Roulette Odds 4 Reds In A Row
2021年10月8日Register here: http://gg.gg/w5xuc
*Roulette Odds Table
*Odds On Roulette
*Roulette Odds 4 Reds In A Row 4
*Roulette Odds 4 Reds In A Row 5
*Roulette Color Odds
*Roulette Odds Red
With the exception of a tiny handful of punters, no player can beat the tables consistently without taking advantage of great streaks of luck. Beating roulette with streak bets can be both fun and profitable.
The chance of red four times in a row is 5.6% but if you come to the table immediately after a red number has appeared, the probability that you will witness a further three reds (making a total of four times in a row) is 11.5% – purely because you will be observing only a series of three rounds.
Perhaps the most famous example of the gambler’s fallacy occurred in a game of roulette at the Monte Carlo Casino on August 18, 1913, when the ball fell in black 26 times in a row. This was an extremely uncommon occurrence: the probability of a sequence of either red or black occurring 26 times in a row is ( 18 / 37 ) 26-1 or around 1 in 66.6. Roulette Odds Explained. Even though there are 37/38 pockets on a roulette wheel (depending on whether you play European or American roulette), odds are calculated based on 35 pockets. That is the maximum payout, secured by winning a straight up bet. Naturally, this means that the house always has an.
Roulette players tend to love their game so much that they adhere to what Nick “The Greek” Dandalos said, “The only thing as good as gambling and winning, is gambling and losing.”
In an effort to find any way possible to supplement or enhance their chances at the roulette table, enthusiasts tend to use betting systems that increase their time at the table without necessarily increasing their overall odds of winning.
The simple fact of roulette is that the house has the advantage and will win one more bet out of every 37 spins than the player. However, roulette betting systems can take advantage of streaks and provide a player with the opportunity to increase their bankrolls in specific situations.
CASINOBONUSRTPRATINGREVIEW1T&C APPLYPAYOUT10READ REVIEWPLAY NOW2T&C APPLYPAYOUT9.6READ REVIEWPLAY NOW3T&C APPLYPAYOUT9.3READ REVIEWPLAY NOW4T&C APPLYPAYOUT9READ REVIEWPLAY NOW5T&C APPLYPAYOUT9.4READ REVIEWPLAY NOWBetting Against a Streak
The Martingale system is one of the best known and easiest to apply betting systems. The player simply doubles their bet each time they lose and waits for a winner. Many roulette players employ this system with an even money bet (odd or even, red or black, first 18 or second 18) after seeing the opposite side hit three times in a row.
The logic that the other side is now due to hit isn’t actually logic at all, it’s the classic gambler’s fallacy that the odds of an event happening are going to even-out quickly by returning a series of events that are the opposite of what has just been observed.
On each spin of the ball, the chance of red coming up is 48.7%. On a single zero wheel, if a bet on red wins .487% of the time, the odds of two consecutive red spins is .237, three in a row is .115, four in a row is .0562, and five in a row is .0274. That means a streak of five happens about once in 36.5 spins, and we’ll round up to 37.
So, if you see four straight black spins, does that mean the next spin has to be red (or green) except in one case out of 37? No. The odds are still 48.7% that black is coming back. However, that’s what keeps Martingale players using their system. They will win their small bet many times before they run into a streak that wipes out their current table bankroll.
A player will see six consecutive black spins about once in 75 spins, seven consecutive about once in 154 spins. That’s why Martingale players can win so many times, build up a little bankroll, and have it in their head that the system is a winner. Because of those long odds, a player may win 100, 200, or even 300 straight betting opportunities.
River rock casino parking fee. Want to give it a try yourself? Give Lonnie’s roulette simulations a look – click on bet, make the number of spins 100, and see if you get a streak of over 7 on an even-money bet. The results might surprise you!
Dodging the bullet of casino odds is like Russian roulette and you’ve got five chances out of six of winning, and one chance in six that will blow your bankroll out of the water. The problem with the Martingale and many other roulette systems is that you are constantly risking a small fortune to win a single bet. Give it a shot, win a few bets, and consider yourself lucky. Don’t expect it to always work.Betting With a Streak
Instead of betting against a streak, some players have had great success by anticipating a streak of even-money outcomes continuing. With this system, the player simply chooses an even-money bet (odd/even, red/black, 1-18/19-36) and places a single five-unit wager. When it loses it is replaced with a new five unit wager. When it wins, let the winnings ride until it wins five straight times for a payoff of 155 units.
