P(H=1)Coin9 x P(H=1)Coin10 x P(H=1)Coin11 x P(H=1)Coin12 x P(H=1)Coin13 x P(H=1)Coin14 x P(H=1)Coin15 x P(H=1)Coin16 x P(H=1)Coin17 x P(H=1)Coin18 x P(H=1)Coin19 x P(H=1)Coin20 x ...
= 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x ...
Probability of getting one of a kind when flipping 82 Coins
Probability of getting one of a kind is close to 0.000000000000000000000000414, about 0.000000000000000000000041% percent.
There are 2 ways to get one of a kind (all Heads or all Tails). The probability of getting all of any kind is then caclulated by adding the probability of getting all Heads or all Tails. Since, probabilities of getting all Heads or all Tails are the same, we can just multiply one of them by 2. So, multiplying the probability of getting all Heads by 2 will give us the probability of getting all of any kind.
Probability of getting all of a kind
= Pobability of getting all Heads + Pobability of getting all Tails
= 2 x Pobability of getting all Heads
= 2 x (1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x ...)
Javascript code to Flip 82 Coins
// Flip 82 Coins
// define the range of numbers to pick from
var lowest = 1; // lowest possible side of the coin // 1 for heads
var highest = 2; // highest possible side of the coin // 2 for tails
var number_of_coins = 82; // how many coins to flip
// allow the coin to land on the side
// change to false if sidelanding not desired
var allow_sidelanding = true;
var PROBABILITY_OF_SIDELANDING = 6000; // set the probability of sidelanding to 1 in 6000
var this_flip = []; // array to store the results of this flip
for (var j = 1; j <= number_of_coins; j++) {
// loop for the number of coins
// for each coin, generate a number between lowest and highest i.e. 1 or 2
var coin_face = Math.floor(Math.random() * (highest-lowest+1) + lowest);
// if sidelanding is allowed
if (allow_sidelanding) {
// first check for sidelanding scenario with given probability
var edge = Math.floor(Math.random() * (PROBABILITY_OF_SIDELANDING-1+1) + 1);
// if sidelanding is probable, override the coin face with edge value i.e. 3
// i.e. if we generated a 1 when generating a number between 1 and PROBABILITY_OF_SIDELANDING
if (edge === 1) {
coin_face = 3;
}
}
this_flip.push(coin_face); //store this in the array
}
// print all the generated flips
for (j = 0; j < this_flip.length; j++) {
// loop through the coin array
//print each coin flip value followed by a space
document.write(this_flip[j]);
document.write(" ");
}
/*
Sample output
*/