Flip 12 Coins

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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

  • Flip 12 Coins
    Total Possible Combinations 13

    Number of combinations are calculated using the formula
    [ (2+12-1) choose (12) ] or [ 13 choose 12 ]
    where 2 is the number of sides of the coin and 12 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 12 coins
  • Flip 12 Coins
    Total Possible Permutations 4,096

    Number of permutations are calculated using the formula [ 2^12 ]
    where 2 is the number of sides of the coin and 12 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 12 coins

Probabilities of this 12 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 12 times

In technical terms, this is equivalent of getting atleast one Heads.

Probability of getting atleast one Heads is close to 1., about 99.98% 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 ]
            = 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 ]
You can generate all possible permutations and combinations of 12 Coins containing atleast one Heads using the following permutations and combinations generator All possible combinations of 12 coins with atleast one Heads All possible permutations of 12 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 12 coins.

Probability of not getting a Heads when flipping 12 Coins

Probability of not getting any Heads is close to 0.000244, about 0.024% 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
            = 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

Probability of getting exactly 1 Heads or 11 Tails when flipping 12 Coins

Probability of getting exactly 1 Heads is close to 0.00293, about 0.29% percent.

You can calculate this using counting or by using the binomial distribution

Using Counting

You can think of 12 coins as spots to be filled. There are 4,096 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 12 spots? Or, in how many ways can we pick 1 out of 12 spots? The answer can be caclulated using the formula [ 12 choose 1 ]. So, there are 12 ways to choose 1 out of 12 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 12 spots / Total permutations of 12 coins
                = (12 choose 1) / (2^12)                
                = 12 / 4096

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=12 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
                = (12 choose 1) (1/2)1 (1/2)12-1
                = (12 choose 1) (1/2)12
                = 12 x (1/4096)

You can generate all possible permutations and combinations of 12 Coins containing exactly 1 Heads using the following permutations and combinations generator All possible combinations of 12 coins with exactly 1 Heads All possible permutations of 12 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 6 Heads or 6 Tails when flipping 12 Coins

Probability of getting exactly 6 Heads is close to 0.226, about 22.56% percent.

You can calculate this using counting or by using the binomial distribution

Using Counting

You can think of 12 coins as spots to be filled. There are 4,096 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 6 Heads. In other words, in how many different ways can we place 6 Heads in 12 spots? Or, in how many ways can we pick 6 out of 12 spots? The answer can be caclulated using the formula [ 12 choose 6 ]. So, there are 924 ways to choose 6 out of 12 spots. The probability is calculated by dividing the number of ways to pick 6 spots by the number of total permuations. 

P (H=6)
                = Ways to pick 6 out of 12 spots / Total permutations of 12 coins
                = (12 choose 6) / (2^12)                
                = 924 / 4096

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 6 Heads, we need k=6 successful events out of total n=12 events. Each event has a probability p=1/2 of occuring.

Probability of k=6 successful events is then calculated as follows. 
P(k=6)
                = (n choose k) pk (1-p)n-k
                = (12 choose 6) (1/2)6 (1/2)12-6
                = (12 choose 6) (1/2)12
                = 924 x (1/4096)

You can generate all possible permutations and combinations of 12 Coins containing exactly 6 Heads using the following permutations and combinations generator All possible combinations of 12 coins with exactly 6 Heads All possible permutations of 12 coins with exactly 6 Heads

To verify the answer, you can divide the number of permutations containing exactly 6 Heads by number of total possible permutations.

Probability of getting exactly 11 Heads or 1 Tails when flipping 12 Coins

Probability of getting exactly 11 Heads is close to 0.00293, about 0.29% percent.

You can calculate this using counting or by using the binomial distribution

Using Counting

You can think of 12 coins as spots to be filled. There are 4,096 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 11 Heads. In other words, in how many different ways can we place 11 Heads in 12 spots? Or, in how many ways can we pick 11 out of 12 spots? The answer can be caclulated using the formula [ 12 choose 11 ]. So, there are 12 ways to choose 11 out of 12 spots. The probability is calculated by dividing the number of ways to pick 11 spots by the number of total permuations. 

P (H=11)
                = Ways to pick 11 out of 12 spots / Total permutations of 12 coins
                = (12 choose 11) / (2^12)                
                = 12 / 4096

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 11 Heads, we need k=11 successful events out of total n=12 events. Each event has a probability p=1/2 of occuring.

Probability of k=11 successful events is then calculated as follows. 
P(k=11)
                = (n choose k) pk (1-p)n-k
                = (12 choose 11) (1/2)11 (1/2)12-11
                = (12 choose 11) (1/2)12
                = 12 x (1/4096)

You can generate all possible permutations and combinations of 12 Coins containing exactly 11 Heads using the following permutations and combinations generator All possible combinations of 12 coins with exactly 11 Heads All possible permutations of 12 coins with exactly 11 Heads

To verify the answer, you can divide the number of permutations containing exactly 11 Heads by number of total possible permutations.

Probability of getting all Heads when flipping 12 Coins

Probability of getting all Heads is close to 0.000244, about 0.024% percent.

This is calculated by multiplying together all the probabilities of getting a Heads for each coin.

P(Heads=12)
            = 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
            = 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

Probability of getting one of a kind when flipping 12 Coins

Probability of getting one of a kind is close to 0.000488, about 0.049% 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)

Javascript code to Flip 12 Coins


    // Flip 12 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 = 12;    // 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 

    

    */

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