Probability of L Consecutive Heads in N Coin Tosses

In summary, the probability of a run of exactly L consecutive heads (or tails) appearing in N independent tosses of a coin can be calculated using the formula P(N,L) = C^{l}_{n} / 2^{n}, where C is the combination function and n is the total number of tosses. However, this formula may not work for all values of N and L, so it is necessary to look into probability texts or other resources for more accurate calculations.
  • #1
grad
12
0
What is the probability that a run of exactly L consecutive heads (or tails) appears in N independent tosses of a coin?

Please help me with this one... I 've searched everywhere but I can't find a general answer, for example P(L,N) = ...
 
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  • #2
[tex]2^{-n}[/tex]
 
  • #3
fatra2 said:
[tex]2^{-n}[/tex]

That's not correct.

Let's suppose we toss a coin 3 times (N = 3) and we want a run of exactly 2 heads (L = 2). Then the combinations that include runs of HH are only two: THH and HHT
The total combinations are 2N=3=8

So, P(2,3) = 2/8 = 1/4

Your answer gives 2-N= 1/8
 
  • #4
Hi there,

I am sorry. I misunderstood your question. I thought you asked what is the probably of having N consecutive heads, on N toss.

I'll look into it a bit deeper, and give you a more precise answer.

Cheers
 
  • #5
fatra2 said:
Hi there,

I am sorry. I misunderstood your question. I thought you asked what is the probably of having N consecutive heads, on N toss.

I'll look into it a bit deeper, and give you a more precise answer.

Cheers

Thank you
 
  • #6
You can look in probability texts for discussions of "runs". There is other information here

http://mathworld.wolfram.com/Run.html

- with a "formula" that gives probabilities as coefficients from a particular generating function.
 
  • #7
statdad said:
You can look in probability texts for discussions of "runs". There is other information here

http://mathworld.wolfram.com/Run.html

- with a "formula" that gives probabilities as coefficients from a particular generating function.

Well, I 've already seen that but I don't understand how these formulas work. Could you explain a little bit more if you can understand them?
 
  • #8
Hi , i think this is the answer P(n,l) = C[tex]^{l}_{n}[/tex] / 2[tex]^{n}[/tex]
 
  • #9
vlad1234 said:
Hi , i think this is the answer P(n,l) = C[tex]^{l}_{n}[/tex] / 2[tex]^{n}[/tex]

Thank you for your answer but what exactly is C?
 
  • #10
Oh , i made a mistake , sorry. Let me fix it . C[tex]^{l}_{n}[/tex] = [tex]\left(^{l}_{n}\right)[/tex]
 
  • #11
HOpe I'm not wrong this time .

P ( N , L ) =
[tex]\left\{2/2^{N} , if N = L + 1 [/tex]

[tex]\left\{( 2^{ N-L+1} + 2^{N-L-2} + 2 ) / 2^{N} , if N = L + 2 [/tex]

[tex]\left\{( 2^{N-L-1} + 2^{N-L-2} + \Sigma^{N-L-2}_{K=1}( 2^{N-L-2-K} * 2^{K} ) + 2^{N-L-1} ) / 2^{N}[/tex]
 
  • #12
vlad1234 said:
HOpe I'm not wrong this time .

P ( N , L ) =
[tex]\left\{2/2^{N} , if N = L + 1 [/tex]

[tex]\left\{( 2^{ N-L+1} + 2^{N-L-2} + 2 ) / 2^{N} , if N = L + 2 [/tex]

[tex]\left\{( 2^{N-L-1} + 2^{N-L-2} + \Sigma^{N-L-2}_{K=1}( 2^{N-L-2-K} * 2^{K} ) + 2^{N-L-1} ) / 2^{N}[/tex]

Sorry mate but both your solutions are wrong. You can easily prove this if you try to find P(2,5) or P(1,5) or whatever you want...

Does anybody know if such a formula even exists?
 

Related to Probability of L Consecutive Heads in N Coin Tosses

1. What does "Probability of L Consecutive Heads in N Coin Tosses" mean?

The probability of L consecutive heads in N coin tosses refers to the likelihood of getting a specific number of consecutive heads in a series of N coin flips. For example, if L=3 and N=10, it means we are interested in the probability of getting exactly 3 heads in a row when flipping a coin 10 times.

2. How is the probability of L consecutive heads in N coin tosses calculated?

The probability of L consecutive heads in N coin tosses can be calculated using the formula P = (1/2)^L * (1/2)^(N-L), where L is the number of consecutive heads we are interested in and N is the total number of coin tosses.

3. Why is the probability of L consecutive heads in N coin tosses important?

This probability is important because it helps us understand the likelihood of a certain outcome in a series of coin tosses. It can also be used in various real-world scenarios, such as predicting the likelihood of a certain event occurring or making decisions based on probability.

4. How does the probability of L consecutive heads in N coin tosses change as N increases?

As N increases, the probability of L consecutive heads decreases. This is because as the number of coin tosses increases, the chances of getting a specific sequence of outcomes (such as L consecutive heads) becomes smaller. However, the overall probability of getting any sequence of outcomes will always be 1, as there is a 100% chance of getting some combination of heads and tails in N coin tosses.

5. Can the probability of L consecutive heads in N coin tosses ever be greater than 1?

No, the probability of L consecutive heads in N coin tosses can never be greater than 1. This is because probability is a measure of likelihood and can never exceed 100%. In other words, there is no scenario where getting L consecutive heads in N coin tosses is guaranteed to happen.

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