Flip 50 Coins

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Flip coin(s) times

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 50 Coins
    Total Possible Combinations 51

    Number of combinations are calculated using the formula
    [ (2+50-1) choose (50) ] or [ 51 choose 50 ]
    where 2 is the number of sides of the coin and 50 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 50 coins
  • Flip 50 Coins
    Total Possible Permutations 1,125,899,906,842,624
    (around 1 quadrillion)


    Number of permutations are calculated using the formula [ 2^50 ]
    where 2 is the number of sides of the coin and 50 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 50 coins

Probabilities of this 50 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 50 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 50 Coins containing atleast one Heads using the following permutations and combinations generator All possible combinations of 50 coins with atleast one Heads All possible permutations of 50 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 50 coins.

Probability of not getting a Heads when flipping 50 Coins

Probability of not getting any Heads is close to 0.000000000000000888, about 0.000000000000089% 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 49 Tails when flipping 50 Coins

Probability of getting exactly 1 Heads is close to 0.0000000000000444, about 0.0000000000044% percent.

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

Using Counting

You can think of 50 coins as spots to be filled. There are 1,125,899,906,842,624 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 50 spots? Or, in how many ways can we pick 1 out of 50 spots? The answer can be caclulated using the formula [ 50 choose 1 ]. So, there are 50 ways to choose 1 out of 50 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 50 spots / Total permutations of 50 coins
                = (50 choose 1) / (2^50)                
                = 50 / 1125899906842624                

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=50 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
                = (50 choose 1) (1/2)1 (1/2)50-1
                = (50 choose 1) (1/2)50
                = 50 x (1/1125899906842624)
                

You can generate all possible permutations and combinations of 50 Coins containing exactly 1 Heads using the following permutations and combinations generator All possible combinations of 50 coins with exactly 1 Heads All possible permutations of 50 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 25 Heads or 25 Tails when flipping 50 Coins

Probability of getting exactly 25 Heads is close to 0.112, about 11.23% percent.

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

Using Counting

You can think of 50 coins as spots to be filled. There are 1,125,899,906,842,624 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 25 Heads. In other words, in how many different ways can we place 25 Heads in 50 spots? Or, in how many ways can we pick 25 out of 50 spots? The answer can be caclulated using the formula [ 50 choose 25 ]. So, there are 126410606437752 ways to choose 25 out of 50 spots. The probability is calculated by dividing the number of ways to pick 25 spots by the number of total permuations.

P (H=25)
                = Ways to pick 25 out of 50 spots / Total permutations of 50 coins
                = (50 choose 25) / (2^50)                
                = 126410606437752 / 1125899906842624                

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 25 Heads, we need k=25 successful events out of total n=50 events. Each event has a probability p=1/2 of occuring.

Probability of k=25 successful events is then calculated as follows.
P(k=25)
                = (n choose k) pk (1-p)n-k
                = (50 choose 25) (1/2)25 (1/2)50-25
                = (50 choose 25) (1/2)50
                = 126410606437752 x (1/1125899906842624)
                

You can generate all possible permutations and combinations of 50 Coins containing exactly 25 Heads using the following permutations and combinations generator All possible combinations of 50 coins with exactly 25 Heads All possible permutations of 50 coins with exactly 25 Heads

To verify the answer, you can divide the number of permutations containing exactly 25 Heads by number of total possible permutations.

Probability of getting exactly 49 Heads or 1 Tails when flipping 50 Coins

Probability of getting exactly 49 Heads is close to 0.0000000000000444, about 0.0000000000044% percent.

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

Using Counting

You can think of 50 coins as spots to be filled. There are 1,125,899,906,842,624 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 49 Heads. In other words, in how many different ways can we place 49 Heads in 50 spots? Or, in how many ways can we pick 49 out of 50 spots? The answer can be caclulated using the formula [ 50 choose 49 ]. So, there are 50 ways to choose 49 out of 50 spots. The probability is calculated by dividing the number of ways to pick 49 spots by the number of total permuations.

P (H=49)
                = Ways to pick 49 out of 50 spots / Total permutations of 50 coins
                = (50 choose 49) / (2^50)                
                = 50 / 1125899906842624                

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 49 Heads, we need k=49 successful events out of total n=50 events. Each event has a probability p=1/2 of occuring.

Probability of k=49 successful events is then calculated as follows.
P(k=49)
                = (n choose k) pk (1-p)n-k
                = (50 choose 49) (1/2)49 (1/2)50-49
                = (50 choose 49) (1/2)50
                = 50 x (1/1125899906842624)
                

You can generate all possible permutations and combinations of 50 Coins containing exactly 49 Heads using the following permutations and combinations generator All possible combinations of 50 coins with exactly 49 Heads All possible permutations of 50 coins with exactly 49 Heads

To verify the answer, you can divide the number of permutations containing exactly 49 Heads by number of total possible permutations.

Probability of getting all Heads when flipping 50 Coins

Probability of getting all Heads is close to 0.000000000000000888, about 0.000000000000089% percent.

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

P(Heads=50)
            = 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 50 Coins

Probability of getting one of a kind is close to 0.00000000000000178, about 0.00000000000018% 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 50 Coins


    // Flip 50 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 = 50;    // 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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