Start with **\$100** and a uniformly shuffled bag of **two green balls and one red ball**. - Before each draw, bet any amount of your current bankroll on green. - Green adds your stake; red subtracts it. - Draw without replacement. You see the ball even if you bet nothing. The best first bet depends on the objective: $$ \begin{aligned} \max\;\mathbb E[W]&:\quad\text{bet }0,\qquad\mathbb E[W]=\tfrac{700}{3}.\\ \max\;\min W&:\quad\text{bet }50,\qquad W=200\text{ on every runout}. \end{aligned} $$ Here $W$ is **final bankroll**, including the starting \$100. Austin Henley posed the original puzzle during a conversation at [Khaki AI](https://khaki.email/). The distinction between those two objectives is what makes it interesting. ## Explore the game Choose an objective, then a possible draw. The wager updates after each observation. Blue balls are optional: drawing one returns the stake and ends the game. Three useful comparisons: - **Original bag, three draws:** maximum average \$233⅓; maximum guarantee \$200. - **Same bag, one draw:** maximizing the average means going all-in. - **Three greens, one red, one blue, five draws:** go all-in for the best average, but no policy can guarantee a profit—blue might appear first. ## Maximum average The original bag has three equally likely runouts: $$ P(RGG)=P(GRG)=P(GGR)=\tfrac13. $$ Wait until the red appears, then go all-in on each remaining green: | Runout | Bankroll after each draw | Finish | | :--- | :--- | ---: | | `RGG` | 100 → 200 → 400 | \$400 | | `GRG` | 100 → 100 → 200 | \$200 | | `GGR` | 100 → 100 → 100 | \$100 | The average is $(400+200+100)/3=700/3$. Why pass up a ⅔ chance of green? A dollar lost on an early red would otherwise be doubled twice. **The value of a bet includes its effect on future wagers.** To prove optimality, keep both uncertain decisions in the calculation. Let $X$ be the first dollar bet and $Y$ the dollar bet after a first green: $$ 0\le X\le100,\qquad 0\le Y\le100+X. $$ All other decisions have known colors: bet everything on green and nothing on red. These choices weakly improve every ending, so they lose no optimum. $$ \begin{aligned} W_{RGG}&=4(100-X),\\ W_{GRG}&=2(100+X-Y),\\ W_{GGR}&=100+X+Y. \end{aligned} $$ Hence $$ \mathbb E[W]=\frac{700-X-Y}{3}\le\frac{700}{3}, $$ with equality at **$X=Y=0$**. Even the 50–50 second bet reduces the whole game's expectation: losing there removes money that could double on the final green. ## Maximum guarantee Set **$X=Y=50$**. The second bet is one-third of the \$150 bankroll after a green. Every branch ends at \$200: ```mermaid flowchart TD S["$100 · bet $50"] -->|"G · 2/3"| G["$150 · bet $50"] S -->|"R · 1/3"| R["$50 · bet $50"] G -->|"G · 1/2"| GG["$200 · bet $0"] G -->|"R · 1/2"| GR["$100 · bet $100"] R -->|"G · certain"| RG["$100 · bet $100"] GG -->|"R · certain"| A["GGR: $200"] GR -->|"G · certain"| B["GRG: $200"] RG -->|"G · certain"| C["RGG: $200"] ``` A higher guarantee is impossible. Requiring all three endings to be at least $m$ gives $$ \begin{aligned} RGG:&\quad X\le100-\tfrac m4,\\ GRG,\;GGR:&\quad X\ge\tfrac{3m}{4}-100. \end{aligned} \qquad\Longrightarrow\qquad m\le200. $$ The first-bet-only strategy ($Y=0$) peaks at a guarantee of \$160. Adapting the second bet raises it to \$200. ## The average–guarantee frontier Suppose a ticket home costs \$150. You can guarantee it with $X=Y=25$: the endings are \$300, \$200, and \$150, averaging \$216⅔. More generally, for a floor $100\le m\le200$, $$ \boxed{\max_{W\ge m}\mathbb E[W]=\frac{800-m}{3}}, \qquad X=Y=\frac{m-100}{2}. $$ **Each extra dollar of guarantee costs one-third of a dollar in average bankroll.** | Goal | $X$ | $Y$ | Guarantee | Average | | :--- | ---: | ---: | ---: | ---: | | Maximum average | 0 | 0 | \$100 | \$233⅓ | | \$150 ticket | 25 | 25 | \$150 | \$216⅔ | | Maximum guarantee | 50 | 50 | \$200 | \$200 |

Deriving the frontier

The `GGR` constraint gives $X+Y\ge m-100$, so $\mathbb E[W]\le(800-m)/3$. Choosing $X=Y=(m-100)/2$ produces endings $600-2m$, $200$, and $m$. For $m\in[100,200]$, their minimum is $m$: the upper bound is attained.

