November 14th, 2010 at 2:58:47 PM
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I am trying to figure out how many spins would u say is it where a number statistically would have to come up

November 14th, 2010 at 3:39:11 PM
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There is no number of spins that guarantees that a particular number will come up. The probability never goes to 1, it will only approach it.

For 300 spins the probability that a particular number will not show up is (37/38)^300 = 0.0335%. That's pretty small, but not zero. If you repeated this 300 spin experiment 3000 times, you'd only expect to completely miss your number on one of the trials.

For 300 spins the probability that a particular number will not show up is (37/38)^300 = 0.0335%. That's pretty small, but not zero. If you repeated this 300 spin experiment 3000 times, you'd only expect to completely miss your number on one of the trials.

November 14th, 2010 at 4:01:26 PM
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Quote:atrainI am trying to figure out how many spins would u say is it where a number statistically would have to come up

It is all about the degree of certainty.

For a 38 number wheel.

From http://wizardofvegas.com/member/nope27/blog/:

"The below table is for the probability of NOT hitting "at least 1" number in x spins.

formula used from:http://wizardofodds.com/askthewizard/roulette.html

question #2 at WoO.

Wizards' Example: for "at least 1" number NOT hitting in 200 spins

Sum i=1 to 37 [(-1)^(i+1) × combin(38,i) × ((38-i)/38)^38] = 16.9845715651245%"

I have expanded the table below from the above blog.

I have also run 1 million spin simulations and have seen many numbers that did not appear in 500 spins.

I would say from the table 1000 spins would be a hard one to ever witness.

spins | prob | 1 in | expected # of spins |
---|---|---|---|

50 | 99.9999975% | . | . |

60 | 99.9991% | . | . |

70 | 99.967896% | . | . |

80 | 99.678210% | . | . |

90 | 98.452658% | . | . |

100 | 95.339700% | . | . |

110 | 89.728676% | . | . |

120 | 81.746470% | . | . |

130 | 72.133375% | . | . |

140 | 61.855739% | . | . |

150 | 51.774802% | 1.9 | 290 |

160 | 42.490296% | 2.4 | 377 |

170 | 34.326903% | 2.9 | 495 |

180 | 27.391862% | 3.7 | 657 |

190 | 21.649355% | 4.6 | 878 |

200 | 16.984572% | 5.9 | 1,178 |

210 | 13.249361% | 7.5 | 1,585 |

220 | 10.290675% | 9.7 | 2,138 |

230 | 7.966128% | 12.6 | 2,887 |

240 | 6.151027% | 16.3 | 3,902 |

250 | 4.740313% | 21.1 | 5,274 |

260 | 3.647758% | 27.4 | 7,128 |

270 | 2.803866% | 35.7 | 9,630 |

280 | 2.153364% | 46.4 | 13,003 |

290 | 1.652705% | 60.5 | 17,547 |

300 | 1.267822% | 78.9 | 23,663 |

310 | 0.972205% | 102.9 | 31,886 |

320 | 0.745304% | 134.2 | 42,935 |

330 | 0.571235% | 175.1 | 57,770 |

340 | 0.437748% | 228.4 | 77,670 |

350 | 0.335412% | 298.1 | 104,349 |

360 | 0.256975% | 389.1 | 140,091 |

370 | 0.196867% | 508.0 | 187,944 |

380 | 0.150810% | 663.1 | 251,973 |

390 | 0.115523% | 865.6 | 337,594 |

400 | 0.088490% | 1,130.1 | 452,028 |

500 | 0.006149% | 16,261.5 | 8,130,771 |

600 | 0.000427% | 234,067.2 | 140,440,306 |

700 | 0.000030% | 3,369,203.4 | 2,358,442,356 |

800 | 0.000002061985% | 48,496,952.1 | 38,797,561,717 |

900 | 0.000000143251% | 698,074,378.6 | 628,266,940,697 |

1000 | 0.000000009952% | 10,048,215,791.0 | 10,048,215,791,014 |

October 8th, 2011 at 4:50:46 AM
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For some number or other not to come up, I would say the probability is zero. At least one number has to come up.

For a particular number not to come up in 300 spins I would agree with PapaChubby: (37/38)^300 = 0.0335%.

For at least one number not to come up, I think that guido111 has a good grip on the problem based on a solution given by the Wizard for the case of 200 spins, however the Wizard's logic may not be easy for everyone to follow. Here is another approach based on counting the ways the balls can be distributed to the numbers.

