# What is the best strategy for roulette?

You start with $\$10$. You have a$\dfrac {18}{38}$chance of winning, and if you win you get back double the money you spent. The minimum bet is$\$1$. How should you split your bets so that you make $\$20$the fastest? This question was given to me by a friend, who in turn got the question from another student. So unfortunately I don't know the context or the exact wording. The only way I thought of interpreting the question is to see which strategy has the best expected value. If you bet$\$10$ directly, your expected value is $E_1 = 10 \cdot \dfrac {18}{38} - 10 \cdot \dfrac {20}{38}$. If you bet $\$5$twice, your expected value is$E_2 = 2\left( 5 \cdot \dfrac {18}{38} - 5 \cdot \dfrac {20}{38} \right) = E_1.$I don't see how splitting the bets in different ways would ever make a difference. • A detailed analysis of the optimal betting strategy to double your money on roulette was done with my question here: math.stackexchange.com/questions/1994169/… Basically, you bet as much as possible on the highest odds adding in surrounding bets as needed to exactly double your money. – doug Mar 5 '17 at 23:29 • @doug Thanks for the link! – Ovi Mar 6 '17 at 2:01 ## 2 Answers First of all,$\frac{18}{38}=\frac{9}{19}$. Also winning gives you double, so$E_1$is really$10\cdot 2\cdot\frac{9}{19}-10\cdot\frac{10}{19}=\frac{80}{19}$. Now: The function for your expected value is$E(m)=2m\cdot\frac{9}{19}-m\cdot\frac{10}{19}=\frac{8m}{19}$, where m is the money you bet. However,$E(m)$is additive; this means that$E(a+b)=E(a)+E(b)$for any two real numbers a and b (obviously though satisfying the given conditions). So you are right that splitting the bets makes no difference. Edit: Indeed, it technically does matter, but only as far as your risk. So if you bet all 10 dollars on your first bet, then you might win immediately, but you might also lose immediately. • Well winning does give you double, but if you bet$10$and win you get$20$, so your net gain is$10$, and this has a$\dfrac {9}{19}$chance of happening. And if you lose, your net loss is$10$, with a chance of$\dfrac {10}{19}\$. – Ovi Nov 15 '16 at 1:00

Expected win

$$E_1 = 10 \cdot \dfrac {18}{38} - 10 \cdot \dfrac {20}{38}$$

makes sense if you repeat this process large number of times, but since you do it once, your chance of losing = 20/38

your chance of winning = 18/38

And if you split the bet into two 5 dollar,

your chance of losing it all = (20/38)*(20/38)

your chance of winning it all = (18/38)*(18/38)

So you chances of winning and losing both decrease. Because now we have some intermediary states 1 win and 1 loss( which is start state) so splitting it in two bets decreases the risk of going broke, but it also decreases the chance of winning.