Ride The Bus

This program solves Schedule I's Ride The Bus casino game. It provides a CLI application intended to be used while playing the casino game that indicates which decision should be made at each step of the game to ensure maximum profits (highest expected value).

Cheatsheet

If played optimally:

• Overall Expected Value: 1.2250
• Average # of plays to profit $10k @ $500/play: 89

Stage	Strategy	Examples
Pick Red/Black	Just pick either (it's 50/50)	N/A
Pick Higher/Lower	Cashout on 7,8,9,10; Otherwise use intuition	5: Higher J: Lower 8: Cashout
Pick Inside/Outside	(Where A-B is bigger rank minus smaller rank) Inside if (A-B) >= 9; Outside if (A-B) <= 2; Otherwise Cashout	2..J: Inside 2..10: Cashout 6..8: Outside 6..9: Cashout
Pick Suit	Always Cashout	N/A

Principle

This program calculates the expected value (EV) of the game by exhaustively analyzing every possible outcome. Here's the breakdown:

• For each decision:
  • Enumerate the choices and the cards, forming a cartesian product of choice+card
  • For each choice+card, you evalulate the next decision and use the EV of the optimal choice
    • If there is no next decision, then you just score the choice+card
  • Average all of the choice+card possibilies to get the EV of just the choice (since the cards are equally random events)
  • Select the optimal choice by finding the choice with the maximum EV
• This process is bottom-up recursive
  • Leaf decisions (in this case, Pick Suit or Cashout) are evaluated first
  • Results then propogate upward through the decision tree
• The result: Every position possible in the game is solved and scored

Observations

Ride The Bus's expected value is 1.2250 if played optimally. This means that for every dollar you put into the game, you expect to make 22.5 cents back.

Since the maximum play for Ride The Bus in Schedule I is $500, on average you expect to profit $112.50 for each play. To profit $10,000, you would expect to play 89 times.

Pick Red/Black

Obviously, you can pick either red or black and have the same odds since it's essentially a 50/50 chance.

However, more interestingly, you can also reason that this game must have an expected value of >=1.0 from just this decision. This step doubles your money 50% of the time, so it's obviously =1.0 for just this step, and since you can Cashout you always expect to at least make your money back if played optimally.

Pick Higher/Lower

On this option, you have to pick whether the unseen card will be higher or lower than the last card shown.

With some intuition, you can pretty much figure out the odds. If the shown card is high, go lower; low, go higher. Near the middle of the ranks, the odds start becoming 50/50 again, and since the profits only go from 2x->3x, you should Cashout.

This program calculates that cards 7, 8, 9, and 10 have expected values below 3.0 for either inside/outside, so you should Cashout if you see one of these cards.

Pick Inside/Outside

For this option, you choose whether the unseen card's rank is inside or outside of the two seen cards. Inside is inclusive, meaning that if the unseen card is either A or B, then it is considered inside.

This option was the least intuitive to me, I thought that you would want pick Inside/Outside unless the range covers close to half of the cards. According to this program, however, you expect to profit by picking Inside/Outside when the range is either very small or very large. For example, 2..4's optimal choice is Outside, but 2..5's optimal choice is Cashout.

More concretely, a range of <=2 cards should be picked as Outside (e.g. 8..10), >=9 should be picked as Inside (e.g. 3..Q), and all other cases should Cashout. This means, most of the time, the best bet is to Cashout.

With some more thought, it's apparent why you would want to Cashout most of the time. The profits for this choice only go from 3x->4x, and so you would like to have very good odds for either Inside/Outside to be able to make more money on average than simply cashingout, else you risk losing the entire bet.

Pick Suit

This one is much more obvious, but I wish I realized it sooner.

At first, it seems great to try for 10x profits, but because this choice has roughly a 1/4 chance for each suit, you end up losing your money 3/4 of the time and only multiplying your money by x2.5 otherwise.

Only if this choice had quadrupled your profits (i.e. it went from 4x->16x), then choosing a suit you haven't seen would have a slightly higher expected value than Cashout (thanks to sampling w/o replacement).

Prolog

As a programming exercise for myself, I made a port of the solver in Prolog. This version uses pretty much the exact same thought process as the Rust code, it's somewhat intended as a verification that the rust code was built correctly (i.e. making sure EVs match).

If you want to get started

$ swipl -s ride-the-bus.pl
# find the best choice expected value
?- best_choice_ev(1.0, [], pickColor, BestChoice, BestEV).
# BestChoice = black,
# BestEV = 1.2249773755656113 .

Drag Race (Rust vs Prolog)

Not much of a surprise, but the Rust version runs faster. Significantly faster. On my M2 Mac Air:

Benchmark	Time to find EV
Rust w/ Optimizations	1.05s
Rust No Optimizations	6.65s
Prolog	19.83s

Why is this the case?

Although Prolog has many optimizations for unifying values extremely quickly, the nature of this problem means we must exhaustively search every game to find the final EV. In other words, we exponentially increase our search by using findall and maplist.

In addition, I'm not good a Prolog developer whatsoever. If someone more familiar with the language took a crack, they would definitely find a better/faster solution.

Building/Running Rust

To start, you must have rust installed.

Afterwards, you can either build an executable or run it directly from cargo:

# run the program
cargo run --release

# build the program
cargo build --release
# executable found in ./target/release