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 10000 Coins containing atleast one Heads using the following permutations and combinations generator All possible combinations of 10000 coins with atleast one Heads All possible permutations of 10000 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 10000 coins.
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 is close to 0., about 0.% percent.
You can calculate this using counting or by using the binomial distribution
You can think of 10000 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 10000 spots? Or, in how many ways can we pick 1 out of 10000 spots? The answer can be caclulated using the formula [ 10000 choose 1 ]. So, there are 10000 ways to choose 1 out of 10000 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 10000 spots / Total permutations of 10000 coins = (10000 choose 1) / (2^10000) = 10000 / INF
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=10000 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 = (10000 choose 1) (1/2)1 (1/2)10000-1 = (10000 choose 1) (1/2)10000 = 10000 x (1/INF)
You can generate all possible permutations and combinations of 10000 Coins containing exactly 1 Heads using the following permutations and combinations generator All possible combinations of 10000 coins with exactly 1 Heads All possible permutations of 10000 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 5000 Heads is close to NaN, about NAN% percent.
You can calculate this using counting or by using the binomial distribution
You can think of 10000 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 5000 Heads. In other words, in how many different ways can we place 5000 Heads in 10000 spots? Or, in how many ways can we pick 5000 out of 10000 spots? The answer can be caclulated using the formula [ 10000 choose 5000 ]. So, there are INF ways to choose 5000 out of 10000 spots. The probability is calculated by dividing the number of ways to pick 5000 spots by the number of total permuations.
P (H=5000) = Ways to pick 5000 out of 10000 spots / Total permutations of 10000 coins = (10000 choose 5000) / (2^10000) = INF / INF
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 5000 Heads, we need k=5000 successful events out of total n=10000 events. Each event has a probability p=1/2 of occuring.
Probability of k=5000 successful events is then calculated as follows.P(k=5000) = (n choose k) pk (1-p)n-k = (10000 choose 5000) (1/2)5000 (1/2)10000-5000 = (10000 choose 5000) (1/2)10000 = INF x (1/INF)
You can generate all possible permutations and combinations of 10000 Coins containing exactly 5000 Heads using the following permutations and combinations generator All possible combinations of 10000 coins with exactly 5000 Heads All possible permutations of 10000 coins with exactly 5000 Heads
To verify the answer, you can divide the number of permutations containing exactly 5000 Heads by number of total possible permutations.
Probability of getting exactly 9999 Heads is close to NaN, about NAN% percent.
You can calculate this using counting or by using the binomial distribution
You can think of 10000 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 9999 Heads. In other words, in how many different ways can we place 9999 Heads in 10000 spots? Or, in how many ways can we pick 9999 out of 10000 spots? The answer can be caclulated using the formula [ 10000 choose 9999 ]. So, there are INF ways to choose 9999 out of 10000 spots. The probability is calculated by dividing the number of ways to pick 9999 spots by the number of total permuations.
P (H=9999) = Ways to pick 9999 out of 10000 spots / Total permutations of 10000 coins = (10000 choose 9999) / (2^10000) = INF / INF
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 9999 Heads, we need k=9999 successful events out of total n=10000 events. Each event has a probability p=1/2 of occuring.
Probability of k=9999 successful events is then calculated as follows.P(k=9999) = (n choose k) pk (1-p)n-k = (10000 choose 9999) (1/2)9999 (1/2)10000-9999 = (10000 choose 9999) (1/2)10000 = INF x (1/INF)
You can generate all possible permutations and combinations of 10000 Coins containing exactly 9999 Heads using the following permutations and combinations generator All possible combinations of 10000 coins with exactly 9999 Heads All possible permutations of 10000 coins with exactly 9999 Heads
To verify the answer, you can divide the number of permutations containing exactly 9999 Heads by number of total possible permutations.
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=10000) = 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 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 ...)
// Flip 10000 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 = 10000; // 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 */