Guaranteed Free Money

A casino offers you $\$55$ worth of “free play vouchers.” You specifically receive three $\$10$ vouchers and one $\$25$ voucher.

You can play any or all vouchers on either side of an even-money game (think red vs. black in roulette, without those pesky green pockets) as many times as you want (or can). You keep the vouchers wagered on any winning bet and get a corresponding cash amount equal to the vouchers for the win. But you lose the vouchers wagered on any losing bet, with no cash award. Vouchers cannot be split into smaller amounts.

What is the guaranteed minimum amount of money you can surely win, no matter how bad your luck? And what betting strategy always gets you at least that amount?

Let $S = (S_1, S_2, W)$ be the state of the system where you currently have $S_1$ $\$10$ vouchers, $S_2$ $\$25$ vouchers and $W$ winnings in cash. Further define $V(S) = V(S_1,S_2,W)$ to be the guaranteed minimum amount of money you can surely win if you start with state $S = (S_1,S_2,W).$

We can define $V(1,0,W) = V(0,1,W) = W,$ since regardless of whether you only have one voucher left, since you cannot split the vouchers, the fair game could go against your bet and you end up with no more money than you started.

Let's now consider $V(1,1,W).$ If you were to bet one $\$10$ voucher on say heads and one $\$25$ voucher on tails, then with a $50\%$ probability you would win $\$10$ and move to the state $S^\prime = (1,0,W+10)$ and with $50\%$ probability you would win $\$25$ and move to state $\tilde{S} = (0,1,W+25)$. Since $V$ measures the guaranteed minimum winnings we have $$\tilde{V}(1,1,W) = \min \{ V(1,0,W+10), V(0,1,W+25) \} = \min \{ W + 10, W + 25 \} = W+10.$$ If we were to pick another strategy, say only wager one $\$10$ voucher on heads, then we would get $$\bar{V}(1,1,W) = \min \{ V(1,1,W+10), V(0,1,W) \} = W,$$ which is sub-optimal with respect to the strategy of betting say $\$10$ on H and $\$25% on tails. So we see that that optimal strategy gives $$V(1,1,W) = W+10.$$

For $V(2,0,W),$ the optimal guaranteed strategy is to bet one voucher on heads and another on tails, since this gives $\tilde{V}(2,0,W) = V(1,0,W+10) = W+10,$ which is better than if you only wager on one side and get $\hat{V}(2,0,W) = W.$ So we get $$V(2,0,W) = \max \{ \tilde{V}(2,0,W), \hat{V}(2,0,W) \} = W+10.$$ Similarly, we get that $$V(3,0,W) = W+20,$$ by successively wagering one voucher on heads and another on tails, twice.

For $V(2,1,W)$, an optimal guaranteed strategy is to bet, say, two $\$10$ vouchers on heads and the $\$25$ voucher on tails. This gives $$\tilde{V}(2,1,W) = \min \{ V(2,0,W+20), V(0,1,W+25) \} = \min \{ W+30, W+25 \}= W+25.$$ This is better than first betting one $\$10$ on heads and another on tails, and then $$\bar{V}(2,1,W) = V(1,1,W+10) = W+20.$$ Both of these are better than betting only one $\$10$ voucher on heads and the $\$25$ voucher on tails which gives $$\hat{V}(2,1,W) = \min \{ V(2,0,W+10), V(1,1,W+25) \} = \min \{ W +20, W+25 \} = W + 20.$$ So we have $$V(2,1,W) = \max \{ \tilde{V}(2,1,W), \hat{V}(2,1,W), \bar{V}(2,1,W) \} = W+25.$$

Finally, for $V(3,1,W),$ we have an optimal guaranteed strategy to bet one $\$10$ on heads and another on tails, and get $$\tilde{V}(3,1,W) = V(2,1,W+10) = W+35.$$ Other strategies involve betting two $\$10$ vouchers on heads and the $\$25$ voucher on tails, which equivalently gives $$\bar{V}(3,1,W) = \min \{ V(3,0,W+20), V(1,1,W+25) \} = W+35.$$ Sub-optimal strategies includes bets on only one side, or betting only on $\$10$ voucher on heads and the $\$25$ voucher on tails, which given $$\hat{V}(3,1,W) = \min \{ V(3,0,W+10), V(2,1,W+25)\} = W+30.$$ So we have $$V(3,1,W) = \max \{ \tilde{V}(3,1,W), \bar{V}(3,1,W), \hat{V}(3,1,W) \} = W+35.$$ Therefore, the guaranteed minimum amount of money you can surely win is $V(3,1,0) = \$35,$ which you can get by first betting one $\$10$ voucher on each side of the game, then in the second round betting the two remaining $\$10$ vouchers on one side of the game and the $\$25$ voucher on the other.