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| Author | Roulette |
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They need to fix the engine.
https://www.lordswm.com/allroul.php
just check probability for this event....
no red, try to calculate...
to have red is chance 16 (red)/38 (fields)=0.421052632 (42.1052632 %)
to have 13 in a row black or 0 (not red) = 0.421052632^13 =1.3*10^-5 = 0,0013 %
so this event should happen 1/0.0013% = 76493 (once in 76493 rolls).
or in other words once in 531.20 days... i seen this before... and i`m not playing this game almost for 2 years.. | oh!!!!!!!!!!!
i dont know tis thing..srry!!!
i have no idea about this | | Just because that's the probability doesn't mean that you have to have that many rolls in between. The odds of rolling a 6 twice in a row are on the order of 1/36, but you and I both know that you don't have to roll the dice 36 times to get a 6 twice in a row. Sometimes the actual numbers just don't match the probability, but over a very long time, they become pretty close. Perhaps you simply haven't watched long enough. | Actually probability to roll 6 twice in a row probability is 1/38 if in previous roll have been 6 if no then probability is 1/38*1/38
Ir roulette have 36 nummber fields + 2 zero fields so total 38... | | Actually there are 18 fields of each colour plus 2 zeros. So the probability of 'not red' is 20/38 in the power of 13 is 0.024 % so the event happens once (expectedly) every 4205 rolls, so once in 29 days. | This just shows the concept of probability isn't very well understood.
How about this? Currently, the last 18 spins have shown a set of numbers {27,5,3,5,7,28,0,17,17,4,18,3,18,1,9,3,6,9}, in this particular order.
The probability of exactly this happening is (1/38)^18 = 0.000000000000000000000000003663 (to 30 decimal places), which is expectedly once every 2.72964 * 10^28 spins, which is equivalent to once in 519,000,000,000,000,000,000,000 years. And yet it just happened! | | We're not talking about single events, but about repetitions. My calculation was just suppose to correct his and to show him that the events are not as unlikely as he thought. | Lol
I think Skuwak thought is as a joke and it was good ;-D
He wanted to say that whatever happens in last 18 spins has this tiny probability, either it is one number repeating 18 times or 18 very different numbers.
Of coarse 5:fusei is correct and I'll just add, that even if there were 50 reds in a row, we couldn't complain cause it's just a chance ... and for those who play roulette - There's no winning strategy, believe me ;-) | | Escape, escape for the deadly cold hand of the roulette. Probability of red/ black is about 47.5%, you can discard the whole "roulette history" if you want to bet on red or black. The chance that the set of 7 red ends with 8 black is same as the chance that the current set will continue. | The probability of 13 red numbers in a row is
(18/38)^13 = 0.000060444 = 0.604*10^(-4)
Skuwak is right, the probability of ANY given 18 numbers in a given order is
(1/38)^18 = 0.366*10^(-28)
For example, the list of numbers [27,5,3,5,7,28,0,17,17,4,18,3,18,1,9,3,6,9] has THE SAME probability as the list of numbers [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0] and should occur only once in (38^18)/(6*24*365) = 0.5193371407*10^(24) years (in average).
It means that you will actually never see the same 18 numbers again :) |
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