Roulette Law of Lg Numbers
January 6th, 2016 at 1:06:12 PM
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Hi, Wiz,
Newbie Roulette player here.
I understand most of the concepts of probability, std dev, etc. However, still don't have the knowledge to calc this scenario:
In 0/00 roulette - just betting streets only:
a. In 38 spins - what is/are the ODDS that ALL streets #1 - #12 appear precisely 3x - no less and no more?
b. Based on the Law of Lg # concept --> 38 spins represent "small #" of independent trials. Is it reasonable (or illogical) to intuitively "guess" that Scenario (a) is a very low probability scenario?
Thank you for all your time, teaching, sharing!
Roulette Newbie
Newbie Roulette player here.
I understand most of the concepts of probability, std dev, etc. However, still don't have the knowledge to calc this scenario:
In 0/00 roulette - just betting streets only:
a. In 38 spins - what is/are the ODDS that ALL streets #1 - #12 appear precisely 3x - no less and no more?
b. Based on the Law of Lg # concept --> 38 spins represent "small #" of independent trials. Is it reasonable (or illogical) to intuitively "guess" that Scenario (a) is a very low probability scenario?
Thank you for all your time, teaching, sharing!
Roulette Newbie
January 6th, 2016 at 1:19:34 PM
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It's much to near 5pm for me to do a... but for b. you should realize that being such a small sampling size it could vastly differ from your EV. The smaller the sampling size the larger the SD's. Scenario a should be taken exactly as that, nothing more nothing less. If you see the same "street" all 38 times, that doesn't mean a blip in the radar when it comes to "the long run," however improbable it may be in the short run. This is why if you see a roulette board with 20 black numbers in a row it doesn't mean diddily squat as the next spin is still a completely independent trial from the rest.
Now if you see black 20,000 times in a row... Well then I'd start betting on black as the wheel must be biased, the dealer is shooting, etc, etc, etc, etc.
Welcome to the forums!
Now if you see black 20,000 times in a row... Well then I'd start betting on black as the wheel must be biased, the dealer is shooting, etc, etc, etc, etc.
Welcome to the forums!
Playing it correctly means you've already won.
January 6th, 2016 at 2:58:15 PM
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a. You mean 36 spins, right?
The answer I get for 36 spins is 0.000000241098932121718
b. You don't need any probability law to guess that a. is a small number. That's IOTTCO.
Think of it like being a necklace with 12 different colored beads, and you have 3 of each bead. You are asked to string the necklace. Altogether you can make combin(36,3)*combin(33,3)* ... *combin(6,3)*combin(3,3) different necklaces.
The answer I get for 36 spins is 0.000000241098932121718
b. You don't need any probability law to guess that a. is a small number. That's IOTTCO.
Think of it like being a necklace with 12 different colored beads, and you have 3 of each bead. You are asked to string the necklace. Altogether you can make combin(36,3)*combin(33,3)* ... *combin(6,3)*combin(3,3) different necklaces.
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January 6th, 2016 at 3:08:39 PM
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1. I am not 100% sure about this - simulation comes close to this number - but I think it's this:
Color each of the 12 3-number streets a different (non-green) color; let 0 and 00 be green.
There are 38! / (3!12 x 2!) ways to arrange 38 colored items where 12 colors have 3 items and a 13th color has the remaining 2.
For each of the 12 3-color items, there are 27 sets of values (e.g. for 1-3, you can have 1-1-1, 1-1-2,. 1-1-3, 1-2-1. 1-2-2. 1-2-3, 1-3-1, and so on); for the 2-color item, there is 4 sets of values (0-0, 0-00, 00-0, 00-00).
The total number of sets of 38 spins that meet the conditions is 38! / (3!12 x 2!) x (2712 x 4).
There are a total of 3838 sets of 38 spins, so the probability is the first value divided by the second one, or about 1 / 14,916,813.
2. Yes, it is reasonable to "guess" that this is a low probability scenario.
You also have to remember one thing when applying the Law of Large Numbers; although the probabilities tend to get closer to their expected values as the numbers get larger, the total amount that you need to bet to get there is also much, much larger.
Example: you bet on heads on coin tosses. In the first 100 tosses, heads comes up 45%; you are behind $10. In the next 900 tosses, heads comes up 449 times; in those 1000 tosses, heads has come up 49.4% of the time, but you are now behind $12. Imagine this on a scale with millions of tosses.
Color each of the 12 3-number streets a different (non-green) color; let 0 and 00 be green.
