Corollaries, by considering different deck shufflings, such as perfectly interleaved as perfectly separated:

  9.08 ~
         52/52 × 52/51 × 50/50 ÷ 50/49 × ... 2/2 × 2/1 
       = 52/51 × 50/49 × ... × 2/1 
       = 2^52 × 26!² / 52!
       = (52/52 × 50/51 × ... ×  2/27) × (52/26 × 50/25 × ... × 2/1)

and these equalities can also be directly verified algebraically

This also points to a non-"many worlds"/portfolio version of the prod of zero-variance.

Every bet is e/d, where e is current edge and d is current deck size. So every outcome multiplies the stack by (d + e × (-1)^i)/d, where is ±1, depending on win or lose.

Note that the product of all the values of d is constant, so we can ignore the denominator.

Since we know (from the OP proof) that the product of these numbers is constant for all shuffles of the deck, we can split a shuffled deck anywhere such that both parts are balanced red=blue, and the total (multiplicative) return over each part of the deck is constant across all shuffling of that part of the deck. (There are at least two ways to prove this part!)

This is gives a further hint toward another fascinating fact: over any span of the deck between points where the deck is balanced, the numerators of the bet results double-cover all the even numbers between the starting and ending deck size.

To see why:

* A loss after a loss has a numerator (deck minus edge) of 2 less than the previous bet, as the deck size decreased by 1 and the edge has inccreased by 1.

* A win after a win also has a numerator (deck plus edge) of 2 less than the previous bet, as the deck size decreased by 1 and the edge has decreased by 1.

* A win after a loss, causes a big swing in the numerator, exactly back to the largest not yet double-covered numerator that started the streak that just ended. Then the new win streak continues making the second cover of even numerators, until... a loss after a win jumps the numerator back to continuing the sequence of decreasing even numberators, which will get their second cover later when the later wins come.

Since the deck is balanced, the number of wins always equals the number of losses, as long as we consider the 0 wager on a balanced subdeck to be a loss, since it increases the edge like non-degenerate losses do.

(When the deck is balanced, edge is 0, so the return of no-bet is same as a win is same as a loss)

You can visualize the numerator changes like so: a crane is driving from 52 to 0. Its arm is pointing either forward or backward, and there is a counterweight of the same length pointing in the opposite direction. At each step, the crane arm is either pointing toward 0 and stretches another step toward 0, or points backward to 52 and shrinks (toward 0 milestone and toward 0 arm length), or it swings to the other direction. Whenever the crane stretches toward 0, the counterweight stretches backward, its end not moving relative to the ground.

Because the deck is balanced at start and empty deck is balanced, the crane starts and ends with a 0-stretch arm. The front side is either the frame arm stepping 2 steps forward at a time relative to the ground, or holding still while the backside crane arm shrinks closer, and the crane arm occasionally flips back and forth pointing forward or ackward. And vice versa for the counterweight.

Over the course of the drive, the crane arm end reaches every even milestone once pointing forward and once again pointing backward.

Intuition for the bet size:

When the deck has d cards left, it is sensible to make d bets of 1/d your stack, where each bet is that one specific card is next. If there are r reds and b=r+e blues, r of these bets simply cancel out r other bets, leaving e (times 1/d) remaining to be a nontrivial bet.

Interesting as a mathematical puzzle - but note that it's difficult to find cooperative, solvent counter-parties for "I can't lose" betting games.