Roulette math help
November 7th, 2015 at 8:37:33 AM
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So my friend likes to play roulette. He plays two numbers on American roulette. He plays the following progression....
Spins 1 and 2 the bet is $10 on each of two numbers.
Spins 3 and 4 the bet is $15 on each of two numbers.
Spins 5 and 6 the bet is $20 on each of two numbers.
Every two spins the bets increase by $5...
I have seen him reach $80 bets by somewhere around spin 24 and down about $3,000 before hitting and then stops for the day. The daily wins are usually between $400 to $1000.
After spin 20 or so he increases his bet an additional $5 ever spin or two to keep a minimum win of about $300 for the day. The table limit on inside numbers is $500. How often can he win with this strategy? Can someone show me the math to win one bet before quitting?
Spins 1 and 2 the bet is $10 on each of two numbers.
Spins 3 and 4 the bet is $15 on each of two numbers.
Spins 5 and 6 the bet is $20 on each of two numbers.
Every two spins the bets increase by $5...
I have seen him reach $80 bets by somewhere around spin 24 and down about $3,000 before hitting and then stops for the day. The daily wins are usually between $400 to $1000.
After spin 20 or so he increases his bet an additional $5 ever spin or two to keep a minimum win of about $300 for the day. The table limit on inside numbers is $500. How often can he win with this strategy? Can someone show me the math to win one bet before quitting?
gambling problem? split tens!
November 7th, 2015 at 9:16:38 AM
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The probability of a particular bet losing is 36/38 = 18/19, so the probability of N bets in a row losing is (18/19)N, and the probability of winning a bet before losing N in a row is (1 - (18/19)N).
In order to keep the amount won at $300 or more as long as possible, the bets have to be increased by $5 each on every spin starting with the 29th spin, then alternating increases of 10 and 5 starting with the 34th, then, starting with the 43rd, the bets on each number are 180, 190, 205, 215, 230, 240, and 250. At that point, he has lost 49 in a row, is behind $8800, and even a maximum bet will still leave him behind $300.
The probability of losing 49 two-number bets in a row on a double-zero wheel is 1 in 14.144. Note that the most that can be won in a single "run" is $790, which happens with wins on either the 11th or 19th spins, but a run of 49 losses in a row loses $8800.
In order to keep the amount won at $300 or more as long as possible, the bets have to be increased by $5 each on every spin starting with the 29th spin, then alternating increases of 10 and 5 starting with the 34th, then, starting with the 43rd, the bets on each number are 180, 190, 205, 215, 230, 240, and 250. At that point, he has lost 49 in a row, is behind $8800, and even a maximum bet will still leave him behind $300.
The probability of losing 49 two-number bets in a row on a double-zero wheel is 1 in 14.144. Note that the most that can be won in a single "run" is $790, which happens with wins on either the 11th or 19th spins, but a run of 49 losses in a row loses $8800.