Mississippi cruises. River: Delve into culture and meet the locals at quaint riverside towns.; Cruise ship: Head to the top deck and watch the sun peek over golden trees.; Boat: Sail in style from a bygone era on romantic paddle-wheel boats.; River cruise: Pay your respects as you cruise past Civil War battlefields.; New Orleans: Dance down the streets to the sounds of live music bands. The history of casino cruises and riverboat casinos in the United States goes back to the beginning of the 19th century when the Mississippi River was a major trade center for farmers and merchants. 2 – Day Cruise Our Signature Cruise. LeClaire, Iowa to Dubuque, Iowa & Return. Includes: All Meals, 1 Night Lodging, Entry to the National Mississippi River Museum, Dubuque Shuttle, Entertainment & Activities Onboard. From $409 /person Double Occupancy. Mississippi River Cruising Perfected Cruise the legendary Mississippi River aboard the newest riverboats in the region and enjoy perfect comfort and modern amenities as you travel to each historical destination. The fabric of the region is brought to life through our delicious regional cuisine, live onboard music, and customized excursions. Mississippi riverboat gambling cruises. Come learn the history of the Betsy Ann, Mississippi Gulf Coast, Barrier Islands, and more when you book a Mississippi Riverboat Tour & Cruise in Biloxi today!
If every bet was a win or a loss, the house would win its 2.7% and that would be it. However, due to several small streaks happening before a streak of five straight appears, the bettor will not lose 19 of every 37 spins. However, it can be a good stretch before a winning streak of five straight appears.
An added twist to this strategy is to pull back a single chip after the first three wins of 5, 10, and 20, so 35 rides on the bet. If this wins, two chips are pulled back so 60 rides. If the bet wins again, either the total of 120 is pulled back, or four chips are pulled back and 100 rides. If the 100 bet wins, the entire 200 is now pulled back and the opposite side of the bet (even – now odd, etc.) is made for a single 5-unit wager.
Experiment with your own mixture of these bets to find a happy medium. If you don’t have a wheel at home, try the free online roulette game on the American, European or French roulette pages.
Barring that, you can always take a deck of cards and make your own set-up of 18 even numbers, 18 odd numbers, and an ace to represent the zero, and shuffle away to try a streak system.
Thread Rating: PeterHi,
I’m currently researching some roulette data. I would like to know how I can calculate the expected number of times a sequence of red comes up. So for example, in 1500 roulette spins, how often do I expect a sequence of 2 reds in a row to come up, how often do I expect a sequence of three reds in a row to come up and so on. Hope somebody can help me with that.
Cheers,
PeterOnceDear
Administrator
Thanks for this post from:
Hi,
In 1500 spins, there are 1499 possible pairs of spins
For any pair of spins, the chances of both coming up red is (18/38) * (18/38) or 0.2243767313 assuming double zero wheel.
So you can expect to see 2 consecutive reds 1499 * 0.2243767313 times or 336 times.
Of course a sequence of 3 reds is treated as two sequences of 2
If you want to work it out for sequences of 3, then..
In 1500 spins, there are 1498 possible triplets of spins
For any triplet of spins, the chances of all three coming up red is (18/38) * (18/38) * (18/38) or 0.10628371482 assuming double zero wheel.
So you can expect to see 3 consecutive reds 1498 * 0.10628371482 times or 159 times.
Of course a sequence of 5 reds is treated as three sequences of 3.
Meanwhile.. Remember, you don’t have a winning system. You never will. :o)Take care out there. Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed.Roulette Odds TableSM777
Hi,
I’m currently researching some roulette data. I would like to know how I can calculate the expected number of times a sequence of red comes up. So for example, in 1500 roulette spins, how often do I expect a sequence of 2 reds in a row to come up, how often do I expect a sequence of three reds in a row to come up and so on. Hope somebody can help me with that.
