The mechanism of the random walk is very different from selecting various pre-shuffled decks of cards. The decks are not transformed from one to the next to the next.

However the end result is the same. The order in which the deck is drawn is mutually independent of the order of the cards within the deck.

(answering this question does not mean that I volunteer to peer review this--I would advise not putting to much effort into this line of questioning until you have spent some time learning more about probability calculations. They're tricky at first, but there are only so many types of scenarios that commonly arise that once you have learned a few of the tricks, it's not so bad)

Thank you for your reply it seems lonely on here at the moment, I believe I have the maths proof now and a good explanation,

posted elsewhere on a maths forum

If we had two variants 1 and 2, and were to conceal their identities, then randomly shuffled the 2 variants, we would have a 1/2 chance of the value 1, being the left aligned of the two variants.

If we were to add a second set of the two variants and randomized them, this would also have a 1/2 chance of 1 being the left aligned.

Now if I was to say, that 1 is a winner and 2 is a loss and your value you would receive is always the left position, we know that if we choose either set, our chance of 1, being aligned to the left by either set is 1/2.

At this point I considered the above and used (n) to represent the above unknown variants in diagram form.

set (a)nn

set (b)nn

The first thing I noticed was each set expanded across a X-axis. Each set remaining a 1/2 chance of 1 being the variant left aligned of each sets X-axis.

Secondly we notice that using multiple sets creates a Y-axis. Rows randomly shuffled that once stopped, creates columns of variants.

The question, what is the chance of receiving a 1 from the aligned left column if we were to pick any one of the sets after the shuffle?

(baring in mind, the right column is removed from the equation)

n

n

My maths to this point-

P(1)/X=(1/2)/t1

P(1)/Y=(?/2)/t2

With a range of entropy minimum 0/2 to a maximum of 2/2.

further maths-

nn

nn

P(1)/^Y=(?/2)/(d/t2)

I believe that if science can not replace ? with a true value, then this alone proves I am correct