http://www.edwardothorp.com/wp-conte...rEachRound.pdf

In page 6, [Proof] section...How does the mapping prove the probability distributions for the two segments are identical?

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Proof. Consider two games played on the same pack of n cards. Game
1, or G1, has m rounds with strategy sets S1, · · · , Sm. On round k, player i
plays simple strategy S. A typical ordering is x1, · · · , xn. In Game 2, or G2,
there is just one round with only one player. This player also uses strategy
S. A typical ordering is y1, · · · , yn. Let f : x = x1, · · · , xn ? y = y1, · · · , yn
be the mapping which selects the cards used by player i, round k of G1, and
those cards which the dealer needs to use in order to settle the hand of player
i in G1, and maps them into A = y1, · · · , yt. If player i busts, the dealer in
G1 doesn’t have to draw for player i, but may have to draw for other players
who don’t bust. These latter cards are not selected for A by f . Also if player
i has a natural the dealer settles with player i without drawing any cards.
Any cards he might draw because of other players are not selected for A.
The order of cards in A is the same as the order in which they were used in
G1, so that they exactly suffice and the player has the same result in G2 as
in G1. Preserve the order of the remaining cards in x and map them into
B = yt+1, · · · , yn.

This mapping is defined on every x because of (A3m) and the fact the
strategy sets are deterministic. It is well-defined, i.e. each x maps onto one
and only one y. Further, f is onto, i.e. every y corresponds to some x. To
see this, pick any y = (y1, · · · , yn). Use game 2 to select an initial segment
A. Call the rest B. Now use A,B to play game 1 as follows. On round k draw
cards in order from A whenever the S player or the dealer need cards prior to
settling the S player’s hand. Also draw cards in order from B whenever other
players or the dealer on round k, or anyone on other rounds, need cards.

Since f is onto and the number of x’s and y’s are the same, namely n!,
then f is one-to-one whence the set of f (x) = y are simply a shuffling or
rearrangement of the x’s. Therefore the list of sub orderings in G1 which
map into A’s in G2 is identical to the list of A’s except for rearrangement.
Therefore the probability distributions for S are identical in G1 and G2. Thus
the probability distribution for S is always the same as that for round 1 with
only one player, using S, versus the dealer. This proves the theorem.