I am a complete novice when it comes to mathematics, however I am trying to calculate the most effective way to play a card game within a virtual casino on another online game I play.
Dealer draws a card 2 thru Ace. You must then guess whether your card will be higher or lower than the dealer card. Aces are high, two's are low.
Rules:
Standard 52 card deck.
Cards are NOT replaced into the deck, until the deck is shuffled.
Deck is ALWAYS shuffled every 9 or 10 'hands'. 1 hand = 1 dealer card + 1 player card.
Payout rules:
Betting is in fixed amounts of 10, 100, 1,000, 10k, 100k, 1m and 10m.
Payout multiplier changes daily and ranges from 15% to 40%.
Every time you beat the dealer, you increase the starting bet by the payout multiplier. For example: 1m becomes 1.15m becomes 1.3225m
A draw, IE dealer has 8 and you draw 8, counts as neither a win or a loss. The game proceeds to the next round without any addition to the pot.
You may cash out at the end of a round for the full pot amount, or after you have seen the dealer card for a 50% reduction in pot.
I have made some basic calculations in excel that counts cards for you and tells you when you should pick higher or lower.
However, I'd like to calculate at what multiplier does the game have a positive EV.
I also think that there are more advanced strategies available where you play 10 hands until the deck is stacked in your favor; for example a minimal amount of 7,8,9's etc.
I'd love to find someone who can help me solve this problem!
Sorry if my post formatting is off or if I have broken netiquette here. This is my first post.
Many thanks,
TJM
Best Answer
Believe it or not, it's the ties that make this problem really complicated to analyze. For that reason I'm going to ignore ties for now, and assume that all 52 cards have a unique order.
Suppose that the dealer has picked a card out of $n$, and we know (through card counting) that there are $l$ lower cards, and $n-1$ cards left. Then our winning chance is (assuming we pick our best choice of higher/lower):
$$p = \frac{\max(l, n-1-l)}{n-1}$$
Suppose we have seen the dealer's card and know our $p$ to win. Now there are two choices:
Now as long as $m > 1$ we should always choose option one. That is because $p$ can never be below $\frac{1}{2}$ (which is easy to see by symmetry - if one choice is < 50% chance to win, the other must be > 50%).
Our average win% in a round with $n$ cards left is:
$$\frac{1}{n}\sum_{l=0}^{n-1}\frac{\max(l, n-1-l)}{n-1} = \frac{1}{n(n-1)}\sum_{l=0}^{n-1}\max(l, n-1-l) = \frac{6n^2-4n-1+(-1)^n}{8n(n-1)}$$
Which for even $n$ simplifies to $\dfrac{3n - 2}{4n - 4}$.
Which gives the following winning percentages for the rounds:
We can see that even going into the deep rounds has a negligible effect on our chance to win. Also note that ties only improve on this, a tie is as if the round never happened while slightly improving our chances.
Now, when is it worth to play this game? If you only play a single round and then cash out it's worth it when
$$m > \frac{4\times 52 - 4}{3\times 52 - 2} = 1.324675$$
The most optimal play is to always play the full 9 or 10 rounds and then cash out.
For $9$ rounds your winning chances are $0.08057$, which makes it worth it if:
$$m^9 > \frac{1}{0.08057}$$ $$m > 1.322927$$
TL;DR: advanced strategizing isn't really applicable here. Wait for a day where the multiplier is $\geq 1.35$ to make a noticeable profit. In fact, assuming you can play the game as fast as you want, don't even bother card counting, just repeatedly play only round 1 as the advantage gained from card counting is so minimal, and faster throughput of games makes more money.