What are the odds that a certain number will come up in 100 spins?
January 30th, 2013 at 9:11:11 AM
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If I were to pick a random number off the roulette wheel and to spin the wheel 100 times what are the odds the number would come up?
January 30th, 2013 at 9:15:44 AM
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Just over 93% on a US wheel.
Just over 93.5% on a European '0' wheel.
Just over 93.5% on a European '0' wheel.
January 30th, 2013 at 9:18:58 AM
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I disagree. I have been waiting over 500 spins for number 37 to come up in an online casino.
Shed not for her
the bitter tear
Nor give the heart
to vain regret
Tis but the casket
that lies here,
The gem that filled it
Sparkles yet
January 30th, 2013 at 9:27:25 AM
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i bet your loving that...
January 30th, 2013 at 9:31:18 AM
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I am betting a martingale, but starting to reach infinity. I have applied for TARP funds .
Shed not for her
the bitter tear
Nor give the heart
to vain regret
Tis but the casket
that lies here,
The gem that filled it
Sparkles yet
January 30th, 2013 at 11:50:53 AM
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you better sell you're house too, and your car... and maybe even your wife and kids
January 30th, 2013 at 11:57:15 AM
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@switch
hey thanks, can you tell me how you got that number? is there a mathimatical equation?
hey thanks, can you tell me how you got that number? is there a mathimatical equation?
January 30th, 2013 at 12:07:45 PM
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The probability that a number will come up at least once is 1.0 minus the probability that the number will not come up at all. The two outcomes are complimentary, so their probabilities sum to 1.0
P(at least once) = 1.0 - P(none)
The "none" outcome implies that the number does not come up on the first spin, AND the number does not come up on the second spin, AND... does not come up on the last spin. When you are ANDing independent events, the probability multiplies.
P(none) = P(not on 1st spin) * P(not on second spin) * ... * P(not on 100th spin)
The probability that a number will not appear on any particular spin is 37/38 on a double zero wheel.
P(none) = 37/38 * 37/38 * ... 37/38 = (37/38) ^100
So...
P(at least once) = 1.0 - (37/38)^100 = 93.053%
P(at least once) = 1.0 - P(none)
The "none" outcome implies that the number does not come up on the first spin, AND the number does not come up on the second spin, AND... does not come up on the last spin. When you are ANDing independent events, the probability multiplies.
P(none) = P(not on 1st spin) * P(not on second spin) * ... * P(not on 100th spin)
The probability that a number will not appear on any particular spin is 37/38 on a double zero wheel.
P(none) = 37/38 * 37/38 * ... 37/38 = (37/38) ^100
So...
P(at least once) = 1.0 - (37/38)^100 = 93.053%
January 30th, 2013 at 12:09:56 PM
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September 22nd, 2010 at 12:04:50 pmQuote: poiklp@switch
hey thanks, can you tell me how you got that number? is there a mathematical equation?
handy Roulette tables and formula
https://wizardofvegas.com/member/nope27/blog/
winsome johnny (not Win some johnny)
