The Traveling AP Problem — break-even distance & optimal trip length

No published formal treatment of this turned up — it appears to live as forum intuition rather than a citable model. Structurally it's a renewal-reward problem (each additional day at Game B pays out at a rate discounted by backoff-survival probability) nested inside a fixed-cost travel decision (like a facility-location / traveling-salesman cost trade-off). Adjust the inputs on the left; the two decision outputs update live. Model detail is in the panel at the bottom.

Game Economics

Travel & Trip Parameters

Risk & Fatigue

Net advantage at current inputs
$0
Trip profit minus what Game A would have paid over the same days, net of all travel costs.
Optimal trip length @ this distance
Break-even max distance @ this trip length
Effective hourly EV, Game B (fatigue-adjusted)
Averaged across the daily session length, given the fatigue decay curve.

Net advantage vs. trip length (at current distance)

Net advantage vs. distance (at current trip length)

Formal model
Decision variables D = one-way distance to Game B (miles) N = trip length (days) Per-day survival (backoff hazard) Let m = mean playable days before backoff (extended by rotating across c playable casinos at the destination, m_eff = m · c). Hazard h = 1 / m_eff S(d) = (1 − h)^(d−1) probability day d is still reached Fatigue-adjusted hourly EV Session length H, fatigue threshold τ, decay rate k. If H ≤ τ: EV_eff = EV_travel If H > τ: EV_eff = EV_travel · [ τ + (H−τ)·avg_decay ] / H avg_decay = (1 − e^(−k(H−τ))) / (k(H−τ)) (hourly EV holds until the fatigue threshold, then decays exponentially for the remaining hours of the session; avg_decay is the mean of that exponential tail) Travel cost & time (round trip) Driving: Cost(D) = 2·D·cost_per_mile Time(D) = 2·D / speed Flying: Cost(D) = 2·(base + D·$/mile) Time(D) = 2·(overhead + D/cruise) Net advantage of the trip vs. grinding Game A for N days NetAdv(D,N) = Σ[d=1..N] S(d) · EV_eff · H_travel gaming profit, backoff-discounted − EV_local · H_local · N forgone Game A profit, same N days − Cost(D) fuel/airfare − EV_local · Time(D) opportunity cost of travel hours − max(0, N−1) · (hotel − comps) lodging net of comps − N · (meals + ground) daily living cost at destination Two outputs solved from NetAdv(D,N) 1. Break-even distance D*: NetAdv(D,N) is strictly decreasing in D (only the cost terms depend on D), so D* is found by bisection on NetAdv(D*,N) = 0. 2. Optimal trip length N*: argmax over N of NetAdv(D,N), found by direct scan (1–60 days). The curve is generally single-peaked because each added day's marginal payoff — S(d)·EV_eff·H_travel − EV_local·H_local − (hotel−comps) − meals − ground — shrinks geometrically with backoff survival while its marginal cost stays roughly flat, guaranteeing the marginal contribution eventually turns negative. What this leaves out Bankroll/Kelly constraints and ruin risk, correlation between backoff probability and bet spread, non-linear comp accrual, and the fact that EV and variance both scale with bet size — this model treats EV_travel and EV_local as given hourly rates rather than deriving them from a Kelly-sized bankroll. Those would be the natural next layer.