Allocating money to different bets on loaded coins I hope this question was not asked before. It is a very interesting problem I came across and for which I have not yet found a solution. If you know where to find it, please feel free to redirect me. Of course I would love to have a solution in this thread. The problem goes as follows.


We are given three different "loaded coins" whose probabilities of turning up heads are known (and possibly different) and denoted by $p_1, p_2, p_3$, and an amount of money, $b$, that has to be fully expended betting on these coins. All the coins are flipped. For each coin turning up heads you get your money back, for each coin turning up tails you lose all the money placed on that particular coin. If I want to maximize the probability of getting at least an amount $a \leq b$ back, how do I allocate the money?


Thank you a lot!
 A: We can normalize total bet size to 1. Let $T$ be the target fraction. Let $X_1$, $X_2$, $X_3$ be fractions of our total bet, respectively for each coin.
There are 8 distinct outcomes.




Outcome
Probability
Money Returned




1
$(1-P_1)(1-P_2)(1-P_3)$
$0$


2
$P_1(1-P_2)(1-P_3)$
$X_1$


3
$(1-P_1)P_2(1-P_3)$
$X_2$


4
$(1-P_1)(1-P_2)P_3$
$X_3$


5
$P_1P_2(1-P_3)$
$X_1+X_2$


6
$P_1(1-P_2)P_3$
$X_1+X_3$


7
$(1-P_1)P_2P_3$
$X_2+X_3$


8
$P_1P_2P_3$
$1$




A successful outcome returns money at or above $T$. Outcome 8 is always a success. We need to maximize the sum of the above probabilities of successes.
The first step is to calculate and rank these probabilities. A basic heuristics is to aim for more of the higher probabilities first.
Lemma 1a. When $T>\frac{1}{2}$, only one of outcomes 2-4 is successful.
Lemma 1b. When $\frac{1}{3}<T\leq\frac{1}{2}$, only two of outcomes 2-4 are successful.
Lemma 1c. When $T \leq \frac{1}{3}$, outcomes 2-4 are all successful.
Lemma 2a. When $T>\frac{2}{3}$, only two of outcomes 5-7 are successful.
Lemma 2b. When $T\leq\frac{2}{3}$, all of outcomes 5-7 are successful.
When assigning bets, it's ok to simply wager just enough to meet the required minimum for successful outcomes 2-4 for the level of $T$, and the rest arbitrarily. The four levels are as follows.
When $T >\frac{2}{3}$, by 1a and 2a, a success includes one of outcomes 2-4 and two of outcomes 5-7. There are $\binom{3}{1} \times \binom{3}{2} = 9$ such combinations. Find sum of the probabilities in each combination. Assign according to the highest total sum.
When $\frac{1}{2}<T\leq\frac{2}{3}$, by 1a and 2b, a success includes one of outcomes 2-4 and all of outcomes 5-7. Assign according to the highest probability in outcomes 2-4.
When $\frac{1}{3}<T\leq\frac{1}{2}$, by 1b and 2b, a success includes two of outcomes 2-4 and all of outcomes 5-7. Assign by rejecting the lowest probability in outcomes 2-4.
Finally, as @M. Nestor pointed out, when $T\leq\frac{1}{3}$, all outcomes are successful. This is shown by 1c and 2b.

When the game is played relatively few times, it's good to maximize probability of reaching a threshold. If the game is played many many times, then it would be better to maximize the expected value. That's more suited for a separate thread, but briefly speaking it can be expressed as a linear programming question of two variables.
