Authored by Ion Saliu on October 31, 2000.
In Reply to: help with roulette chances posted by player on October 30, 2000.
: Ion, I've noticed that you are quite proficient in figuring odds on roulette. Could you please tell me the chances on a single and double zero roulette wheel of 1 column or dozen not showing for 8,9,10,11,12, and 13 times? If possible could you give the answer in 1 in a thousand or 1 in 87 or whatever it works out to be? Thank you for your help.
o I understand your question this way.
What is the probability of one dozen or one column to SKIP N consecutive spins?
In such a case, we work with the complementary event of one dozen/column. The probability for one dozen/column to appear is 12/38 = 0.3158 (31.58%). The probability of the complementary event is p=(38-12)/38 = 26/38 = 0.6842 (68.42%).
We can use now this last number and raise it to various powers. Or, we can use FORMULA.EXE, the binomial distribution formula (BDF) to automate the process.
Here are the results for one dozen/column to miss N consecutive roulette spins.
5 times in a row: p = 15% (it happens 15 times in 100 spins; 1 in 6.667 spins);
6 times in a row: p = 10.3% (1 in 9.7 spins);
7 times in a row: p = 7% (1 in 14.3 spins);
8 times in a row: p = 4.8% (1 in 20.8 spins);
9 times in a row: p = 3.3% (1 in 30.3 spins);
10 times in a row: p = 2.2% (1 in 45.5 spins);
11 times in a row: p = 1.5% (1 in 66.7 spins);
12 times in a row: p = 1.1% (1 in 90.9 spins);
13 times in a row: p = 0.7% (it happens 7 times in 1000 spins; 1 in 142.9 spins).
Good luck, Player(s)!
Ion Saliu
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