Cheers,
Peter
Odds On RouletteDon’t worry about it, and move onto your next project. Your idea won’t beat roulette.ThatDonGuyIt depends on whether or not how many times you count a sequence of, say, 5 reds as ’a sequence of 2 reds.’
Zero, since you are counting how many times red comes up exactly twice in a row?
One?
Two, since it is ’two pairs of reds’ (1 & 2, 3 & 4)?
Four, since spins 1 & 2 are consecutive, as are 2 & 3, 3 & 4, and 4 & 5?
But as SM777 has pointed out, if you are looking for some mysterious way to beat roulette, don’t bother. Remember, after two (or four, or 257, or zero) consecutive reds, the probability that the next spin will be red on a double-zero wheel is still 9/19 (and black is 9/19, and green is 1/19).mustangsally
I would like to know how I can calculate the expected number of times a sequence of red comes up.it is easy.
just remember dealing with averages or expected numbers is not the same as the probability.
2 totally different animals
*****
for 3 or more run (in Excel for example)
parameters:
p=(18/38)
length=3
trials=1500
q=(20/38)
THE FORMULA (without proof - that is internet stuff)
=(p^length)*(1+((trials-length)*q))
for exactly length of 3
calculate 4 and 3Roulette Odds 4 Reds In A Row 4
subtract 4 from 3
here is my Excel in Google if want to see (easy)
https://goo.gl/98yjKp
Quote: Peter
So for example, in 1500 roulette spins,
how often do I expect a sequence of 2 reds in a row to come up,
how often do I expect a sequence of three reds in a row to come up and so on. here is a table of data.
I simulated this 1 million times (1 million sets of 1500 spins)
and calculated it also (rounded to 4 decimals)
super close I do say
.simsimsimcalccalc. lengthfreqexact 33 or moreexpected #exact 3length 1197014850197.0149374.1835374.1856197.05791 29328373593.2837177.1686177.127793.28112 34416857044.168683.884983.846744.15633 42091215620.912239.716339.690320.90224 598986169.898618.804218.78819.89445 646894284.68948.90568.89374.68376 722177192.21774.21614.21002.21717 810508691.05091.99841.99291.04958 94989500.49900.94750.94340.49689 102360310.23600.44860.44660.235210 111119770.11200.21260.21140.111311 12531590.05320.10060.10010.052712 13250120.02500.04740.04740.024913 14119000.01190.02240.02240.011814 1555200.00550.01050.01060.005615 1626020.00260.00500.00500.002616 1712320.00120.00240.00240.001317 186130.00060.00120.00110.000618 192720.00030.00050.00050.000319 201400.00010.00030.00030.000120 21590.00010.00010.00010.000121 22300.00000.00010.00010.000022 23200.00000.00000.00000.000023 24110.00000.00000.00000.000024 2550.00000.00000.00000.000025 2650.00000.00000.00000.000026 2700.00000.00000.00000.000027 2800.00000.00000.00000.000028 2910.00000.00000.00000.000029
have fun
hope this helps some(sum)
SallyPeterThanks all for the answers. Don’t worry, I’m not looking to find a way to beat the wheel, I’m actually trying to show a sample size I have is random and I like to take a look at all kinds of different statistics from it. In this case the distribution of number of red numbers in a row.
Roulette Odds 4 Reds In A Row 5I guess I wasn’t totally clear: I’m looking for ’exactly two in a row’, ’exactly three in a row’, etc. So a series of four reds in a row will be counted as just that, four in a row. Looks like the best way is to start with 1500 in a row (in a sample size of 1500) and calculate 1499 in a row as suggested with substracting 1500 in a row and work my way down to one.
Or apparently using the below formula. Could you point me in the right direction for the proof? I haven’t been able to find anything about this subject on the web, hence I got here.
THE FORMULA (without proof - that is internet stuff)
=(p^length)*(1+((trials-length)*q))
Thanks,
Roulette Color OddsPeterruss451Roulette Odds RedI have to ask.. Why 1500 instead of an ’even’ number like 1,000, or 10,000.
At least it would be easier to turn into percentages.
