WheelJarWheel Odds Chart
Every option holds one equal slice, so the chance per spin is 100 divided by the number of options — 10 names is 10% each, and no spin remembers the one before it. The numbers people actually get wrong are the ones that follow from that, and the table below works all of them out for wheel sizes from 2 to 50.
Five figures from it, each computed from the same uniform draw the spinner performs:
- Ten names is 10% a spin, a 36° slice, and a 1% chance of the same name twice in a row.
- A repeat inside the first five spins is normal, not broken: with ten names it happens 69.8% of the time.
- A wheel of ten needs about 29 spins on average before every name has won at least once.
- A class of thirty needs about 120 spins for the same thing — which is why a classroom picker removes each name instead.
- Doubling the options quarters the chance of a repeat pair: 20 options give 0.25% twice in a row against 1% at ten.
Every figure on this page is generated at build time from the same arithmetic the wheel runs: a uniform draw over equal slices, with the winner left on the wheel unless you switch that off.
| Options | Slice | Chance per spin | Same winner twice in a row | A repeat within 5 spins | Spins to give everyone a turn |
|---|---|---|---|---|---|
| 2 | 180.0° | 50.00% | 25.00% | 100.0% | 3 |
| 3 | 120.0° | 33.33% | 11.11% | 100.0% | 6 |
| 4 | 90.0° | 25.00% | 6.25% | 100.0% | 8 |
| 5 | 72.0° | 20.00% | 4.00% | 96.2% | 11 |
| 6 | 60.0° | 16.67% | 2.78% | 90.7% | 15 |
| 8 | 45.0° | 12.50% | 1.56% | 79.5% | 22 |
| 10 | 36.0° | 10.00% | 1.00% | 69.8% | 29 |
| 12 | 30.0° | 8.33% | 0.69% | 61.8% | 37 |
| 15 | 24.0° | 6.67% | 0.44% | 52.5% | 50 |
| 20 | 18.0° | 5.00% | 0.25% | 41.9% | 72 |
| 25 | 14.4° | 4.00% | 0.16% | 34.7% | 95 |
| 30 | 12.0° | 3.33% | 0.11% | 29.6% | 120 |
| 40 | 9.0° | 2.50% | 0.06% | 22.9% | 171 |
| 50 | 7.2° | 2.00% | 0.04% | 18.6% | 225 |
Why does a wheel need more spins than it has names?
Because the wheel has no memory. Once eight of your ten names have won, the next spin still gives each of the ten a 10% chance, so the two names left need on average five spins each to appear — and the last one alone accounts for ten. Adding that up across every stage is the coupon collector’s sum, n multiplied by 1 + ½ + ⅓ + … + 1/n, and it grows faster than the list does.
That is the whole argument for remove winner after each spin on the random name picker. With removal, ten names take exactly ten spins and a class of thirty takes thirty. Without it, thirty names take about 120 — four lessons of spinning before the quietest pupil gets a turn.
Is a repeat proof the wheel is rigged?
No, and the table’s fifth column is the reason. Drawing five times from ten options repeats about 70% of the time, because there are ten chances for two of those five draws to collide. People read a repeat as a fault in the wheel, when it is the arithmetic of independent draws — the same effect that makes shared birthdays in a small room far likelier than they feel.
If you want repeats to be impossible rather than unlikely, that is a different tool, not a fairer wheel: the list randomizer shuffles once and hands out an order, the team generator deals names round-robin, and the round robin generator pairs everyone exactly once.
How the numbers are worked out
The slice is 360 degrees divided by the option count. The chance per spin is 1/n, and the chance of the same option winning twice in a row is 1/n² — two independent spins multiplied, not added. The repeat column is the birthday calculation, 1 − (n−1)(n−2)(n−3)(n−4)/n⁴, which is the chance that five draws are not all different, subtracted from one. The last column is n·H(n) rounded to whole spins.
All of it assumes what the tools on this site actually do: equal slices, an unbiased draw from
crypto.getRandomValues with rejection sampling, and no memory between spins. Weighted entries
change the first assumption on purpose — the weighted random picker
prints the resulting percentages for exactly that reason.
How the randomness works
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