Easy math puzzles

Quote: Ace2

You go to Vegas over a long weekend to play a single-deck blackjack game with a 0% edge ($500 minimum). You decide to play 1,200 hands but you will quit if you bust or double your initial bankroll before reaching 1200 hands.

What size bankroll should you bring to give yourself a 1/3 chance of busting, 1/3 chance of doubling and 1/3 chance of finishing the 1200 hands without busting or doubling?

Assume a standard deviation of 1.1547, flat betting $500 one hand at a time and perfect basic strategy to realize the 0% edge
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Applying my conjecture that the probability of reaching a point at any time during a session is double the probability of ending the session at/beyond that point , we need the z-score corresponding to the probability of 1/3 * 1/2, which is 0.967. Then take 1200^.5 * 1.1547 * 0.967 to get the answer of 38.7 units * $500 = $19,350.

Verification: Knowing the standard deviation and edge, you can easily calculate that this game is statistically equivalent to a bet with a 3/7 probability of winning 7 for 3. Markoving 1200 bets shows that a bankroll of 38 units * $500 = $19,000 gives bust/double/finish probabilities of 33.4%/33.4%/33.2%. I believe this is the closest you can get to 1/3.

So I'd bring $20,000