120
elementary orderings.

Never take 1

^{st}card. Upon 2^{nd}card, if difference (2^{nd}minus 1^{st}) is
·
-400: Take it (6 outcomes, E=N+0)

·
-300: Take it (12 outcomes, E=N+50; 50% chance
of N, 50% of N+100)

·
400: Wait
for N+100 (because you now know N) (6 outcomes, E=N+100)

·
300: Define
the first card as X. Take X+100 when it
comes, or take X-100 if that comes first.
(12 outcomes, E=N+100)

·
-200, -100, 100, 200: Choose next card that is better than *both*
of the first two, with two exceptions:

·
If the sequential differences are +200, +100,
-200, or if they are -200, +300, -200, then see the 4

^{th}card. It is in between the 1^{st}and 2^{nd}card (not below both) but you get a better expectation than waiting for the 5^{th}card.
·
If you come to know the optimal remaining choice
(because you’ve seen values 400 apart), take it. This occurs when you see
sequential diffs of -100/+400, or +100/+300, you know what N+200 is and that it
is your optimal choice. Or if you see
+200/+200 or -200/+400, you know what N+100 is and that it is your optimal
choice.

·
This strategy (including exceptions) gives you E=N+88.9
when the first difference is +/-200, 18 outcomes each. And gives you E=N+108.3 when the first
difference is +/-100 (24 outcomes each).

1

^{st}Difference Outcomes: E(as amount over N)
-400 6 0

-300 12 50

400 6 100

300 12 100

+/-200 36 88.9

+/-100 48 108.3

Total
overage = 10,800 out of 120 cases, average overage of 90 (N+90)

- Never take 1
^{st}card (C1). Upon C2, calculate difference (D1=C2-C1). - If D1 in (-400,-300) then choose C2
- Else if D1=400 then wait for N+100 and choose it
- Else if D1=300 then choose the next card that is equal to C1+100 or C1-100.
- Else if D1 in (-200,-100,100,200) then see C3:
- If C3 < min (C1,C2) then choose it, else see C4
- If C4 < min (C1,C2) then choose it,
- Else if C3 is N+400 (i.e., C3 – min (C1,C2) = 400) then the optimal remaining card is known (could be N+200 or N+100). It could be C4 (choose it now) or C5 (wait for it).*
- Else if the sequential differences (D1/D2/D3) are +200/+100/-200, or -200/+300/-200, then choose C4.**

* [Note: this affects 8/120 elementary
orderings, with differences of -200/+400, or -100/+400, or +100/+300, or
+200/+200]

**
[Note: This affects 4/120 elementary
outcomes. The basic strategy of “wait
for 1 better than both of C1/C2” results in N twice and N+400 twice. Choosing C4 results in N+100 twice and N+200
twice.]

There are
120 (5!) possible orderings. Among those,
you pay

N: 53 times

N+100: 42

N+200: 15

N+300: 4

N+400: 6