Your answer is considerably funnier than mine, but I think the answer is highly dependent on the game, and a bit more complicated to calculate.
The 1/2% edge is on the expected amount that is to be wagered, not on the amount of the starting bankroll, so the probability of winning is not necessarily 99 1/2%. Even if it were 99 1/2%, that would mean a loss every 200 days rather than every 996 days.
The simplest game that I could think of that meets the criteria is a game where a win pays exactly what you wager and the probability of a win is 0.4975. This gives an expectation of 0.4975 x Bet - 0.5025 x Bet = -0.0050 x Bet.
We have to adjust the Martingale to a limited Martingale since we are never allowed to bet more than the lesser of $25,000 or our remaining bankroll. If we always bet what's necessary to win our goal subject to the limits in the previous sentence, we can set up a recursive relationship for the probabilities of the remaining bankroll after each bet, then iterate it a few thousand times. This can be facilitated by an Excel spreadsheet.
We will bet ($1,001,000 - Bankroll) as long as we have at least $976,000, bet $25,000 when our bankroll is between $25,000 and $975,000, and bet our entire bankroll when it is below $25,000. It turns out that our probability of a win of $1000 is about 99.8763%, producing a loss of the bankroll about 1 day in every 808.5 days.
The assumption here is that the $1000 won on most days is spent for living expenses or the other things you mentioned but not added to the bankroll. The calculation would be a bit more involved if we were able to add the $1000 to our bankroll with each win, and would lengthen the expected duration before bankroll loss somewhat.
Frank could have figured this out if he had remembered a trick to add 1 or 2% in bounceback cash to this play. This would make the game positive so he could have used his positive expectancy math to get an answer and then simply subtract out the pseudo-bounceback.
The whole calculation might be considerably more difficult with a video poker game with a 1/2% house edge due to the numerous payoffs. We might also need to consider different strategies with different bankrolls. For example if we started with $1,000,000 and bet $1000, when dealt 'AQJT' J in a Jacks or Better game, we might consider keeping the pair of jacks over the 4-card royal draw, since drawing to the jacks would give us a better probability of reaching or exceeding our goal of $1,001,000. (It would be even more complicated if we were planning on adding winnings to our bankroll for future days! Then, we would likely need to hold the 4-card royal.)
Over the entire range of the bankroll, this would be extremely tough to work out any near-perfect strategy.
________________________________
From: nightoftheiguana2000 <nightoftheiguana2000@yahoo.com>
To: vpFREE@yahoogroups.com
Sent: Thursday, May 5, 2011 2:44 PM
Subject: [vpFREE] Re: A Hypothetical Question
Sounds approximately right. On average with a Martingale, you win $1,000 on 995 days for $995,000, then on the 996th day (black Friday) you lose your original million plus the rake, but you still have $995,000 in previous winnings and the casino gives you a car or house or boat or something plus 10% loss rebate plus of course the super secret top whale "I lost a million in one day" card and free cruises and shrimp cocktails for life. Plus the cocktail waitress or pool boy (your choice) suddenly has a high school crush on you. You got the original million from a cash out house refi which is now under water so you mail in the keys. You write a book called "The Secret World of Martingale Hustlers and How they Beat the Casinos for Millions!!!". Woo Hoo!
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