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 - 1000 times
Total Possible Combinations INF
Number of combinations are calculated using the formula
[ (2+2000-1) choose (2000) ] or [ 2001 choose 2000 ]
where 2 is the number of sides of the coin and 2000 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 2000 coins -
Flip 2 Coins - 1000 times
Total Possible Permutations INF
Number of permutations are calculated using the formula [ 2^2000 ]
where 2 is the number of sides of the coin and 2000 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 2000 coins
Probabilities of this 2000 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 2000 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 2 Coins - 1000 times containing atleast one Heads using the following permutations and combinations generator
All possible combinations of 2000 coins with atleast one Heads
All possible permutations of 2000 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 2000 coins.
Probability of not getting a Heads when flipping 2 Coins - 1000 times
Probability of not getting any Heads is close to 0., about 0.% 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 1999 Tails when flipping 2 Coins - 1000 times
Probability of getting exactly 1 Heads is close to 0., about 0.% percent.
You can calculate this using counting or by using the binomial distribution
Using Counting
You can think of 2000 coins as spots to be filled. There are INF 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 2000 spots? Or, in how many ways can we pick 1 out of 2000 spots? The answer can be caclulated using the formula [ 2000 choose 1 ]. So, there are 2000 ways to choose 1 out of 2000 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 2000 spots / Total permutations of 2000 coins
= (2000 choose 1) / (2^2000)
= 2000 / INF
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=2000 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
= (2000 choose 1) (1/2)1 (1/2)2000-1
= (2000 choose 1) (1/2)2000
= 2000 x (1/INF)
You can generate all possible permutations and combinations of 2 Coins - 1000 times containing exactly 1 Heads using the following permutations and combinations generator All possible combinations of 2000 coins with exactly 1 Heads All possible permutations of 2000 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 1000 Heads or 1000 Tails when flipping 2 Coins - 1000 times
Probability of getting exactly 1000 Heads is close to NaN, about NAN% percent.
You can calculate this using counting or by using the binomial distribution
Using Counting
You can think of 2000 coins as spots to be filled. There are INF 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 1000 Heads. In other words, in how many different ways can we place 1000 Heads in 2000 spots? Or, in how many ways can we pick 1000 out of 2000 spots? The answer can be caclulated using the formula [ 2000 choose 1000 ]. So, there are INF ways to choose 1000 out of 2000 spots. The probability is calculated by dividing the number of ways to pick 1000 spots by the number of total permuations.
P (H=1000)
= Ways to pick 1000 out of 2000 spots / Total permutations of 2000 coins
= (2000 choose 1000) / (2^2000)
= INF / INF
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 1000 Heads, we need k=1000 successful events out of total n=2000 events. Each event has a probability p=1/2 of occuring.
Probability of k=1000 successful events is then calculated as follows.P(k=1000)
= (n choose k) pk (1-p)n-k
= (2000 choose 1000) (1/2)1000 (1/2)2000-1000
= (2000 choose 1000) (1/2)2000
= INF x (1/INF)
You can generate all possible permutations and combinations of 2 Coins - 1000 times containing exactly 1000 Heads using the following permutations and combinations generator All possible combinations of 2000 coins with exactly 1000 Heads All possible permutations of 2000 coins with exactly 1000 Heads
To verify the answer, you can divide the number of permutations containing exactly 1000 Heads by number of total possible permutations.
Probability of getting exactly 1999 Heads or 1 Tails when flipping 2 Coins - 1000 times
Probability of getting exactly 1999 Heads is close to NaN, about NAN% percent.
You can calculate this using counting or by using the binomial distribution
Using Counting
You can think of 2000 coins as spots to be filled. There are INF 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 1999 Heads. In other words, in how many different ways can we place 1999 Heads in 2000 spots? Or, in how many ways can we pick 1999 out of 2000 spots? The answer can be caclulated using the formula [ 2000 choose 1999 ]. So, there are INF ways to choose 1999 out of 2000 spots. The probability is calculated by dividing the number of ways to pick 1999 spots by the number of total permuations.
P (H=1999)
= Ways to pick 1999 out of 2000 spots / Total permutations of 2000 coins
= (2000 choose 1999) / (2^2000)
= INF / INF
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 1999 Heads, we need k=1999 successful events out of total n=2000 events. Each event has a probability p=1/2 of occuring.
Probability of k=1999 successful events is then calculated as follows.P(k=1999)
= (n choose k) pk (1-p)n-k
= (2000 choose 1999) (1/2)1999 (1/2)2000-1999
= (2000 choose 1999) (1/2)2000
= INF x (1/INF)
You can generate all possible permutations and combinations of 2 Coins - 1000 times containing exactly 1999 Heads using the following permutations and combinations generator All possible combinations of 2000 coins with exactly 1999 Heads All possible permutations of 2000 coins with exactly 1999 Heads
To verify the answer, you can divide the number of permutations containing exactly 1999 Heads by number of total possible permutations.
Probability of getting all Heads when flipping 2 Coins - 1000 times
Probability of getting all Heads is close to 0., about 0.% percent.
This is calculated by multiplying together all the probabilities of getting a Heads for each coin.
P(Heads=2000)
= 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 2 Coins - 1000 times
Probability of getting one of a kind is close to 0., about 0.% 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 2 Coins - 1000 times
// Flip 2 Coins - 1000 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 = 1000; // 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
*/