sorry, i was mistaken let me rephrase the question, what are the odds that one would play 9 numbers inside on a 38 roulette wheel and only hit one number in 55 spins? 1 in ????????????
2007-12-09 09:29:28 · 3 answers · asked by Paul T 1 in Games & Recreation ➔ Gambling
1 in 167,495
The first answer almost had it right. It gave you the odds of hitting one number on one particular spin and no numbers on the other 54 spins. Since the one hit could occur on any of the 55 spins you need to multiply the answer from his calculation by 55. Adding that correcting, simplifying the equations he gave a little, and converting the result to "1 in x" format the final equation becomes
x = 1 / ((29 ^ 54 x 9 x 55) / (38 ^ 55))
2007-12-10 03:40:45 · answer #1 · answered by zman492 7 · 0⤊ 0⤋
The probability to lose 54 spins is (29/38) ^ 54
The probability to win one spin is (9/38)
Multiply them together and you have the answer
The probability for EXACTLY one number is miniscule
During any series of 55 spins you will probably get 10 or 11 numbers.
2007-12-09 18:15:46 · answer #2 · answered by youngmoigle 5 · 0⤊ 0⤋
It's a Circle and bases upon the Moon...Play roulette on the quarter and three quarter days.. Although I gave up gambling I love to watch the wheel to see what could be or not to be in the wealth of man. 85/1 on full moon..17/1 on other moons.
2007-12-10 00:19:48 · answer #3 · answered by Brother Enrique 3 · 0⤊ 0⤋