September 8th, 2010 at 1:51:23 PM
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Betting on the 1st and 2nd dozen (both amounts same). Betting at a single zero table.
So, the odds of one dozen or ther other hitting is 24/37 or .64865 Correct?
So, the question is: what are the odds of either the 1st and/or 2nd dozen to hit in six rolls
Am I right at the answer of (.64865) to the fifth power approximately 11.5% of the time this will happen.
So, the odds of one dozen or ther other hitting is 24/37 or .64865 Correct?
So, the question is: what are the odds of either the 1st and/or 2nd dozen to hit in six rolls
Am I right at the answer of (.64865) to the fifth power approximately 11.5% of the time this will happen.
September 8th, 2010 at 8:42:26 PM
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Quote: timcoBetting on the 1st and 2nd dozen (both amounts same). Betting at a single zero table.
So, the odds of one dozen or ther other hitting is 24/37 or .64865 Correct?
So, the question is: what are the odds of either the 1st and/or 2nd dozen to hit in six rolls
Am I right at the answer of (.64865) to the fifth power approximately 11.5% of the time this will happen.
No.
Easier to calculate how a bet can NOT win then subtract that answer from 1.
13/37 chance not hitting within ONE spin 0.351351
13/37^6 NOT hitting in 6 spins is 0.001881264
Hitting within 6 spins = 1 - 0.001881264
answer = 99.8119%
excel can set up a nice (cumulative frequency) table very quickly to give you answers for different spins.