If 6 fair coins are tossed, find probability of getting more heads than tails
Answers
Answered by
1
I am going to assume that your question is “If 6 fair coins are tossed, what is the probability that most will be tails.”
Bjarke’s answer is correct, but I want to make sure I clarify some of the points he made about symmetry and the binomial distribution.
We know that there are 7 general cases of what could happen if we toss 6 fair coins.
6 heads, 0 tails
5 heads, 1 tail
4 heads, 2 tails
3 heads, 3 tails
2 heads, 4 tails
1 head, 5 tails
0 head, 6 tails
Only cases 5, 6, and 7 satisfy our proposition, so we want to find out the probability that these cases happen. One naive method of doing this is just calculating the probability of case 5, 6, and 7 each and summing up these probabilities. This works, but it is a lot of unnecessary work.
The key insight in this problem is noticing that case 1 and case 7 are essentially the same thing, since the ()=()
. Similarly, case 2 mirrors case 6 and case 3 mirrors case 5. This means the probability of achieving the first three cases is equal to probability of getting case 5, 6, and 7. This means that if we can somehow find the probability of all the cases except for case 4, we can just divide by 2 and get our answer. But hey, that’s just 1−(4)/2
And this key insight is why the symmetry argument is useful in this problem, because now we can just calculate the probability of case 4 occurring to solve this problem.
So now let’s calculate (4)
. The total possible combinations of heads and tails for 6 coins is 2^6= 64
. This is our denominator. For the numerator, we want to calculate the number of combinations of 3 heads and 3 tails. There are many ways to solve this problem. One intuitive way of looking at it is choosing 3 spots for heads from 6 spots, and then putting tails everywhere else. This ends up being (6 3)=20
. Another cool way to look at this is ask yourself how many ways can you rearrange the letters HHHTTT. This yields the exact same result. Therefore, our total probability for case 4 is 20/64=5/12
Now this calculation we did here is exactly what happens in the binomial distribution. The binomial distribution takes in
events that are either success or failure (probability of success is )and states that the probability of exactly
k
of them occurring is (=)=( k)p^k(1−)^n-k
In this example
, =3
, =0.5
, and =6
Therefore the probability of getting most of your coins to be tails is 1−5/16 / 2 = 11/32
Bjarke’s answer is correct, but I want to make sure I clarify some of the points he made about symmetry and the binomial distribution.
We know that there are 7 general cases of what could happen if we toss 6 fair coins.
6 heads, 0 tails
5 heads, 1 tail
4 heads, 2 tails
3 heads, 3 tails
2 heads, 4 tails
1 head, 5 tails
0 head, 6 tails
Only cases 5, 6, and 7 satisfy our proposition, so we want to find out the probability that these cases happen. One naive method of doing this is just calculating the probability of case 5, 6, and 7 each and summing up these probabilities. This works, but it is a lot of unnecessary work.
The key insight in this problem is noticing that case 1 and case 7 are essentially the same thing, since the ()=()
. Similarly, case 2 mirrors case 6 and case 3 mirrors case 5. This means the probability of achieving the first three cases is equal to probability of getting case 5, 6, and 7. This means that if we can somehow find the probability of all the cases except for case 4, we can just divide by 2 and get our answer. But hey, that’s just 1−(4)/2
And this key insight is why the symmetry argument is useful in this problem, because now we can just calculate the probability of case 4 occurring to solve this problem.
So now let’s calculate (4)
. The total possible combinations of heads and tails for 6 coins is 2^6= 64
. This is our denominator. For the numerator, we want to calculate the number of combinations of 3 heads and 3 tails. There are many ways to solve this problem. One intuitive way of looking at it is choosing 3 spots for heads from 6 spots, and then putting tails everywhere else. This ends up being (6 3)=20
. Another cool way to look at this is ask yourself how many ways can you rearrange the letters HHHTTT. This yields the exact same result. Therefore, our total probability for case 4 is 20/64=5/12
Now this calculation we did here is exactly what happens in the binomial distribution. The binomial distribution takes in
events that are either success or failure (probability of success is )and states that the probability of exactly
k
of them occurring is (=)=( k)p^k(1−)^n-k
In this example
, =3
, =0.5
, and =6
Therefore the probability of getting most of your coins to be tails is 1−5/16 / 2 = 11/32
Similar questions