There are 38! / (3!12 x 2!) ways to arrange 38 colored items where 12 colors have 3 items and a 13th color has the remaining 2.
For each of the 12 3-color items, there are 27 sets of values (e.g. for 1-3, you can have 1-1-1, 1-1-2,. 1-1-3, 1-2-1. 1-2-2. 1-2-3, 1-3-1, and so on); for the 2-color item, there is 4 sets of values (0-0, 0-00, 00-0, 00-00).
The total number of sets of 38 spins that meet the conditions is 38! / (3!12 x 2!) x (2712 x 4).
There are a total of 3838 sets of 38 spins, so the probability is the first value divided by the second one, or about 1 / 14,916,813.
2. Yes, it is reasonable to "guess" that this is a low probability scenario.
You also have to remember one thing when applying the Law of Large Numbers; although the probabilities tend to get closer to their expected values as the numbers get larger, the total amount that you need to bet to get there is also much, much larger.
Example: you bet on heads on coin tosses. In the first 100 tosses, heads comes up 45%; you are behind $10. In the next 900 tosses, heads comes up 449 times; in those 1000 tosses, heads has come up 49.4% of the time, but you are now behind $12. Imagine this on a scale with millions of tosses.
January 6th, 2016 at 4:07:16 PM
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For the benefit of others, I think the poster is asking what is the probability that in 38 spins on a double-zero wheel, what is the probability the ball will land as follows:
1. In the grouping 1, 13, 25 three times.
2. In the grouping 2, 14, 26 three times.
3. Each subsequent group of three numbers three times.
4. In the grouping 0, 00 twice.
Please note this is different from Teliot's interpretation.
It would not surprise me if the inspiration for this question was another faulty display board.
1. In the grouping 1, 13, 25 three times.
2. In the grouping 2, 14, 26 three times.
3. Each subsequent group of three numbers three times.
4. In the grouping 0, 00 twice.
Please note this is different from Teliot's interpretation.
0.000000067038449 = 1 in 14,916,813.
combin(38,3)*combin(35,3)*combin(32,3)*...*combin(5,3)*3^36*2^2/36^38
It would not surprise me if the inspiration for this question was another faulty display board.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 7th, 2016 at 9:36:58 PM
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Hi, Wizard & Forum Members,
Wizard is correct - I mean 38 numbers, with each street (1-12) appearing 3x, and also 0, 00.
It does not matter the order/sequence - but occurs exactly in 38 spins.
Members - thank you very much for your input and answers! Great help to think through.
Roulette Newbie
Wizard is correct - I mean 38 numbers, with each street (1-12) appearing 3x, and also 0, 00.
It does not matter the order/sequence - but occurs exactly in 38 spins.
Members - thank you very much for your input and answers! Great help to think through.
Roulette Newbie
January 8th, 2016 at 8:01:49 AM
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Hello, Wizard & Members,
Taking one step more "global" - can somebody help with this scenario in 0/00 roulette:
1. Assume 38 spins exactly.
2. Base population = 12 streets (triplet numbers).
3. Plus 0, 00 (American wheel).
4. If spin 38 time exactly --> what is the probability that each street will have occurred precisely 3 times in the 38 spins?
I realize it is quite similar to my original base question (that one looked at numbers 1-36 individually, plus 0,00).
This new question of using 12 streets - probably have a significantly higher probability of occurrence vs original scenario.
However, intuitively, I am guessing still that even using 12 streets as units (instead of 36 individual numbers), the probability is still very small. Perhaps < 6 std dev (I think even more?)?
Can somebody here in this community do the calc? Compare 36 indiv #s vs 12 streets?
Greatly appreciate all the thinking and inputs here! Thank you.
Roulette Newbie
Taking one step more "global" - can somebody help with this scenario in 0/00 roulette:
1. Assume 38 spins exactly.
2. Base population = 12 streets (triplet numbers).
3. Plus 0, 00 (American wheel).
4. If spin 38 time exactly --> what is the probability that each street will have occurred precisely 3 times in the 38 spins?
I realize it is quite similar to my original base question (that one looked at numbers 1-36 individually, plus 0,00).
This new question of using 12 streets - probably have a significantly higher probability of occurrence vs original scenario.
However, intuitively, I am guessing still that even using 12 streets as units (instead of 36 individual numbers), the probability is still very small. Perhaps < 6 std dev (I think even more?)?
Can somebody here in this community do the calc? Compare 36 indiv #s vs 12 streets?
Greatly appreciate all the thinking and inputs here! Thank you.
Roulette Newbie