Russ
Register here: http://gg.gg/w5xuc
https://diarynote-jp.indered.space
*Roulette Odds Table
*Odds On Roulette
*Roulette Odds 4 Reds In A Row 4
*Roulette Odds 4 Reds In A Row 5
*Roulette Color Odds
*Roulette Odds Red
With the exception of a tiny handful of punters, no player can beat the tables consistently without taking advantage of great streaks of luck. Beating roulette with streak bets can be both fun and profitable.
The chance of red four times in a row is 5.6% but if you come to the table immediately after a red number has appeared, the probability that you will witness a further three reds (making a total of four times in a row) is 11.5% – purely because you will be observing only a series of three rounds.
Perhaps the most famous example of the gambler’s fallacy occurred in a game of roulette at the Monte Carlo Casino on August 18, 1913, when the ball fell in black 26 times in a row. This was an extremely uncommon occurrence: the probability of a sequence of either red or black occurring 26 times in a row is ( 18 / 37 ) 26-1 or around 1 in 66.6. Roulette Odds Explained. Even though there are 37/38 pockets on a roulette wheel (depending on whether you play European or American roulette), odds are calculated based on 35 pockets. That is the maximum payout, secured by winning a straight up bet. Naturally, this means that the house always has an.
Roulette players tend to love their game so much that they adhere to what Nick “The Greek” Dandalos said, “The only thing as good as gambling and winning, is gambling and losing.”
In an effort to find any way possible to supplement or enhance their chances at the roulette table, enthusiasts tend to use betting systems that increase their time at the table without necessarily increasing their overall odds of winning.
The simple fact of roulette is that the house has the advantage and will win one more bet out of every 37 spins than the player. However, roulette betting systems can take advantage of streaks and provide a player with the opportunity to increase their bankrolls in specific situations.
CASINOBONUSRTPRATINGREVIEW1T&C APPLYPAYOUT10READ REVIEWPLAY NOW2T&C APPLYPAYOUT9.6READ REVIEWPLAY NOW3T&C APPLYPAYOUT9.3READ REVIEWPLAY NOW4T&C APPLYPAYOUT9READ REVIEWPLAY NOW5T&C APPLYPAYOUT9.4READ REVIEWPLAY NOWBetting Against a Streak
The Martingale system is one of the best known and easiest to apply betting systems. The player simply doubles their bet each time they lose and waits for a winner. Many roulette players employ this system with an even money bet (odd or even, red or black, first 18 or second 18) after seeing the opposite side hit three times in a row.
The logic that the other side is now due to hit isn’t actually logic at all, it’s the classic gambler’s fallacy that the odds of an event happening are going to even-out quickly by returning a series of events that are the opposite of what has just been observed.
On each spin of the ball, the chance of red coming up is 48.7%. On a single zero wheel, if a bet on red wins .487% of the time, the odds of two consecutive red spins is .237, three in a row is .115, four in a row is .0562, and five in a row is .0274. That means a streak of five happens about once in 36.5 spins, and we’ll round up to 37.
So, if you see four straight black spins, does that mean the next spin has to be red (or green) except in one case out of 37? No. The odds are still 48.7% that black is coming back. However, that’s what keeps Martingale players using their system. They will win their small bet many times before they run into a streak that wipes out their current table bankroll.
A player will see six consecutive black spins about once in 75 spins, seven consecutive about once in 154 spins. That’s why Martingale players can win so many times, build up a little bankroll, and have it in their head that the system is a winner. Because of those long odds, a player may win 100, 200, or even 300 straight betting opportunities.
River rock casino parking fee. Want to give it a try yourself? Give Lonnie’s roulette simulations a look – click on bet, make the number of spins 100, and see if you get a streak of over 7 on an even-money bet. The results might surprise you!
Dodging the bullet of casino odds is like Russian roulette and you’ve got five chances out of six of winning, and one chance in six that will blow your bankroll out of the water. The problem with the Martingale and many other roulette systems is that you are constantly risking a small fortune to win a single bet. Give it a shot, win a few bets, and consider yourself lucky. Don’t expect it to always work.Betting With a Streak
Instead of betting against a streak, some players have had great success by anticipating a streak of even-money outcomes continuing. With this system, the player simply chooses an even-money bet (odd/even, red/black, 1-18/19-36) and places a single five-unit wager. When it loses it is replaced with a new five unit wager. When it wins, let the winnings ride until it wins five straight times for a payoff of 155 units.
