Monty Hall Problem

Everyone

Every game played here, added up.

Across all 100 finished games

63.00%

of first picks were not the car. Theory says 66.67%. Every one of those games was a game where switching would have won.

Finished runs1
Unfinished0
Completion rate100.0%
Still playing1

Best run so far

37 cars

The most cars won in a single finished run, out of 100. Switching every time averages 67, so beating that by much is luck, not skill.

How the runs spread out

Each finished run produced one number: how many of its 100 first picks were wrong. Here is every one of those numbers stacked up. The curve is what the maths predicts, Binomial(100, 2/3), centred on 67 with a standard deviation of 4.7. The bars should follow it.

1 0 0: 01: 02: 03: 04: 05: 06: 07: 08: 09: 010: 011: 012: 013: 014: 015: 016: 017: 018: 019: 020: 021: 022: 023: 024: 025: 026: 027: 028: 029: 030: 031: 032: 033: 034: 035: 036: 037: 038: 039: 040: 041: 042: 043: 044: 045: 046: 047: 048: 049: 050: 051: 052: 053: 054: 055: 056: 057: 058: 059: 060: 061: 062: 063: 164: 065: 066: 067: 068: 069: 070: 071: 072: 073: 074: 075: 076: 077: 078: 079: 080: 081: 082: 083: 084: 085: 086: 087: 088: 089: 090: 091: 092: 093: 094: 095: 096: 097: 098: 099: 0100: 0 67 expected 0102030405060708090100
Finished runs What theory predicts

Runs left unfinished

0 runs stopped early, after 0 games between them. A run counts as left behind once it sits untouched for 30 minutes. Runs still being played are not counted here. These are shown as a rate, in 5 point steps, because the runs are different lengths.

Nothing to show yet.
Unfinished runs

Pooled across every unfinished game

Nothing to show yet.

Recomputed every 30 seconds. Last at 07:52:47 UTC.