What is the probability of me losing a 3/1 bet 10 times in a row I've tried working out myself but the equations needed go straight over my head
2026-03-25 07:42:43.1774424563
Curiosity of roulette system
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1
The probability of loosing one time is $2/3$, then to loose $10$ times in a row is $(2/3)^{10}$ (since the games are supposed to be uncorrelated). If calculated this equals about $1.7\%$.
For the record there is no such bet in actual roulette. There are 37 or 38 numbers (including $0$ and possibly $00$). The $3/1$ bets (such as dozen or column bet) actually only have a $12/37$ or $12/38$ chance of winning and $25/37$ or $26/38$ risk of loosing (giving the house a slight advantage). Then the risk of loosing ten times becomes slightly higher $(25/37)^{10} \approx 2.0\%$ or $(26/38)^{10} \approx 2.2\%$.