We are discussing probability and odds in my elementary math class. The students came up with two scenarios. They are as follows...
In a spinner wheel game based on the days of the week, students bet a coin or coins on the day(s) they hope the spinner stops on. There are seven outcomes.
Faced with their final two quarters, what has the better chance of winning?
a. Bet one quarter in one game. Bet a second quarter in a second game.
OR
b. Bet two quarters on two different days in one game.
Is one preferable over the other?
Your probability of winning at least once is (slightly) better if you use option (b).
In (a), the probability of failure in any game is $\frac{6}{7}$, so the probability of $2$ failures in a row is $\left(\frac{6}{7}\right)^2$, which is $\frac{36}{49}$.
So the probability of at least one success is $1-\frac{36}{49}$, which is $\frac{13}{49}$.
In (b), the probability of a success is $\frac{2}{7}$. This is $\frac{14}{49}$, a little larger than the probability in (a).
Remark: The expected number of wins with either strategy is $\frac{2}{7}$. The slightly smaller probability of at least one success in (a) is compensated for by the fact that it is possible with (a) to have two wins. This is not possible with strategy (b).