Quote:
Originally Posted by BeaucoupFish
Bonus question:
How would you try to convince someone that thinks "if I flip 99 heads in a row, the next flip is more likely to be tails, because the Law of Large Numbers means coin flips converge on 50%".
I was surprised to hear it from this person, but at the same time, I was more surprised at how difficult it was to find a persuasive explanation, instead of just declaring "every expert resource calls this the Gambler's Fallacy", which is not particularly persuasive.
Telling them that each coin flip is independent hasn't changed their mind, I think it's because they are imagining that the "convergence to 50%" is changing the probabilities over time, I think they are viewing the Law of Large Numbers as not simply descriptive but prescriptive somehow.
It might just be an intuition they have that I can't see how to change, but it's been interesting to talk about. They are otherwise very smart and rational, but not familiar with gambling.
I had another idea about this that I like although it might not help people who don't know much probability. You can represent each of the i flips, i=1,2,... as random variables Xi which take on the values 1 for heads and 0 for tails. Then n flips are represented as a random n-tuple (X1,X2,...,Xn). An example outcome of the n flips might then look like (1,0,0,1,1,1,0,...,1,0,0).
The number of heads after n flips can be represented by the sum of the Xi's. i.e. S(n) = X1+X2+...+Xn. Notice S(n) is a random variable. It only takes on a specific value when the coin is actually flipped n times. Before you flip the coin n times you don't know what S(n) will be.
Just as you have a random number of heads in n flips, S(n), you also have a random average of the 0,1 Xi outcomes in n flips given by S(n)/n. Let's denote this random average by M(n) = S(n)/n . M(n) is called the "Sample Mean" for the Xi to distinguish it from the "Mean" of Xi. "Mean" is just another word for "Average".
We know the "Mean" for Xi is 1/2. We know this before we do the flips. The Mean is a fixed number depending only on the coin being fair and not on any specific outcome of n flips. But the Sample Mean M(n) is a random variable like S(n). We don't know what M(n) will be before we do the n flips. However, we can talk about probabilities for the Sample Mean M(n). For example, how likely is it that M(n) will come out to be close to the actual Mean, 1/2 ?
Now, what does it mean to say that the RANDOM Sample Mean, M(n) = S(n)/n "converges" to 1/2 as n gets large. Loosely speaking, it means that we know before the coins are flipped that M(n) will be close to 1/2 for large n
with high probability. Or, a little more precisely, decide how close to 1/2 you want to get, for example somewhere within 1/2 +or- 0.0001. Also pick a probability as close to certainty that you want, say 99.999%. Then before we do the flips there is a large number N we can pick so that as long as we flip the coin more than N times
the probability M(n) will be within 0.0001 of 1/2 will be greater than 99.999%. i.e. P(1/2 - 0.0001 < M(n) < 1/2 + 0.0001) > 99.999% for all n>N. And if we want M(n) closer to 1/2 with higher probability we just need to pick N larger.
Now here's the point to all this. We can handicap or advantage the number of heads S(n) by and number we want. We don't have to actually "see" 99 straight heads to start the sequence of flips. We can just specify an initial heads advantage to S(n) by adding 99 to it. i.e. look at the heads-advantaged random variable S(n)+99. Certainly, just choosing to add 99 to S(n) from the start should have no effect on S(n) or X1. And despite advantaging S(n) by artificially adding 99 heads to it from the start we will still see the same convergence for the advantaged S(n). i.e. the advantaged sample mean will still converge to 1/2.
(S(n) + 99)/n will still converge to 1/2 just like S(n)/n does.
In fact, you can advantage S(n) by any fixed amount and the same is true despite there being no effect on S(n) by the artificial advantage you give it.
The random advantaged mean (S(n)+1000000000)/n still converges to 1/2, not because S(n) decides to have more tails to balance out the artificial advantage it doesn't even know we gave it, but because the artificial advantage is drowned out by the law of large numbers for S(n)/n.
Large numbers can get very very large. Consider the googol. Then consider the googolplex.
PairTheBoard
Last edited by PairTheBoard; 05-07-2021 at 10:40 AM.
Reason: typo