Mississippi cruises. River: Delve into culture and meet the locals at quaint riverside towns.; Cruise ship: Head to the top deck and watch the sun peek over golden trees.; Boat: Sail in style from a bygone era on romantic paddle-wheel boats.; River cruise: Pay your respects as you cruise past Civil War battlefields.; New Orleans: Dance down the streets to the sounds of live music bands. The history of casino cruises and riverboat casinos in the United States goes back to the beginning of the 19th century when the Mississippi River was a major trade center for farmers and merchants. 2 – Day Cruise Our Signature Cruise. LeClaire, Iowa to Dubuque, Iowa & Return. Includes: All Meals, 1 Night Lodging, Entry to the National Mississippi River Museum, Dubuque Shuttle, Entertainment & Activities Onboard. From $409 /person Double Occupancy. Mississippi River Cruising Perfected Cruise the legendary Mississippi River aboard the newest riverboats in the region and enjoy perfect comfort and modern amenities as you travel to each historical destination. The fabric of the region is brought to life through our delicious regional cuisine, live onboard music, and customized excursions. Mississippi riverboat gambling cruises. Come learn the history of the Betsy Ann, Mississippi Gulf Coast, Barrier Islands, and more when you book a Mississippi Riverboat Tour & Cruise in Biloxi today!
If every bet was a win or a loss, the house would win its 2.7% and that would be it. However, due to several small streaks happening before a streak of five straight appears, the bettor will not lose 19 of every 37 spins. However, it can be a good stretch before a winning streak of five straight appears.
An added twist to this strategy is to pull back a single chip after the first three wins of 5, 10, and 20, so 35 rides on the bet. If this wins, two chips are pulled back so 60 rides. If the bet wins again, either the total of 120 is pulled back, or four chips are pulled back and 100 rides. If the 100 bet wins, the entire 200 is now pulled back and the opposite side of the bet (even – now odd, etc.) is made for a single 5-unit wager.
Experiment with your own mixture of these bets to find a happy medium. If you don’t have a wheel at home, try the free online roulette game on the American, European or French roulette pages.
Barring that, you can always take a deck of cards and make your own set-up of 18 even numbers, 18 odd numbers, and an ace to represent the zero, and shuffle away to try a streak system.
Thread Rating: PeterHi,
I’m currently researching some roulette data. I would like to know how I can calculate the expected number of times a sequence of red comes up. So for example, in 1500 roulette spins, how often do I expect a sequence of 2 reds in a row to come up, how often do I expect a sequence of three reds in a row to come up and so on. Hope somebody can help me with that.
Cheers,
PeterOnceDear
Administrator
Thanks for this post from:
Hi,
In 1500 spins, there are 1499 possible pairs of spins
For any pair of spins, the chances of both coming up red is (18/38) * (18/38) or 0.2243767313 assuming double zero wheel.
So you can expect to see 2 consecutive reds 1499 * 0.2243767313 times or 336 times.
Of course a sequence of 3 reds is treated as two sequences of 2
If you want to work it out for sequences of 3, then..
In 1500 spins, there are 1498 possible triplets of spins
For any triplet of spins, the chances of all three coming up red is (18/38) * (18/38) * (18/38) or 0.10628371482 assuming double zero wheel.
So you can expect to see 3 consecutive reds 1498 * 0.10628371482 times or 159 times.
Of course a sequence of 5 reds is treated as three sequences of 3.
Meanwhile.. Remember, you don’t have a winning system. You never will. :o)Take care out there. Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed.Roulette Odds TableSM777
Hi,
I’m currently researching some roulette data. I would like to know how I can calculate the expected number of times a sequence of red comes up. So for example, in 1500 roulette spins, how often do I expect a sequence of 2 reds in a row to come up, how often do I expect a sequence of three reds in a row to come up and so on. Hope somebody can help me with that.
