Not losing money in a fair betting scheme? If a bookie makes the payoff odds correspond to the true probabilities, why is it that we can always put down our money in such a way that we don't lose any money (or gain any)?
Edit: What I said above is not expressing exactly what I meant. There was a formula I saw that was $\frac{1}{r_i+1}$, where $r_i$ is from the odds $r_i:1$ for a specific event. What the paper I was reading said is that if $\sum\frac{1}{r_i+1}$ was less than one, you can always divvy up your money so you make a profit no matter what. So, in the case of horse racing, it said that bookies always make sure that sum is more than one.
 A: Assume that you are betting on $n$ different events, and the odds given by the bookie $r_1,\dots,r_n$ are such that
$$
\sum_{i=1}^n\frac{1}{r_i+1}\le1.
$$
Then, bet $1/(1+r_i)$ on each of the outcomes $i=1,\dots,n$. Assuming that you are quoting fractional odds, then you receive 
$$
\frac{r_j}{1+r_j}+\frac1{1+r_j}=1,
$$
when the winning outcome is $j$. Additionally, you lose $\sum_{i=1}^n\frac{1}{r_i+1}$ every time as a cost. The final gain is thus always
$$
+1-\sum_{i=1}^n\frac{1}{r_i+1},
$$
which is nonnegative by assumption.
A: Bet in the events such that the betting amounts for event i are in the following ratio:
$ x_1 :x_2...:x_n=\frac{1}{r_1+1}:\frac{1}{r_2+1}:...:\frac{1}{r_n+1} $. 
Suppose you invest a total of $\frac{1}{r_1+1}+\frac{1}{r_2+1}+...+\frac{1}{r_n+1}$, which is less than one. This simplifies the amount of bet on each event on ebent i as $\frac{1}{r_i+1}$.
Now, your expected payoff= $p_1.(1+r_1)\frac{1}{r_1+1}+....+p_n.(1+r_n) \frac{1}{r_n+1}=p_1+...+p_n=1$.
So, expected earning is more than your investment, hence, profit.
A: First note that in betting jargon, a bet pays off at a rate of “$a$ to $b$” if a bet of $b$ yields a returned amount of $a+b$.  For example, if you bet \$1, and on a winning bet you keep your dollar, plus get \$2 more, the bet is paying off at “2 to 1”. But another way to look at the same situation, mathematically simpler, is that you pay \$1 up front to enter the game, and this payment is gone forever, but if you win, the bookie pays you a prize of \$3.  To convert these odds rates, such as “8 to 5”, to prize amounts, you add 1, which represents the amount of your original bet that is returned. (This is where the $+1$ in entering in the formula you cite.) So odds of 8 to 5 mean that the prize amount for a \$1 entry fee is $\frac85+1 = \frac{13}5$. "Even money" is bookie jargon for odds of 1 to 1.
Now suppose the game has $k$ possible outcomes  (horses, say) occurring with probabilities $p_1, p_2,\ldots, p_k$ respectively, where $$\sum p_i = 1.$$  Suppose further that when event $i$ ($1\le i\le k$) occurs, the 
prize amount paid by the kindly bookie for a \$1 entry fee is $\frac1{p_i}$.
(That is, if the probability of the event occurring was $\frac ab$, the bookie will pay off a prize amount of $\frac ba$, or, in the language of odds, will pay off bets at $b-a$ to $a$.)
For example, suppose you bet \$1 on event 7, which occurs with probability $\frac14$.  Then if event 7 actually occurs, which it does one-quarter of the time, the bookie keeps your \$1 and  pays you $\frac1{\frac14} = \$4$.  (That is payoff odds of three to one.) If some other event occurs, the bookie still keeps your \$1 and you get nothing in return.
Now you can achieve a guaranteed payoff as follows: For each event $i$, bet $\$p_i$. Your total bet is therefore $\sum p_i = \$1$.  Your payoff $P_i$ if event $i$ occurs is then $$\$p_i \cdot \frac1{p_i} = \$1$$ where the $p_i$ on the left is the amount that you bet on event $i$ and the $\frac1{p_i}$ is the rate at which the bookie pays you. So if event $i$ occurs, you get back an amount exactly equal to the \$1 you bet.
For example, let us suppose that we are betting on a horse race with three horses, $A, B,$ and $C$, who win with probabilities $\frac12, \frac13,$ and $\frac16$ respectively; for each \$1  bet on horse $A$, the bookie will pay a prize of \$2 if horse $A$ wins, which is even money.  Similarly, for each \$1 paid as entry fee to bet on $B$, the bookie will pay \$3 if horse $B$ wins, which is 2 to 1 odds, and for each \$1 paid as entry fee to bet on $C$, the bookie will pay \$6 if horse $C$ wins, which is 5 to 1 odds.
Then your strategy is to bet $50¢$ on $A$, $33\frac13¢$ on $B$, and $16\frac23¢$ on $C$.  The three bets total \$1 and in any event the bookie keeps the whole $\$1$.  But if $A$ wins we get a prize of $50¢\cdot 2 = \$1$; if $B$ wins we get a prize of $3\cdot33\frac13¢$, totalling \$1, and if $C$ wins we get a prize of $6\cdot16\frac23 = \$1$.  So every outcome returns our dollar to us.
For a perhaps more instructive example, consider a roulette wheel with 18 red numbers, 18 black numbers, and a green 0 and 00.  Bets on red or black pay off at a rate of $\frac{38}{18}$ (in betting jargon, this is ten to nine), and bets on the 0 or 00 pay off a prize of $38$ (in betting jargon this is thirty-seven to one.)  Then we can bet $\$\frac{18}{38}$ on each of red and black, and $\$\frac1{38}$ on the 0 and 00.  Note that Las Vegas actually pays off bets on red and  black at even money, which is $\frac 21 < \frac{38}{18}$, and actually pays off bets on 0 and 00 at $36$ (that is, thirty-five to one) rather than at $38$.  
