Three Cards of Unknown Offset

Key idea. Since $n$ is unknown, a card’s face value tells you nothing by itself. Only differences between revealed cards carry information. So we assume the order of the cards is uniformly random, and we restrict attention to strategies that depend only on these differences. Then every payout has the form $n+X$, where $X\in\{0,1,2\}$ is the offset of the card you keep. We maximise $\mathbb{E}[X]$ over the $3!=6$ equally likely orders of the offsets $0,1,2$.

Stopping at once. At the first card you have no information. Stopping gives $\mathbb{E}[X]=1$.

Decision at the second card. Suppose the first card is $a$ and the second is $b$, and write $d=b-a\in\{\pm1,\pm2\}$. Each value of $d$ is consistent with some of the orders below. In each case, compare the offset you get by staying with the offset of the third card.

• $d=2$: the order is $(0,2,1)$. Staying gives $2$ and the third card gives $1$, so stay: $2$.
• $d=-2$: the order is $(2,0,1)$. Staying gives $0$ and the third card gives $1$, so continue: $1$.
• $d=1$: the order is $(0,1,2)$ or $(1,2,0)$. Staying gives $1,2$ (total $3$) and continuing gives $2,0$ (total $2$), so stay: total $3$.
• $d=-1$: the order is $(1,0,2)$ or $(2,1,0)$. Staying gives $0,1$ (total $1$) and continuing gives $2,0$ (total $2$), so continue: total $2$.

Summing the best totals over all six orders gives $2+1+3+2=8$. So after passing the first card, the optimal continuation gives $\mathbb{E}[X]=\tfrac{8}{6}=\tfrac43$. This exceeds the value $1$ of stopping at once, so the optimal strategy is as follows:

• always pass the first card;
• keep the second card if it is higher than the first;
• otherwise take the third card.

Its expected payout is $n+\tfrac43$, so