What is the probability of flipping a coin six times and getting exactly 50 heads

2022. 10. 4. — Coin flip probability calculator lets you calculate the likelihood of obtaining a set number of heads when flipping a coin multiple times.

Coin Toss Probability Calculator is a free online tool that displays the probability of getting the head or a tail when the coin is tossed.

This coin flip probability calculator lets you determine the probability of getting a certain number of heads after you flip a coin a given number of times.

Know the probability of outcomes when coin is tossed using Coin Toss Probability Calculator. This Calculator checks Coin Flip Probability, gives output quickly.

Enter the number of the flips; Insert the number of the heads; Choose the Type of the probability; Hit the calculate button to calculate the coin flip.

A coin toss is an independent event getting heads/tails in one trial does not affect the outcome of other trials. What is a Coin Toss Probability Calculator? ' ...

(n) Coin Tosses with a list of scenario results displayed * Monte Carlo coin toss simulation. This calculator has 7 inputs.

How does the Coin Toss Probability Calculator work?

What 4 concepts are covered in the Coin Toss Probability Calculator?

If two coins are flipped, it can be two heads, two tails, or a head and a tail. The number of possible outcomes gets greater with the increased number of coins.

By theory, we can calculate this probability by dividing number of expected outcomes by total number of outcomes. The formula: Coin Toss Probability Formula.

2011. 5. 12. — Get the free "Coin Toss Probabilities" widget for your website, blog, Wordpress, Blogger, or iGoogle. Find more Statistics & Data Analysis ...

I can tell you the actual math and formulas, but let's keep it simple and abstract.

First of all, how many outcomes are there? If you flip one coin, just two. If you flip two coins, four. If you flip three coins, it's eight - two for the first times two for the second times two for the third. Simple numbers.

Flip 4 coins, and you're at 16 outcomes, a 2-digit number. Flip 10 coins, and and you're at a 4-digit number. 100 coins is a 31-digit number. Yikes! Roughly "a gajillion."

So, to answer the question you asked out-right: are the odds of an even 50-50 split better than all-heads? Well, of the gajillion outcomes for your hundred flips, there's only one where all the coins come up heads. Even if you're also counting "all tails", that's just two measly outcomes out of a gajillion - a drop in the bucket. (For comparison: There are 100 outcomes where you get 99 heads and 1 tails.)

Now think of how many ways you could get 50 heads and 50 tails. You could flip 50 heads in a row and then 50 tails in a row; or vice versa; you could have all the even flips be heads and the odd ones be tails; or vice versa; you could - why do I need to list them out? We've just come up with four ways. There's actually "a bunch" of ways - small compared to the "gajillion" outcomes, but still easily bigger than the single way to get all heads.

(The exact values of "gajillion" and "bunch" are left as an exercise to the reader.)

Now hold on a dang moment. Doesn't increasing the sample size (that is, the number of flips) actually decrease my odds of getting exactly 50-50? If I flip twice, I have four options: heads-heads, heads-tails, tails-heads, and tails-tails. So I've got even odds of coming up with exactly 50-50. If I flip 100 times, I'm far more likely to get 53-47 or 48-52 or something like that, right?

Well, right. Here's the catch: Increasing the sample size doesn't increase your odds that your experimental data will match the exact odds. Rather, it decreases the likelihood that your experimental data will be far off from the exact odds.

If you flip two coins, you have even odds of thinking that all coin flips have the same outcome! If you flip three coins, you have a disaster - you're either going to get "100-0" or "67-33." However, if you flip a larger amount of times - even if you flip an odd amount of times, where it'd be impossible to get an even 50-50 split - the chances that you'll get results close to 50-50 increases, and the odds that you'll get wacky results like all-heads will shoot down.

Want to join the conversation?

• what if there were would be P(of atleast 2HH in 10 flips). We would probably not use the same method. RIGHT?
  

  • Short Answer: You are right, we would not use the same method.

    Long Answer:
    You would use a similar method, which involves what we've been doing. However, instead of just subtracting "no tails" from one, you would also subtract "one heads" from it too.

    P(at least 2 heads) = 1 - P(No heads) - P(One heads)

    Since there are ten repetitions of the experiment, and two possible outcomes per experiment, the number of different outcomes is 2 ^ 10, or 1024.

