Gambler’s Ruin Problem [pdf] (columbia.edu)

Is it because one can get ruined by hitting 0 but a "win" is subjective because there's no absolute upper bound? If one could go in to negative at 50/50 odds, then the game will dance around that $1 starting point forever?

> Its wild how the chance of going broke is your money in bet units above both sides money at 50/50

It's actually quite simple. In a fair game you expect to leave with as much money as you started with. initial money = expected final money = p * all money

The main difference is that Kelly has a multiplicative assumption and the paper has an additive assumption. The Kelly criterion is often computed as a proportion of the initial bankroll, whereas this paper handles non-zero chances of losing a _fixed_, positive amount of money (i.e., a varying proportion of your bankroll, and that proportion increases precisely when you would least want it to -- when you're doing poorly).

From some quick "just several minutes" beer-laden kit bashing in my brain:

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.

This reminds me how I was introduced to this topic, the St Petersburg paradox. It was after a discrete 2 final. The professor heard us talking about a competition from our Quant finance club and decided to ask us this problem. Essentially asking the same thing but with a dice, so with replacement, with bigger values and asking what logic our stop algorithm would be to maximize our return. And what would happen if the gain and loss weren't equal with something like +4,+3,+2,-1,-2,-3 as well aas with -4,-3,-2,+1,+2,+3. We ended up staying an extra 2 hours discussing it.

Very nice puzzle! A quick Python script that solves it as it's difficult to do with a pen and paper:

  #awesome feature from Python 3.9!
  from functools import cache
  
  @cache
  def ev(score, cards_left):
      if cards_left == 0: return 0
      up_cards_no = (cards_left - score) / 2
      p_up = up_cards_no / cards_left
      p_down = 1 - p_up
      return max(score, p_up * ev(score + 1, cards_left - 1) +  p_down * ev(score - 1, cards_left - 1))

  >>> ev(0,52)
  2.6244755489939253

Of course we should continue if the current score is lower than ev. Here is the script to find exact thresholds at which we should stop drawing more cards:

  def find_thresholds():
      print(f"EV of tha game: {ev(0,52)}") # fill the cache up
      for i in range (0,10):
          thres = 52 - i
          while ev(i, thres) > i:
              thres -= 2
          print(f"At score {i} you should stop drawing with {thres} cards left")

  >>> find_thresholds()
  EV of tha game: 2.6244755489939253
  At score 0 you should stop drawing with 0 cards left
  At score 1 you should stop drawing with 3 cards left
  At score 2 you should stop drawing with 8 cards left
  At score 3 you should stop drawing with 17 cards left
  At score 4 you should stop drawing with 28 cards left
  At score 5 you should stop drawing with 41 cards left
  At score 6 you should stop drawing with 46 cards left
  ...

Being up 6 or more we should always stop. The last interesting number is being up 5.

What's the 1% risk rule?