Not sure I followed what you meant by >= $26 in 2a. The max profit one can make is $26, by drawing 26 red cards in a row, never more.
You are correct though that you'd never want to stop on a negative number, as you always can execute 3b
Looking back, I think I got my numbers mixed up (thanks, beer!). $12 or $13 is probably the number I'm looking for, not $26.
Anyway, my logic is if I am at or higher than the median after I have drawn half the deck, then that means there are more black cards than reds in the remaining deck and I have a greater chance each draw of losing a buck than gaining.
So it is in my best interest to pull out at that point, than to try my luck against ever worsening odds.
Re:Gambler’s Ruin Problem [pdf]
by chrchang523 on Sunday October 02, 2022 @12:16AM
(#33052794)
For the same reason, you’d never intentionally stop at $0 (unless the deck is exhausted). You can always get back to $0 by drawing the rest of the deck, and there’s a 50% chance of going up a dollar on the next card.
If your sum is currently positive, then the chance of going up a dollar is below 50%.
The EV to drawing another card is going to be tied to the probabilities of continued random walks eventually landing on higher scores than you currently have. You stop as soon as the EV is negative (or 0 if you’re not a gambler).
That calculation is presumably the tricky part… integrating over Pascal’s Triangle or something. But it must be positive to start: If the first card is red we have already gained, and if it’s black we’re at worst back where we started.
1. Draw 26 cards.
2a. Equal to or greater than $26? Stop.
2b. Less than $26? Draw 13 more cards.
3a. Equal to or greater than $1? Stop.
3b. Equal to or less than $0? Keep drawing until equal to $0, then stop.
I'm assuming I want to always stop at a non-negative value, else I'd pay the house on the way out. If ending with a negative value has no consequences beyond the game itself, then just stop at 2b after drawing.