You decide to employ the following strategy. Suppose that, by counting already-flipped cards, you deduce that there are $R$ red cards and $B$ black cards left in the deck. If $R > B$, you guess red while betting $\frac{R-B}{R+B}$ of your total money. Similarly if $R < B$ you guess black while betting $\frac{B-R}{B+R}$ of your total money. If $R=B$ you don't bet that turn. Show that this strategy *guarantees* a final amount of:
$$\frac{2^{2n}n!n!}{(2n)!}$$
Note: this final amount is much bigger than people think. That sum equals to:
$$\left(1 + \frac{1}{2n-1} \right) \left(1 + \frac{1}{2n-3} \right) \dots \left(1 + \frac{1}{3} \right) \left(1 + \frac{1}{1} \right)$$
which is already greater than 2, and only increases as $n$ increases. This is an extension to the obvious strategy to wait until the last card before we bet the one dollar that we start with, to which we'd be guaranteed a final amount of two.

### First Solution

First we claim the following: if at any point there are $R$ red cards and $B$ black cards left in the deck, and your current total is $x$, employing the above strategy will give you a final amount of: $$\frac{2^{R+B}R!B!}{(R+B)!} x$$ That claim is easily verified if either $R+B=1$ because then we would know exactly what's left in the deck. Now we prove the claim by induction on $R+B$. Clearly if $R=B$ then we don't bet that flip, and we reduced it to the case $R+B-1$. But if $R \neq B$, and WLOG we may assume that $R > B$. We bet $\frac{R-B}{W+B}$ that the next card will be red.Case 1: the next card is red. That means our current total becomes $(1 + \frac{R-B}{R+B})x = \frac{2R}{R+B}x$. But the deck now contains $R-1$ red and $B$ black, so by our induction hypothesis, our final amount will be: $$\frac{2^{R+B-1}(R-1)!B!}{(R+B-1)!} \frac{2R}{R+B}x =\frac{2^{R+B}R!B!}{(R+B)!} x $$

Case 2: the next card is black and we lost our bet. That means our current total is now $(1 - \frac{R-B}{R+B})x = \frac{2B}{R+B}x$ and the deck contains $R$ red and $B-1$ black, so our final amount is: $$\frac{2^{R+B-1}R!(B-1)!}{(R+B-1)!} \frac{2B}{R+B}x =\frac{2^{R+B}R!B!}{(R+B)!} x $$ And the claim is proven. Now it's straightforward to apply that if $R=B=n$, we have: $$\frac{2^{2n}n!n!}{(2n)!} = \frac{(2n)(2n-2)\dots (4)(2)}{(2n-1)(2n-3) \dots (3) (1)}$$ $$= \left(1 + \frac{1}{2n-1} \right) \left(1 + \frac{1}{2n-3} \right) \dots \left(1 + \frac{1}{3} \right) \left(1 + \frac{1}{1} \right)$$

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