You have the same $\$55$ worth of vouchers from the casino in the same denominations. But this time, you’re not interested in guaranteed winnings. Instead, you set your betting strategy so that you will have at least a $50$ percent chance of winning $W$ dollars or more.
What is the maximum possible value of W? In other words, what is the greatest amount of money you can have at least a 50 percent chance of winning from the outset, with an appropriate strategy? And what is that betting strategy?
Ok, well to explore this space, let's first explore the distribution of optimal strategy from the Classic answer. If we assume that we first bet one on each side then almost surely we end up with $2$ remaining $\$10$ vouchers and the $\$25$ voucher remaining, but with $W = 10.$ At this point the bet is $\$20$ on heads and $\$25$ on tails, so with probability $50\%$ you end up with $2$ $\$10$ vouchers and $W=30$. At this point in this strategy you would bet one $\$10$ voucher on heads and the other on tails, to reach the state of a single $\$10$ voucher and $W=40,$ again with a total probability of $50\%$. At this point you would bet your single $\$10$ voucher over and over again until you lose, which means you can attain $W=40+10n$ with probability $$p_n = \mathbb{P}\{W = 40+10n\} = \frac{1}{2^{n+2}}.$$
On the other hand, there is another possibility when you bet the $2$ $\$10$ vouchers against the single $\$25$ voucher, which is that, with probability $50\%$ you end up with on a single $\$25$ voucher and $W=35.$ In this case, you would just choose to wager your single $\$25$ voucher over and over again until you lose. So we have $W = 35+25n$ with probability $$q_n = \mathbb{P} \{W = 35+25n\} = \frac{1}{2^{n+2}}.$$ Putting these altogether we get the full distribution $$\mathbb{P} \{ W = w \} = \begin{cases} \frac{1}{4^{n+1}}, &\text{if $w=35+50n;$}\\ \frac{1}{2\cdot 4^{n+1}} + \frac{2}{32^{n+1}} = \frac{4\cdot 8^n +2}{32^{n+1}}, &\text{if $w = 60+50n$;} \\ \frac{1}{4 \cdot 2^n}, &\text{if $w = 40+10n$ and $n \mod 5 \not\equiv 2$;}\\ 0, \text{otherwise.}\end{cases}$$ In this case, we see that $$W^* = \max \{ w \in \mathbb{N} \mid \mathbb{P} \{ W \geq w \} = \frac{1}{2} \} = 50.$$ Additionally, we confirm that since $\mathbb{P} \{W \geq 35\} = 1,$ as we surmised in the classic answer.
However, as expected, we can do way ......... better. For instance, even if you bet all of your vouchers on heads each time you have a $50\%$ probability of getting at least $\$55,$ since in that greedy strategy your probability distribution if $$\hat{\mathbb{P}} \{ W = w \} = \begin{cases} \frac{1}{2^n}, &\text{ $w = 55n$;} \\ 0, &\text{otherwise.}\end{cases}$$
However, let's combine the greediness of throwing everything on one side and hoping for the best, with the optimal strategy to ensure a high of a minimum strategy as possible. In this case, you bet all $\$55$ on heads, let's say, in the first round. If you lose, well, that sucks. But if you win then you will be at the state $(3,1,55)$ with probability $50\%.$ Again, since we are looking to maximize the the value of $\max \{ w \in \mathbb{N} \mid \mathbb{P}^* \{ W \geq w \} \geq \frac{1}{2} \}$ and we already spent $50\%$ of the probability mass on the first wager, we see that our problem has now transformed into finding the maximum guaranteed winnings based on starting at the state $(3,1,55),$ since guaranteed winning from this point would still have a greater than or equal to $50\%$ overall probability from $(3,1,0).$ Thankfully, we've already solved this problem in the Classic Fiddler, and see that the greatest amount of money you can have at least a $50$ percent chance of winning from the outset is $$V(3,1,55) = 55+35 = \$90,$$ which employs the strategy of first betting everything on one side, say heads, and then if you win the first bet, then employing the strategy from the Classic Fiddler.
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