We assume that both the balls and the numbers are distinguishable. The numbers are distinguishable because they are all different and the balls are distinguished by the order in which they are thrown. There is a function T(m,n) that gives the number of ways that m distinguishable objects may be distributed to n containers such that every container has at least one object. The number of ways of distributing m objects to n containers without restriction is n

It is always possible either for a particular number or for some number or other not to come up based on the inequality T(m,n) < n

The T function is discussed in several books on combinatorics. I shall be glad to provide a bibliography if anyone asks.

For a particular number not to come up in 300 spins I would agree with PapaChubby: (37/38)^300 = 0.0335%.

For at least one number not to come up, I think that guido111 has a good grip on the problem based on a solution given by the Wizard for the case of 200 spins, however the Wizard's logic may not be easy for everyone to follow. Here is another approach based on counting the ways the balls can be distributed to the numbers.

We assume that both the balls and the numbers are distinguishable. The numbers are distinguishable because they are all different and the balls are distinguished by the order in which they are thrown. There is a function T(m,n) that gives the number of ways that m distinguishable objects may be distributed to n containers such that every container has at least one object. The number of ways of distributing m objects to n containers without restriction is n

^{m}so the probability that all numbers will have been hit by at least one ball is T(m,n)/n^{m}. Subtract this from 1 and that is the probability that at least one number will remain unhit. For m = 300 and n = 38 this evaluates to 0.01267822135, which agrees with guido111's result.It is always possible either for a particular number or for some number or other not to come up based on the inequality T(m,n) < n

^{m}.The T function is discussed in several books on combinatorics. I shall be glad to provide a bibliography if anyone asks.

A fool is someone whose pencil wears out before its eraser does. - Marilyn Vos Savant

October 8th, 2011 at 8:13:28 AM
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Its similar to that famed "two people with a common birthday". With 24 people in a room you have a fifty percent chance, simply because you are not designating one particular date that has to be matched.

Its the same with a roulette wheel, if you don't specify a particular number that must hit then there are lots of spins that can go by and there being some surviving number that is unhit is still quite likely.

Ofcourse I don't want to know what number will NOT hit on that next spin, I want to know what number will hit on that next spin.

Its the same with a roulette wheel, if you don't specify a particular number that must hit then there are lots of spins that can go by and there being some surviving number that is unhit is still quite likely.

Ofcourse I don't want to know what number will NOT hit on that next spin, I want to know what number will hit on that next spin.

October 8th, 2011 at 10:47:20 AM
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If you were to use the binomial distribution to approximate the answer over 300 spins, the answer would be 38 x (37/38)^300 = .012742

You are using a different (correct) formula to account for all possibilities.

You are using a different (correct) formula to account for all possibilities.

-----
You want the truth! You can't handle the truth!

October 29th, 2011 at 1:55:08 PM
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(lol) I dont need to look at a chart, I do all the work on my own. TRIAL & ERROR.

I use to play a method (it did quite well for sometime), I tracked all numbers until only ONE left unhit. When I got to that point, I started a 110 progression on that one number. It was around a $3,600 BR. Like I said, I made ALOT of money BUT as usual, it slowly tanked. The 110 combined with how far back it last hit was well over 300 spins and on multiple occasions I might add so I stopped playing it. That particular method slowly lured me away from playing sleepers/due.

Ken

I use to play a method (it did quite well for sometime), I tracked all numbers until only ONE left unhit. When I got to that point, I started a 110 progression on that one number. It was around a $3,600 BR. Like I said, I made ALOT of money BUT as usual, it slowly tanked. The 110 combined with how far back it last hit was well over 300 spins and on multiple occasions I might add so I stopped playing it. That particular method slowly lured me away from playing sleepers/due.

Ken

October 29th, 2011 at 3:51:49 PM
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That's why several gambler's like you have become victims of the 'gambler's fallacy'.

October 29th, 2011 at 4:15:33 PM
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Quote:KeyserThat's why several gambler's like you have become victims of the 'gambler's fallacy'.

.......and you Keyser are low on gas, better stop. Which city this week? I heard the 21/33 is hitting ALOT at a casino in Tucson. Pack up the car and away you go!! (LMAO) Talk about putting all your eggs in ONE basket.

Ken