I'm not sure you understand the other article that I linked.  It's a different argument.  Watch:

Pick two distinct positive integers, A and B.   You can pick any ones.  Even ones that are purposefully designed to defeat my algorithm below.  They don't have to be random.  You can always pick 3 and 5, or 3 and a number over Googleplex, or two numbers over Googleplex.  You don't have to tell me anything at all about the distribution.  They don't have to be random, but can be if you want.  They just have to be distinct positive integers.

1.  I randomly pick either A or B to start with.  I look at the ones I picked (either A or B.)  I have a 50% chance that I am looking at the larger number, 50% chance that I'm looking at the smaller number.

2.  I don't know ANYTHING, whatsoever, about the other number.  (Except that it is distinct from the first one.)

It could be that in every single case you pick the two integers 7 and 9, so if I'm looking at 7 the other number might have a 100% chance of being a 9.  But I don't know this.  I don't know the distribution.  It could be that in every case you pick two numbers over googleplex, or one very small and one very large number, or any other combination you want.

Now here is the question: can I make a decision based on the number I see about whether to switch numbers, and as a result of this decision I no longer have an exactly 50% chance of stopping on the bigger number, but ever-so-slightly larger?

The surprising answer is, "yes I can".

Let's start with a base case so you can understand the algorithm, then we'll generalize.  The base case is that you pick the integers 2 and 3.

I pick one of them (either 2 or 3).  Then I flip a coin that many times (either 2 or 3).  If it comes up all heads, then switch to the other number.  If it doesn't I stay on the first ones.

What are the chances I stay on 2 and what are the chances that I stay on 3?  The chances of flipping 2 heads are 1 in 4 (2^2) whereas the chances of flipping 3 heads are 1 in 8 (2^3).

So the chances of staying on the 2 IF I PICKED IT are 75% and the chances of staying on the 3 IF I PICKED IT are 87.5%.

Question: do you think 75% is equal to 87.5%?  If they are unequal, then I have a greater chance (by what the difference is) of switching from the one with a lower probability of staying than I do of switching from the one with the higher probability of staying.

Even though I started with a 50% chance of picking the 2 and a 50% chance of starting with the 3, I've now changed it to:

50%...
and 75% chance of staying.

versus
50%...
and 87.5% chance of staying.

So there is a:

  I    0.375 chance of picking the 2 and staying on 2     (0.5 * 0.75) Result: 2
  II   0.125 chance of picking the 2 and switching to 3   (0.5 * 0.25)  Result: 3
  III  0.4375 chance of picking the 3 and staying on 3    (0.5 * 0.875) Result: 3
  IV   0.0625 chance of picking the 3 and switching to 2  (0.5 * 0.125) Result: 2
  ------
  The sum of the above is:
  Probability 1 total.  (I+II+III+IV).  As it should be. All cases accounted for.
  0.5625 (II + III) of staying on 3.
  0.4375 (I+IV) of staying on 2.

As you can see, if the original integers were 2 and 3, we have an ever-so-slightly greater chance of staying on the greater one.

That's because 1/2^n is smaller for 3 (1/8) than it is for 2 (1/4).

Do you follow so far?  Do you agree that 0.5625 is not the same as 0.4375?  Do you agree with how I calculated the possibilities?

2.

Now let's generalize.

Instead of 2 and 3, you get to pick ANY two integers.  I don't know what they are.  They can be any positive integers, as long as they're distinct.

The fact that they're distinct means the probability of switching is not the same, exactly.  It will ever-so-slightly favor the smaller integer, REGARDLESS of what it is.

Let's say you pick "1 googleplex" and "1 googleplex-and-one".

Then the probabilities are:

  I    ~0.5 chance of picking GP and staying on GP       (0.5 * 1-1/2^GP)      Result: GP
  II   ~0.0 chance of picking GP+1 and switching to GP+1 (0.5 * 1/2^GP)        Result: GP+1
  III  ~0.5 chance of picking GP+1 and staying on GP+1    (0.5 * 1-1/2^(GP+1)) Result: GP+1
  IV   ~0.0 chance of picking GP+1 and switching to GP  (0.5 * 1/2^(GP+1))     Result: GP

Sum of probabilities:
The sum of the above probabilities are 1.  (~0.5 + ~0.0 + ~0.5 + ~0.0).

The probability of stopping on GP (I + IV) is ~0.5
The probability of stopping on GP+1 (II + III) is ~0.5

So, we're all set.  The probabilities are the same, right?

Well.  Not exactly.

The probability of stopping on GP (I + IV) is (0.5 * 1-1/2^GP) + (0.5 * 1/2^(GP+1)) which is approximately 0.5 + 0.
The probability of stopping on GP+1 (II + III) is (0.5 * 1/2^GP) + (0.5 * 1-1/2^(GP+1)) which is approximately 0 + 0.5

Are these probabilities equal?  They are in a practical sense: 0.5 + 0 is the same as 0 + 0.5.

But in a rigorous sense one read

(0.5 * 1-1/2^GP) + (0.5 * 1/2^(GP+1))

and one read

(0.5 * 1/2^GP) + (0.5 * 1-1/2^(GP+1))

Are these two exactly equal?

They are if 2^GP is the same as 2^(GP+1).  If these two values are not the same, the probabilities are not the same.

Do you think these two probabilities are equal?  Is 2^googleplex the same as 2^(googleplex+1)?

For all practical purposes they are equal.  But, mathematically speaking, they are clearly NOT equal.

So without knowing whether you would pick 2 and 3, or 2 and a googleplex, or googleplex and googleplex+1, we can arrange our choice such that mathematically speaking, we are not exactly likely to stay on the one as the other.  Just as you were slightly more likely to leave 2 and end up on 3, you are slightly more likely to leave Googleplex and end up on Googleplex+1.

It's not a PRACTICAL difference.  It's a mathematical difference.

Basically, to state otherwise would have to state that 2^n is the same as 2^(n+k), where n is the smaller number and k is the difference between the numbers, and k is at least one?

Do you believe it's true that 2^534 is the same as 2^3423  -- because if they are not the same, then the probability of staying is not EXACTLY equal anymore.

I hope you've been able to follow this reasoning.