Probability of consecutive same faces in coin flip

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Probability of consecutive same faces in coin flip

Postby >-) » Tue May 29, 2012 7:29 am UTC

I would like to know, for a coin, or any other mechanism that is completely random and generates two possible outcomes, what is the chance that you will get 'a' amount of the same outcomes 'b' times in a row when the device makes 'n' amount of random flips. So like, if a coin is flipped randomly 'n' times, what is the probability that you would get 'a' heads or tails in a row 'b' times in the entire sequence?

The best I can do right now is just figure out the probability that 'a' heads or tails will happen in a row in a sequence of 'n' flips. And that would be 1-((a-1)2-1/(a-1)2)(n-a+1)?

I tried to do something like (1-((a-1)2-1/(a-1)2)(n-a+1))b , but that didn't work because it made no sense when i plugged some numbers in (for example i somehow got about 20% chance that i could have 5, 5 heads/tails in a row in 20 flips.

Ben-oni
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Re: Probability of consecutive same faces in coin flip

Postby Ben-oni » Tue May 29, 2012 9:21 am UTC

You may have to clarify the question.

For instance, if you get (a+1) of the same result in a row, does that count? If so, does it count twice?

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Re: Probability of consecutive same faces in coin flip

Postby >-) » Tue May 29, 2012 9:30 am UTC

I guess, no, it wouldn't count more than once.

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Re: Probability of consecutive same faces in coin flip

Postby OverBored » Tue May 29, 2012 11:22 am UTC

Would 2a in a row count twice? Or do you demand that sequences are separated by an opposite result?

Are we just counting sequences of one type, or both simultaneously? e.g a = 2: HHTT, is that a count of 2 or 1?
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Re: Probability of consecutive same faces in coin flip

Postby >-) » Tue May 29, 2012 1:39 pm UTC

HHHHHHHH would be 1 sequence of 8 h's and nothing else.

Not completely sure what you mean with the second one, but HHTT would be 2 different strings of 2 outcomes.

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Snark
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Re: Probability of consecutive same faces in coin flip

Postby Snark » Tue May 29, 2012 1:44 pm UTC

Not sure how to tackle this mathematically, but it shouldn't be too difficult to program a brute force search for the solution with a smallish N. And you can use random sampling if N gets too large.

For N = 20 and A = 5, my brute force search program takes ~20 seconds and gives

Code: Select all

75.0129699707  percent chance of 0 sequences of 5 heads
22.8675842285  percent chance of 1 sequences of 5 heads
2.06680297852  percent chance of 2 sequences of 5 heads
0.052547454834  percent chance of 3 sequences of 5 heads
9.53674316406e-05  percent chance of 4 sequences of 5 heads


For N = 20 and A = 5, a sample size of 100,000, and a run time of ~6 seconds, my random sampling program gives

Code: Select all

75.033  percent chance of 0 sequences of 5 heads
22.837  percent chance of 1 sequences of 5 heads
2.084  percent chance of 2 sequences of 5 heads
0.046  percent chance of 3 sequences of 5 heads
0.0  percent chance of 4 sequences of 5 heads

Which is pretty close.

I assumed that 6 A's count as a single sequence but 10 A's count as two sequences.

If you're looking for a mathematical answer, you can obviously ignore this post (unless you want to double-check your formulas if you obtain any). If you just want an answer, I can post the source code as well (though it's not difficult to figure out).
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Re: Probability of consecutive same faces in coin flip

Postby Timefly » Tue May 29, 2012 4:40 pm UTC

Assuming that HHHH counts as 2 sets of 2 heads in a row.

Then the chance of getting at least 'a' heads in a row 'b' times in a sample of 'n' trials is given by the function p(a,b,n) = ((n-b)!/n!)*0.5^(ab)

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Re: Probability of consecutive same faces in coin flip

Postby OverBored » Wed May 30, 2012 12:14 pm UTC

But, consider n = 3, b = a = 2
Clearly impossible since 2 strings of length 2 require at least 4 bits in order to happen, but your formula gives 1/96,

In fact, for fixed a,b, your formula gives a decreasing probability as n increases, which is absurd.
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