Streak probability in European roulette
Where x is the amount in a row you want to see.
Quote: Mission146(18/37)^x = Probability
Where x is the amount in a row you want to see.
Don't understand.(LSD in the 70's)
Can you kindly show us how to calculate five red numbers in a row?
Thank you. (I don't know what that upside down V is or means?)
Quote: NokTangDon't understand.(LSD in the 70's)
Can you kindly show us how to calculate five red numbers in a row?
Thank you. (I don't know what that upside down V is or means?)
^ means 'raised to the power' of or more crudely 'multiplied by itself (this number of extra times - 1)
so for a streak of 1
(18/37)^x = (18/37)^1 = (18/37) multiplied by itself (1-1) more times =(18/37) = Probability
so for a streak of 2
(18/37)^x = (18/37)^2 = (18/37) multiplied by itself (2-1) more times =(18/37) * (18/37) = Probability
so for a streak of 5
(18/37)^x = (18/37)^5 = (18/37) multiplied by itself (5-1) more times =(18/37)*(18/37)*(18/37)*(18/37)*(18/37) = Probability
if you know x then you can have google do the calc for you
https://www.google.co.uk/search?q=%2818/37%29^5&ie=utf-8&oe=utf-8&gws_rd=cr&ei=J65RVuSYMov7UIKwptAK
average number of times is 1/Probability which for a streak of 5 is about 36.7
Quote: OnceDear
average number of times is 1/Probability which for a streak of 5 is about 36.7
Many thanks.
I read the above as indicating that there is a 1 in about 36.7 chance, that when I walk up to a single zero roulette table and bet red, that there will be a streak of red(or black or odd or even or high or low) in a row occurring, starting from that first bet. Correct?(as past history has no bearing on my wager or said streak?, or does it?)
Or does it means I'm asking if I find a table/wheel where it has already occurred five times in a row, would I count those and then be talking about six plus in a row?
Warmest regards again. It's going to be for a living soon so very very important!
EDIT: I find streaks in roulette much more common that say in craps on the pass/don't pass line wager.
If you walk up to the table and put a chip on red there is a 18/37 chance that you will get red. There's then a 18/37 chance that it will be red next time and so on and so on.Quote: NokTangMany thanks.
I read the above as indicating that there is a 1 in about 36.7 chance, that when I walk up to a single zero roulette table and bet red, that there will be a streak of red(or black or odd or even or high or low) in a row occurring, starting from that first bet. Correct?(as past history has no bearing on my wager or said streak?, or does it?)
If you walk up to the table and want to know what the probability is that you can place a chip on red and let any winning bet roll for another red, then probability just before placing the wager is (18/37) * (18/37). Ie the probability that you are about to see a streeak of two reds is (18/37) * (18/37). It's not difficult.
History is of no significance. probability applies to future events.Quote:Or does it means I'm asking if I find a table/wheel where it has already occurred five times in a row, would I count those and then be talking about six plus in a row?
Don't talk nonsense.Quote:Warmest regards again. It's going to be for a living soon so very very important!
Quote:EDIT: I find streaks in roulette much more common that say in craps on the pass/don't pass line wager.
And?
Quote: OnceDear
.Don't talk nonsense.
And?
It's not nonsense. I honestly think a person with control of his/her emotions can in fact make a living at this, not "martingale" but rather, a mathematical chain of events occurring. Betting against a 1 in 36.7 chance of something occurring seems reasonable to me, even in the long run.
And...I was just curious why at the craps table you see less streaks than you do at the roulette wheel. That's all. There must be a reason?
Thank you again.
Quote: NokTangIt's not nonsense. I honestly think a person with control of his/her emotions can in fact make a living at this, not "martingale" but rather, a mathematical chain of events occurring. Betting against a 1 in 36.7 chance of something occurring seems reasonable to me, even in the long run.
And...I was just curious why at the craps table you see less streaks than you do at the roulette wheel. That's all. There must be a reason?
Thank you again.
The streaks on roulette are mostly noticed on even/odd, high/low, and red/black. There's no comparable craps bet/roll with a near-50% chance of resolving each spin. So I think that's why you see more streaks, or at least have the perception of them. Since you only have 20 unique outcomes at craps (doesn't matter for this purpose which die rolls what), and you have 38 unique outcomes on American Roulette, it sort of depends more on what you perceive as a streak than whether one is occurring.
