Roulette Math

This problem and subsequent math came up after a discussion I had with a friend about roulette. Couldn’t you, he wondered, just always bet on red and keep doubling your bet until you win? Aren’t you guaranteed to make money that way?

Short answer: yes. You will earn exactly the amount of money you placed for your initial bet. I though through the math of this, and arrived at this conclusion after a couple hours of pondering and scribbling. Why exactly does this method work?

It’s true because the following is true:

2^k = 1 + \displaystyle\sum_{i=0}^{k-1} 2^i

I will return to roulette in a moment, but first let me offer a proof for the above statement:

Proof: This is a proof by induction. We begin by checking to see if the equation holds for k = 1 and we see that it does:

2^1 = 1 + \displaystyle\sum_{i=0}^0 2^i

2 = 1 + 1 = 2

We continue by assuming the equation holds for any arbitrary k. We then must show that k \Rightarrow k+1:

2^{k+1} = 1 + \displaystyle\sum_{i=0}^{k} 2^i

We do this by adding 2^k to both sides.

2^k + 2^k = 2^{k+1} = 1 + \displaystyle\sum_{i=0}^{k-1} 2^i + 2^k

2^{k+1} = 1 + 2^0 + 2^1 + ... + 2^{k-1} + 2^k

2^{k+1} = 1 + \displaystyle\sum_{i=0}^{k} 2^i

QED

Application To Roulette

I’m dying, btw, to have an actual mathematician check my math and tell me if the above proof is valid. But on to the money! The above theorem essentially says, in english, that any power of two is the sum of the previous powers, plus one. We apply that to gambling thusly:

Imagine you play roulette, betting only on red (or black/even/odd), and begin with a principal amount of cash, p. After k rounds of doubling your bet, betting, and losing, you have lost

p + 2p + p(2^2) + ... + p(2^k) or:

p\displaystyle\sum_{i=0}^k 2^i

If you win on the next round, round k+1, you will win of course p(2^{k+1}). From the theorem, it follows that you have won

p(2^{k+1}) = p + p\displaystyle\sum_{i=0}^k 2^i

If we subtract what we’ve won from what we’ve spent, we find how much we’ve gained:

p + p\displaystyle\sum_{i=0}^k 2^i - p\displaystyle\sum_{i=0}^k 2^i = p

We find that we are left having gained exactly p. Incidentally, I’ve discovered that the theorem I proved above can (probably) be extended to all bases:

n^k = 1 + (n-1)\displaystyle\sum_{i=0}^{k-1} n^i

But I haven’t really found a way of proving it. I’m quite certain that it can be proved using similar logic as before. That is, you add (n-1)-many n^k’s to each side. I just have no way of formalizing it. Any help?

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