    P(No heads) is simple enough to find, just take the probability of tails to the tenth power.
    P(No heads) = (1 / 2) ^ 10 = 1 / 1024

    In order to find P(One Heads) you're going to have to think. If you want only one heads out of ten, there are going to be ten different ways to get one head. Heads could be first, second, third, fourth, fifth, sixth, seventh, eighth, ninth, or tenth, so that's ten different ways you would have just one heads. (Scroll to the bottom of comment to see a picture of what I'm talking about) We have the same number of different possibilities as before, so we keep the denominator the same.
    P(One Heads) = 10 / 1024

    So now we have P(No Heads) and P(One Heads), so we just plug those in to find P(At Least Two Heads):
    P(At Least Two Heads) = 1 - (1 / 1024) - (10 / 1024)
    P(At Least Two Heads) = (1024 - 1 - 10) / 1024
    P(At Least Two Heads) = 1013 / 1024

    P(One Heads) Visual
    1 2 3 4 5 6 7 8 9 10
    H T T T T T T T T T
    T H T T T T T T T T
    T T H T T T T T T T
    T T T H T T T T T T
    T T T T H T T T T T
    T T T T T H T T T T
    T T T T T T H T T T
    T T T T T T T H T T
    T T T T T T T T H T
    T T T T T T T T T H

• Isn't this called complementary counting? Or is it called complimentary counting? What's the difference between the two?

  • It is complementary counting. Compliment has a different meaning than complement.

• 

  So the question of P(at least 2 heads in 10 flips) was asked and the answer was
  P(at least 2 heads) = 1 - P(No heads) - P(1 heads)
  I figure we subtract P(1 heads) because it does not meet our conditions of 2 heads. So I was curious if the rule follows as such:
  P(at least 3 heads) = 1 - P(No heads) - P(1 heads) - P(2 heads)
  And the generalization being
  P(at least n heads) = 1 - P(No heads) + ∑(k=1 to n-1) of -P(k heads)

  • Well done! Yes, your generalization works (though you could just start the summation from k=0, to avoid separating the P(X=0) each time).

    Though be careful about this "rule". Say with ten flips, you wanted the probability of at least 9 heads. With your generalization it would be:

    P(X>=9) = 1 - ∑{k=0 to n-1} P(X=k)

    But this might have you calculate 9 probabilities (0,...,8), when it might be easier to calculate P(X=9) + P(X=10). It's good to know how to manipulate the probability expressions, and knowing that probability sums to 1 is a very useful 'trick'. Using it, or other little rules, we can flip around probability statements to make what we have to calculate easier. For example, some calculators include functions for P(X <= k), and so it is easiest to express the probability in terms of that (like what you did above) when possible. Or say we wanted:

    P( A <= X <= B)

    That is, at least A, but no more than B. We could rewrite this as:

    P(X <= B) - P(X <= A)

    It's just a game of making it easier to calculate for the tools you have available.

• Why do we subtract those from 1?

  • "1" represents the total number of possible events, or 100%. If you want to know what the probability is to get at least one Heads, then that is the same as the probability of all the events (100%, or 1) minus the probability of getting all Tails. If you subtract the possibility of having all tails from the probability of anything happening (100%), then you are left with the probability of all the scenarios where there are Heads involved. Because getting all Tails is the only scenario where the "at least one Heads" requirement is not met, all other scenarios are good and have at least one Heads.

• Suppose that a fair coin is tossed 100 times. Find the probability of observing at least 60 heads. How do you solve for this?

• Why wouldn't you just do .5^10 instead of multiplying all the 1/2s?

  • Yes. Normally you would do P(at least 1H) = 1- (1/2)^10
    Sal was just writing all of the 1/2's out to illustrate what's going on to people who are new to the concept.

• Any kind peeps out there to help me understand this problem in detail please? The problem goes:
  "Toss a coin for 50 times. Calculate the probability of getting head for 30 times.Draw Probability Diagram for the case (for 5 times only)"

  • You will need to use binomial theorem for this question. So the answer is (50 C 30) (0.5^30)(0.5^20) which also equals (50 C 20) (0.5^30)(0.5^20)= (50 C 20)(0.5^50)

    The (n C r) means n!/(r!(n-r)!).

• What is the probability that you will get heads four times in a row when flipping a fair coin

  • A coin has a 50% chance of landing on heads the each time it is thrown. For the first coin toss, the odds of landing heads is 50%. On the second coin toss, take the 50% from the first toss, and multiply it by another 50%. Repeat this for the third and fourth tosses and it should look something like this:

    (1/2)(1/2)(1/2)(1/2) = 1/(2*2*2*2) = 1/(2^4) = 1/16 = 6.25%

    The 1/2 is the chance that the desired outcome occurs, the answer remains the same if the question was "What are the chances of landing tails four times in a row?" or "What are the chances of landing heads the first two times and tails the second two times?"

• I am not really sure when you will use this technique. Can you please explain when using 1- P(not event) would be helpful?

  • Flip a fair coin 13 times. Find the probability of at least 1 head. We could either do:

    P( X ≥ 1 ) = P( X=1 ) + P( X=2 ) + P( X=3 ) + P( X=4 ) + ... + P( X=13 )

    or, we could do:

    P( X ≥ 1 ) = 1 - P( X = 0 )

    Calculating just one probability, P( X=0 ), is much easier than calculating many (in this case, 13) probabilities.

• What does the "Fair" in fair coin represent; what does it mean?

  • Fair means that the coin has a 50-50% chance of getting HEADS or TAILS. They have that because in some other situations, there are "unfair" coins that have more chance of getting one result or another.

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