Problem: Di Birthday Wheel
How many spins before one number go repeat? E few well well pass wetin most pipo dey guess.
Di birthday problem for wheel of 37 pocketsequipmentPocketOne of di 37 small boxes wey get number for di rotor. Dem also dey call am: slot, cup.Read di whole tin →: afta just 8 spins, chance wey pass half dey sey one number don already show two times. Players dey read repeats as patterns; na wetin fair wheel suppose do.
Di Problem
Spin fair European wheel. How many spins until di probabilitymathsProbabilityDi chance of one outcome, from 0 to 1 or from 0% to 100%. One number: 1/37 = 2.70%.Read di whole tin → sey at least two spins don show di same number go pass half? Most players dey guess somewhere around 18 or 19, half of 37. Di answer na 8.
Di Solution
Calculate di probability sey no repeat go happen and remove am from one. Afta k spins, p(all different) = (37/37)(36/37)(35/37)…((38 − k)/37). For k = 8: 1 × 0.973 × 0.946 × 0.919 × 0.892 × 0.865 × 0.838 × 0.811 ≈ 0.44, so p(some repeat) ≈ 0.56. For k = 14 e na 0.93; for k = 20 e na 0.996. Wetin pipo feel dey fail becos di number of pairs of spins dey grow like k², and any pair fit match.
Why E Matter for Di Table
Results board of di last 20 numbers go almost always show at least one repeat, often two or three. Players go see “17 come two times for ten spins” and conclude sey 17 dey hot; di birthday arithmetic tok sey repeat inside ten spins dey happen 74% of di time and di number wey repeat na random. Di pattern na real; di meaning no be.
Di Next Problem
How many spins until one particular number, like 17, don show two times? Dat one na different question wey get much bigger answer: di waiting wey we expect na 74 spins, and di probability of two 17s inside 20 spins na about 10%. Di difference between “some number go repeat” (8 spins) and “dis number go repeat” (74 spins) na di whole difference between pattern wey you notice afta and prediction wey you make before.
Try Am
Run di demo game for ten spins and look di history. Usually, one repeat go dey. Run am again. Di number wey repeat go different. Do am ten times and you don see di birthday problemmathsBirthday problemHow many pipo e go take before two of dem go probably share birthday, or how many spins before one number go probably repeat. For wheel wey get 37 pockets, 8 spins don do for chance wey pass even.Read di whole tin →; do am one thousand times (di simulators fit) and you don measure am.
Formula
Questions wey pipo dey ask
Exactly, with 37 pockets instead of 365 days. 23 pipo dey give 50% for birthdays; 8 spins dey give 56% for di wheel.
50% afta about 19 spins. E still dey normal.
No. Thousands of spins from one wheel go show am; one repeat no fit ever show am.
Play with Maths
Try dem by yourself first. Each solution dey folded; open am when you ready.
One results board wey dey near American wheel dey show di last 16 spins. How likely e be sey no number go show two times for am?
Show di solution
Answer About 2.5%: around 39 boards out of 40 dey show at least one number wey repeat.
Go spin by spin. Di first spin fit be anytin. Di second must miss am: 37 chances for 38. Di third must miss both: 36 for 38. Di sixteenth must miss fifteen numbers: 23 for 38.
Multiply: 38/38 × 37/38 × 36/38 × … × 23/38 ≈ 0.0246.
Board wey no get repeat na di rare one. Board wey get repeat no dey tell you anytin about di wheel.
For ten spins of European wheel, wetin be di chance sey 17 go come out exactly five times? Sey one number, any number, go come out five times or more? And sey one number go come out at least three times for four spins?
Show di solution
Answer 17 exactly five times: 1 for 315,576. One number five times or more: about 1 for 8,333. One number three times or more for four spins: 1 for 349.
17 five times: choose which five of di ten spins, 252 ways, each one with chance (1/37)⁵ × (36/37)⁵. Togeda, 1 for 315,576.
Any number: each of di 37 numbers get chance of 1 for 308,325 to hit five times or more, wey give 37 for 308,325, or 1 for 8,333. Di rare case where two numbers hit five times each dey count two times, but to remove am no change anytin for dis level of precision.
Three times for four spins: for one number, 4 × 36 + 1 = 145 out of di 37⁴ results wey fit happen; for any of di 37 numbers, 145/37³ = 1 for 349.
For double-zerowheelZeroDi one green pocket wey no even-money, dozen or column bet dey cover; na im be di source of di house edge. Dem also dey call am: single zero, 0.Read di whole tin → wheel, di figures wey match na 1 for 359,275, 1 for 9,243 and 1 for 368. Di gap between “17” and “any number” na di birthday problemmathsBirthday problemHow many pipo e go take before two of dem go probably share birthday, or how many spins before one number go probably repeat. For wheel wey get 37 pockets, 8 spins don do for chance wey pass even.Read di whole tin → again: 37 ways dey for you to dey surprised.
Small laugh
Sometin dey repeat early. Dat sometin na random.
Sources
- Feller, W. An Introduction to Probability Theory and Its Applications, Vol. 1.
- Wizard of Odds. “Ask the Wizard”, di questions about roulette.