The rules permit arbitrarily divisible stakes, with no fees or borrowing. Rounding bets to cents changes the optimization problem. Maximizing the *probability* of reaching a target is also a separate objective. ## General bags: the Bellman equations Use state $s=(g,r,b,n)$: remaining greens, reds, blues, and draws. Write $T=g+r+b$ and wager a fraction $f\in[0,1]$ of bankroll $D$. $$ (D,s)\longrightarrow \begin{cases} \bigl(D(1+f),\;(g-1,r,b,n-1)\bigr)&\text{with probability }g/T,\\ \bigl(D(1-f),\;(g,r-1,b,n-1)\bigr)&\text{with probability }r/T,\\ D\quad\text{(stop)}&\text{with probability }b/T. \end{cases} $$ Wealth scales linearly, so memoize only the count state. Let $E(s)$ and $H(s)$ be optimal expected and guaranteed wealth **per starting dollar**. Both equal 1 when $n=0$ or the bag is empty. Subscripts $G,R$ denote continuation values after the corresponding draw. ### Expected wealth $$ E(s)=\max_{0\le f\le1}\frac{g(1+f)E_G+r(1-f)E_R+b}{T}. $$ This is affine in $f$. Its slope determines the wager: $$ f_E^*=\begin{cases} 1&gE_G>rE_R,\\ 0&gE_G0 \bigr\}. $$ Omit absent colors. The optimum lies at an endpoint or an intersection of branch lines. With both green and red present, their equalizing candidate is $$ f_H=\frac{H_R-H_G}{H_R+H_G}, $$ when this lies in $[0,1]$. For one green, one red, and two draws, $H_G=1$ and $H_R=2$: $$ E(f)=\tfrac32-\tfrac f2, \qquad H(f)=\min\{1+f,\;2-2f\}. $$ The mean decreases from the outset; the guarantee peaks at $f=1/3$. If any blue remains, an immediate stop caps the guarantee at **1**, attained by betting zero throughout. ## Closed forms ### Full bag, no blue: maximum average For $N=g+r$, $$ E(g,r,0,N)=\frac{\sum_{k=0}^{g}\binom Nk}{\binom Ng}. $$ **Wait for the last red, then go all-in on every green.** This policy is optimal for any complete bag without blue; it need not be optimal with a shorter horizon or blue stops.

Proof by induction

Substitution into the wager coefficient, using Pascal's identity, gives for $g,r>0$ $$ gE(g-1,r)-rE(g,r-1)=-r<0. $$ Thus zero is optimal while a red remains. The zero-bet recurrence agrees with the formula, as do the boundaries $E(0,r)=1$ and $E(g,0)=2^g$. Induction proves the result.

### Limited horizon, no blue: maximum guarantee For $0\le n\le g+r$, define $S(n,r)=\sum_{k=0}^{\min(r,n)}\binom nk$. Then $$ H(g,r,0,n)=\frac{2^n}{S(n,r)}, \qquad f_H^*=\frac{\binom{n-1}{r}}{S(n,r)}\quad(n>0). $$ Out-of-range binomial coefficients are zero. The original game has $S(3,1)=4$, giving $H=2$ and $f_H^*=1/2$. The number of greens disappears once the horizon is feasible. Extra greens improve probabilities but leave the worst feasible sequences unchanged when $n,r$ stay fixed.

Proof via the reciprocal recurrence

If $r\ge n$, an all-red prefix is possible and $H=1$. If $r=0$, all draws are green and $H=2^n$. Otherwise both colors exist. Inductively, their continuation guarantees $A,C$ satisfy $A\le C$. Equalizing the branches gives $$ H=\frac{2AC}{A+C}, \qquad \frac1H=\frac12\left(\frac1A+\frac1C\right). $$ Pascal's identity gives $S(n,r)=S(n-1,r)+S(n-1,r-1)$, completing the induction. Substitution into $(C-A)/(C+A)$ gives the stated wager fraction.

## Code and checks [Download pikurn.ts](./pikurn.ts) and run it with Node.js 24.2 or newer: ```sh node pikurn.ts ``` No dependencies or build step. It prints both policies and every runout. For a custom bag: ```ts import { solve, enumerate } from './pikurn.ts' const state = { g: 3, r: 1, b: 1, n: 5 } console.log(solve(state, 'expected')) // value: 1.9; fraction: 1 console.table(enumerate(state, s => solve(s, 'expected').fraction, 100)) ```

Read the complete solver and command-line program



The listing and explorer import the same file. The ZIP below includes all visualization code and editable figures.

- **Solver:** memoized count states; at most $(g+1)(r+1)(b+1)(n+1)$ combinations, with constant work per state.
- **Explorer:** exhaustive runouts for bags up to eight balls; displayed trees stop after three draws. Enumeration can grow exponentially.
- **Verification:** independent action-grid recursion, probability conservation, bankroll scaling, and comparisons with both closed forms.

The code uses floating-point arithmetic with a small tie tolerance and rejects numerical overflow. The derivations establish the exact results; large computations remain limited by memory, recursion depth, and number range.


Background: Bellman's On the Theory of Dynamic Programming and Auckland's sampling-without-replacement notes. The Pikurn formulas are derived above. Thanks to Austin Henley, Gregory Croisdale, and Ben Klein for the original discussion.