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08-04-2023 , 02:44 AM
I'm trying to answer questions in the theme of the following:

1) What's the probability that over 100 flips, a fair coin will ever have a score of +8 heads?

2) What's the probability that over 100 flips, a coin with a bias of 53% heads will ever be -4 heads?

3) What's the probability that a coin with a bias of 53% heads will be tied heads-to-tails after 100 flips?

How do I solve these sorts of problems?
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Basic question on variance of a series of trials, like coin flips
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Basic question on variance of a series of trials, like coin flips
08-04-2023 , 01:39 PM
Binomial distribution.

There are on-line calculators that can be used or Excel. When you say +8 you have to specify exactly 8 or >=8. Same for exactly -4 or <=-4.
Basic question on variance of a series of trials, like coin flips Quote
08-04-2023 , 02:04 PM
The "ever" aspect of the first two questions adds a wrinkle but it is not too difficult to overcome.
Basic question on variance of a series of trials, like coin flips Quote
08-04-2023 , 04:13 PM
Q3 is the aforementioned Binomial PMF. There are C(100,50) desirable permutations, each with probability (.53*.47)^50, for an overall probability of .066457

Q1 and Q2 are finite-time gambler's ruin problems and they are indeed wrinkly compared to Q3.

A1 = sum_k[8 to 54] of (8/k)*C(2k-9, k-8)*.5^(2k-8) = .42616

A2 = sum_k[4 to 52] of (4/k)*C(2k-5, k-4)*(.47^k)*.53^(k-4) = .51792

With infinite flips, A1 would be 100% and A2 would be (47/53)^4
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08-04-2023 , 08:11 PM
Quote:
Originally Posted by whosnext
The "ever" aspect of the first two questions adds a wrinkle but it is not too difficult to overcome.
Good point. I ignored the "ever" and heehaw provided the method it include it
Basic question on variance of a series of trials, like coin flips Quote
08-04-2023 , 11:58 PM
Thanks for the responses.
Basic question on variance of a series of trials, like coin flips Quote
08-05-2023 , 07:13 PM
Quote:
Originally Posted by heehaww
With infinite flips, A1 would be 100% and A2 would be (47/53)^4
With infinite flips, are you saying there will never be a 100 flip string where it's -4 heads?
Basic question on variance of a series of trials, like coin flips Quote
08-06-2023 , 09:01 AM
With infinite flips, (47/53)^4 or 61.84% is the chance that the overall score will ever be -4 heads.

If you were to reset the score after every 100 flips, then you'd be 100% to reach a -4, ie it's 100% that there will be a series of 100 flips containing a -4 subtotal. Each disjoint series of 100 flips is independent and therefore has the same .51792 probability of -4, so of course with infinite repeated attempts at a 52% chance it's sure to hit.

When not resetting the score, Heads' lead grows with time and -4 becomes harder to reach. The probability of reaching -4 amounts to a sum of ever-decreasing probabilities, making it different from monkeys typing Shakespeare.
Basic question on variance of a series of trials, like coin flips Quote
08-06-2023 , 12:41 PM
Quote:
Originally Posted by heehaww
With infinite flips, (47/53)^4 or 61.84% is the chance that the overall score will ever be -4 heads.

If you were to reset the score after every 100 flips, then you'd be 100% to reach a -4, ie it's 100% that there will be a series of 100 flips containing a -4 subtotal. Each disjoint series of 100 flips is independent and therefore has the same .51792 probability of -4, so of course with infinite repeated attempts at a 52% chance it's sure to hit.

When not resetting the score, Heads' lead grows with time and -4 becomes harder to reach. The probability of reaching -4 amounts to a sum of ever-decreasing probabilities, making it different from monkeys typing Shakespeare.
Can this be why some number theory conjectures will never be proven? No one seems to know.
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08-14-2023 , 02:39 AM
God knows. But God is a moronic concept invented by diseased people for their own nefarious ends. God is ignorant of mathematics. Numbers are an illusion, like getting to heaven after you die.
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08-15-2023 , 03:45 PM
my sense is that OP's use of "ever" really meant "how often does it happen?"

that's how i read it in relation to the rest of the post.

and definitely, OP should check out "binomial calculators"
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08-23-2023 , 03:47 PM
If one is familiar with normal distribution probabilities (I.e. probability of at least 1SD greater than average is about 2/3, 2SD greater is about 95%, 3SD about 99%), you can get a decent estimate of the binomial probability by using a normal distribution with an average of N/2 and a standard deviation of sqrt(N)/2 if N is large enough. It may not be a gray approximation for N=100, but it gets better for larger N. For biased coins the same approximation can be used except that the average is pN and the standard deviation is sqrt(Np(1-p)) where p is the bias probability (0.53 in the OP).
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