Roulette Math – Part II

Thanks to no one, I’ve finally reached a conclusion to the question I posed at the end of my Roulette Math post. To prove it, proof by induction with two variables will be used. Explicitly,

let $P(n,k)$ be the statement $n^k = 1 + (n-1)\displaystyle\sum_{i=0}^{k-1} n^i$

To prove $P$ ,

Prove $P(1,1)$
Assume $P(n,1)$
Prove that $P(n+1,1)$ is true if $P(n,1)$ is true
Assume $P(n,k)$
Prove that $P(n,k+1)$ is true if $P(n,k)$ is true

Phase 1

$1^1 = 1 + (1-1)\displaystyle\sum_{i=0}^{1-1} n^i = 1 + 0 = 1$ is true

Phases 2, 3

Induction is not necessary here.

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

Phases 4,5

Assume $n^k = 1 + (n-1)\displaystyle\sum_{i=0}^{k-1} n^i$ (phase 4)

All that remains is to show that $P(n,k+1)$ is true if the above is true (phase 5):
Since $n^k = 1 + (n-1)\displaystyle\sum_{i=0}^{k-1} n^i$ by assumption, then $n^k + (n-1)n^k = (n-1)n^k + 1 + (n-1)\displaystyle\sum_{i=0}^{k-1} n^i$

$n^k + (n-1)n^k = n^k(1+(n-1)) = n^k(n) = n^{k+1}$

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

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

$n^{k+1} = 1 + (n-1)n^k + (n-1) + (n-1)n +...+ (n-1)n^{k-1}$

$= 1 + (n-1) + (n-1)n +...+ (n-1)n^{k-1} + (n-1)n^k$

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

Thus, if $P(n,k)$ is true, $P(n,k+1)$ must be true. Phase 5 complete. QED.

Further Application To Roulette

Once again, imagine you’re playing roulette, betting only on red, black, even, or odd (the ones with 2:1 payout). Instead of doubling, you decide that you’d rather triple, quadruple, or even $n$ -uple your bet each round. You continue multiplying your principle bet, $p$ by $n$ until you win. After $k$ rounds, you have lost

$p+p(n)+p(n^2)+...+p(n^k) = p\displaystyle\sum_{i=0}^k n^i$ dollars

If you win in round $k+1$ you will win $p(n^k) = p + p(n-1)\displaystyle\sum_{i=0}^{k} n^i$

Your total earnings are what you’ve won minus what you spent:

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

Which, according to Mathematica, can be simplified to

$\frac{p(1+n^{k+2}-2n^{k+1})}{n-1}$

That means if your initial bet is $100, you quintuple your bet every time, and it takes you 4 rounds to win, you’ll walk away with a bonus $103,700. That is, if you can afford the betting (and if there’s no maximum bet!)

This entry was posted on Tuesday, October 7th, 2008 at 2:19 am and is filed under Math. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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