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   Forums-->Ideas and suggestions-->

Roulette



AuthorRoulette
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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