The Gambler's Ruin: Why the House Always Wins
Everyone knows the vig means the odds aren't perfectly fair. Fewer people know that even in a world where the odds were exactly fair, a coin flip with zero house edge at all, the house would still come out ahead more often than not, for a completely different mathematical reason. It's called gambler's ruin, and it's real probability theory, not a saying.
The fair-odds version of this, first
Picture two people flipping a truly fair coin, one dollar per flip, until one of them hits zero. No house, no vig, a genuinely 50/50 game. You'd think both players have an even shot at busting the other. They don't, unless they started with the exact same amount of money. The player with less capital is more likely to go broke first, purely because they have less room to absorb a losing streak before hitting zero.
Run the numbers on a $1,000 bankroll against an opponent with $100,000. Even at a perfectly fair coin flip, your odds of ever busting that bigger bankroll before going broke yourself are $1,000 ÷ $101,000, about 1%. A 99% chance of going broke first, with a completely fair coin, purely because of the size of the two bankrolls.
Now add a real house edge
A sportsbook isn't even offering you a fair coin. The vig means your true break-even win rate is already higher than 50%, before gambler's ruin enters the picture at all. Stack a real, persistent disadvantage on top of a capital gap this large, and the conclusion stops being 'you're likely to go broke first' and becomes something stronger: against a bankroll that large, any consistent edge against you, no matter how small, pushes your probability of eventual ruin toward certainty the longer you keep playing. That's not a rough estimate. It's the same math as the fair-coin example, just tilted further in the bigger side's favor.