Cheers,
Peter
Odds On RouletteDon’t worry about it, and move onto your next project. Your idea won’t beat roulette.ThatDonGuyIt depends on whether or not how many times you count a sequence of, say, 5 reds as ’a sequence of 2 reds.’
Zero, since you are counting how many times red comes up exactly twice in a row?
One?
Two, since it is ’two pairs of reds’ (1 & 2, 3 & 4)?
Four, since spins 1 & 2 are consecutive, as are 2 & 3, 3 & 4, and 4 & 5?
But as SM777 has pointed out, if you are looking for some mysterious way to beat roulette, don’t bother. Remember, after two (or four, or 257, or zero) consecutive reds, the probability that the next spin will be red on a double-zero wheel is still 9/19 (and black is 9/19, and green is 1/19).mustangsally
I would like to know how I can calculate the expected number of times a sequence of red comes up.it is easy.
just remember dealing with averages or expected numbers is not the same as the probability.
2 totally different animals
*****
for 3 or more run (in Excel for example)
parameters:
p=(18/38)
length=3
trials=1500
q=(20/38)
THE FORMULA (without proof - that is internet stuff)
=(p^length)*(1+((trials-length)*q))
for exactly length of 3
calculate 4 and 3Roulette Odds 4 Reds In A Row 4
subtract 4 from 3
here is my Excel in Google if want to see (easy)
https://goo.gl/98yjKp
Quote: Peter
So for example, in 1500 roulette spins,
how often do I expect a sequence of 2 reds in a row to come up,
how often do I expect a sequence of three reds in a row to come up and so on. here is a table of data.
I simulated this 1 million times (1 million sets of 1500 spins)
and calculated it also (rounded to 4 decimals)
super close I do say
.simsimsimcalccalc. lengthfreqexact 33 or moreexpected #exact 3length 1197014850197.0149374.1835374.1856197.05791 29328373593.2837177.1686177.127793.28112 34416857044.168683.884983.846744.15633 42091215620.912239.716339.690320.90224 598986169.898618.804218.78819.89445 646894284.68948.90568.89374.68376 722177192.21774.21614.21002.21717 810508691.05091.99841.99291.04958 94989500.49900.94750.94340.49689 102360310.23600.44860.44660.235210 111119770.11200.21260.21140.111311 12531590.05320.10060.10010.052712 13250120.02500.04740.04740.024913 14119000.01190.02240.02240.011814 1555200.00550.01050.01060.005615 1626020.00260.00500.00500.002616 1712320.00120.00240.00240.001317 186130.00060.00120.00110.000618 192720.00030.00050.00050.000319 201400.00010.00030.00030.000120 21590.00010.00010.00010.000121 22300.00000.00010.00010.000022 23200.00000.00000.00000.000023 24110.00000.00000.00000.000024 2550.00000.00000.00000.000025 2650.00000.00000.00000.000026 2700.00000.00000.00000.000027 2800.00000.00000.00000.000028 2910.00000.00000.00000.000029
have fun
hope this helps some(sum)
SallyPeterThanks all for the answers. Don’t worry, I’m not looking to find a way to beat the wheel, I’m actually trying to show a sample size I have is random and I like to take a look at all kinds of different statistics from it. In this case the distribution of number of red numbers in a row.
Roulette Odds 4 Reds In A Row 5I guess I wasn’t totally clear: I’m looking for ’exactly two in a row’, ’exactly three in a row’, etc. So a series of four reds in a row will be counted as just that, four in a row. Looks like the best way is to start with 1500 in a row (in a sample size of 1500) and calculate 1499 in a row as suggested with substracting 1500 in a row and work my way down to one.
Or apparently using the below formula. Could you point me in the right direction for the proof? I haven’t been able to find anything about this subject on the web, hence I got here.
THE FORMULA (without proof - that is internet stuff)
=(p^length)*(1+((trials-length)*q))
Thanks,
Roulette Color OddsPeterruss451Roulette Odds RedI have to ask.. Why 1500 instead of an ’even’ number like 1,000, or 10,000.
At least it would be easier to turn into percentages.
Russ
Register here: http://gg.gg/w5xuc
https://diarynote-jp.indered.space
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