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 2 Coins - 3 times
Total Possible Combinations 7
Number of combinations are calculated using the formula
[ (2+6-1) choose (6) ] or [ 7 choose 6 ]
where 2 is the number of sides of the coin and 6 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 6 coins -
Flip 2 Coins - 3 times
Total Possible Permutations 64
Number of permutations are calculated using the formula [ 2^6 ]
where 2 is the number of sides of the coin and 6 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 6 coins
Probabilities of this 6 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 6 times
In technical terms, this is equivalent of getting atleast one Heads.
Probability of getting atleast one Heads is close to 0.984, about 98.44% 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 ]
= 1 - [ 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 ]
You can generate all possible permutations and combinations of 2 Coins - 3 times containing atleast one Heads using the following permutations and combinations generator
All possible combinations of 6 coins with atleast one Heads
All possible permutations of 6 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 6 coins.
Probability of not getting a Heads when flipping 2 Coins - 3 times
Probability of not getting any Heads is close to 0.0156, about 1.56% 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
= 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2
Probability of getting exactly 1 Heads or 5 Tails when flipping 2 Coins - 3 times
Probability of getting exactly 1 Heads is close to 0.0938, about 9.38% percent.
You can calculate this using counting or by using the binomial distribution
Using Counting
You can think of 6 coins as spots to be filled. There are 64 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 6 spots? Or, in how many ways can we pick 1 out of 6 spots? The answer can be caclulated using the formula [ 6 choose 1 ]. So, there are 6 ways to choose 1 out of 6 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 6 spots / Total permutations of 6 coins
= (6 choose 1) / (2^6)
= 6 / 64
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=6 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
= (6 choose 1) (1/2)1 (1/2)6-1
= (6 choose 1) (1/2)6
= 6 x (1/64)
You can generate all possible permutations and combinations of 2 Coins - 3 times containing exactly 1 Heads using the following permutations and combinations generator All possible combinations of 6 coins with exactly 1 Heads All possible permutations of 6 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 3 Heads or 3 Tails when flipping 2 Coins - 3 times
Probability of getting exactly 3 Heads is close to 0.312, about 31.25% percent.
You can calculate this using counting or by using the binomial distribution
Using Counting
You can think of 6 coins as spots to be filled. There are 64 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 3 Heads. In other words, in how many different ways can we place 3 Heads in 6 spots? Or, in how many ways can we pick 3 out of 6 spots? The answer can be caclulated using the formula [ 6 choose 3 ]. So, there are 20 ways to choose 3 out of 6 spots. The probability is calculated by dividing the number of ways to pick 3 spots by the number of total permuations.
P (H=3)
= Ways to pick 3 out of 6 spots / Total permutations of 6 coins
= (6 choose 3) / (2^6)
= 20 / 64
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 3 Heads, we need k=3 successful events out of total n=6 events. Each event has a probability p=1/2 of occuring.
Probability of k=3 successful events is then calculated as follows.P(k=3)
= (n choose k) pk (1-p)n-k
= (6 choose 3) (1/2)3 (1/2)6-3
= (6 choose 3) (1/2)6
= 20 x (1/64)
You can generate all possible permutations and combinations of 2 Coins - 3 times containing exactly 3 Heads using the following permutations and combinations generator All possible combinations of 6 coins with exactly 3 Heads All possible permutations of 6 coins with exactly 3 Heads
To verify the answer, you can divide the number of permutations containing exactly 3 Heads by number of total possible permutations.
Probability of getting exactly 5 Heads or 1 Tails when flipping 2 Coins - 3 times
Probability of getting exactly 5 Heads is close to 0.0938, about 9.38% percent.
You can calculate this using counting or by using the binomial distribution
Using Counting
You can think of 6 coins as spots to be filled. There are 64 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 5 Heads. In other words, in how many different ways can we place 5 Heads in 6 spots? Or, in how many ways can we pick 5 out of 6 spots? The answer can be caclulated using the formula [ 6 choose 5 ]. So, there are 6 ways to choose 5 out of 6 spots. The probability is calculated by dividing the number of ways to pick 5 spots by the number of total permuations.
P (H=5)
= Ways to pick 5 out of 6 spots / Total permutations of 6 coins
= (6 choose 5) / (2^6)
= 6 / 64
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 5 Heads, we need k=5 successful events out of total n=6 events. Each event has a probability p=1/2 of occuring.
Probability of k=5 successful events is then calculated as follows.P(k=5)
= (n choose k) pk (1-p)n-k
= (6 choose 5) (1/2)5 (1/2)6-5
= (6 choose 5) (1/2)6
= 6 x (1/64)
You can generate all possible permutations and combinations of 2 Coins - 3 times containing exactly 5 Heads using the following permutations and combinations generator All possible combinations of 6 coins with exactly 5 Heads All possible permutations of 6 coins with exactly 5 Heads
To verify the answer, you can divide the number of permutations containing exactly 5 Heads by number of total possible permutations.
Probability of getting all Heads when flipping 2 Coins - 3 times
Probability of getting all Heads is close to 0.0156, about 1.56% percent.
This is calculated by multiplying together all the probabilities of getting a Heads for each coin.
P(Heads=6)
= 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
= 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2
Probability of getting one of a kind when flipping 2 Coins - 3 times
Probability of getting one of a kind is close to 0.0312, about 3.13% 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)
Javascript code to Flip 2 Coins - 3 times
// Flip 2 Coins - 3 times
// 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 = 2; // how many coin to flip
var times = 3; // how many times 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 generated_flips = [];
for (var i = 1; i <= times; i++) {
// outer loop for the number of 'times' to flip the coins
var this_flip = []; // array to store the results of this flip
for (var j = 1; j <= number_of_coins; j++) {
// inner loop for the number of coins
// for each coin, generate a number between lowest and highest
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
}
generated_flips.push(this_flip); // store the result of this flip in generated_flips
}
// print all the generated flips
for (i = 0; i < generated_flips.length; i++) {
// loop through the outer times array
for (j = 0; j < generated_flips[i].length; j++) {
// loop through the inner coin array
//print each coin flip value followed by a space
document.write(generated_flips[i][j]);
document.write(" ");
}
//print a line break for each time
document.writeln("<br>");
}
/*
Sample output
*/