Complete nonsense. I urge you to try it, for that's the only way you will be convinced.Quote: NokTangIt's not nonsense. I honestly think a person with control of his/her emotions can in fact make a living at this.
He can win if he owns the casino.
Quote: Mission146(18/37)^x = Probability
Where x is the amount in a row you want to see.
Say im tracking both red and black.
I'm interested in a streak length of 4, and if i was NOT concerned whether it was a black streak or red streak, only in either a black or red streak of that length occurring , would the formula be:
2x((18/37)^4)
Quote: OnceDear
average number of times is 1/Probability which for a streak of 5 is about 36.7
Is this the probability of red 5 streak occuring?
Say im tracking both red and black, would i divide 36.7 by 2?
Quote: DanthemanIs this the probability of red 5 streak occuring?
Say im tracking both red and black, would i divide 36.7 by 2?
I'm not the math guy in this discussion, but it appears the one in apx. 36.7 refers to one(red) or the other(black) occurring five times in a row. Five spins in a row. We were unable to continue an intelligent discussion about it due to personal attacks being launched. A sad day actually....Cheers mate.
Quote: NokTangI'm not the math guy in this discussion, but it appears the one in apx. 36.7 refers to one(red) or the other(black) occurring five times in a row. Five spins in a row. We were unable to continue an intelligent discussion about it due to personal attacks being launched. A sad day actually....Cheers mate.
An intelligent discussion and a discussion involving the emotional control regarding a game of probability are two different things.
Quote: DodsferdAn intelligent discussion and a discussion involving the emotional control regarding a game of probability are two different things.
I'm not clear why you post that?
Having the emotional control to take your $100.usd a day to live on, after laying apx 36 to 1 odds, seems intelligent enough to me. What have I got wrong? I think we all know you "can't" do this hour after hour, it takes "emotional control" to just do it once a day, have your prime rib dinner, do your laundry, watch the news, and sleep in the comfort of AC or heat, with running water etc..
I have a thread here:Quote: DanthemanI am wanting to find out the average number of spins it would take playing European roulette, if betting both black and red, to see a streak length of 2,3,4,5 etc
on average How many trials
easy to use formula for wait time until binomial run (streak) of successes
Now,
data for you
18/37
0 Roulette
(all calculated values here checked with simulations)
run length:average number of spins
a=red
2:6.280864198
3:14.96622085
4:32.81945397
5:69.51776649
b=black
2:6.280864198
3:14.96622085
4:32.81945397
5:69.51776649
c=red or black
2:3.140432099
3:7.483110425
4:16.40972699
5:34.75888325
c=a+b-d
c=a/2 or b/2
d=red and black
2:9.421296297
3:22.449331275
4:49.22918095
5:92.05098036
d=a+b-c
Now lots of partially correct answers in this thread so far.
not picking on any one (or two)
How about this...
5 in a row for red
many say (18/37)^5=about 1 in 37 spins on average
that 37 is not spins
(Unless you count overlapping streaks and that is way too confusing to even attempt)
37 is the number of trials or attempts and each attempt is NOT 1 spin in length.
we need more for the average length of each trial
1+p+p^2+p^3+p^4,p=(18/37) = 3550231/1874161
result: about 1.894304171
(sure we can use 1 plus the sum of a geometric series
but this is different too)
1/(18/37)^5 * 3550231/1874161 = 131358547/1889568 =
69.52
Wolfram Alpha
in agreement with simple simulation results
grouped data
items: 1000000
minimum value: 5.00
first quartile: 23.00
median: 49.00
third quartile: 95.00
maximum value: 925.00
mean value: 69.48 <<<<<
midrange: 465.00
range: 920.00
interquartile range: 72.00
mean abs deviation: 48.35
sample variance (n): 4301.07
sample variance (n-1): 4301.07
sample std dev (n): 65.58
sample std dev (n-1): 65.58
hope this helps out current and future visitors
averages are like apple eating
probability distributions are like a 10 course meal.
I am so hungry.
hope no errors
fun question!
Sally
