I loves me a good paradox, and am surprised to have never encountered this one before, 'cuz it's a doozy! (another simple probability conundrum… and apparently fairly well-known within certain philosophy circles).

Called the "shooting room paradox" Futility Closet recently ran it in Greg Ross's always succinct manner.

I'll re-state it in my own fashion here:

The situation involves a theoretically infinitely large room and infinitely large population of players… and, 2 dice. The first 'player' enters the room and the 2 dice are thrown. IF the result isIn a quick Google search of the problem (some of the 'meatier' papers on it were unfortunately behind paywalls), what I garnered from a few links is that the basic explanation for such divergent perspectives is that the 90% figure is of course an aggregate figure, and even though it'sdouble sixes, the player is shot and game over. Otherwise the player leaves the room unscathed and 9 new players enter. Once again the dice are rolled, and IF the result is double sixes, ALL 9 are shot. If not, they leave, happy and healthy, and 90 new players enter the room….

This pattern continues, with the number of players increasing tenfold with every new round of play. The game simply goes UNTIL double sixes ARE rolled and an entire room group is shot, at which point the game is over.

IF you are in the pool of players how worried should you be for your safety? ...perhaps not very, since your chance of being shot isNEVERmore than 1 in 36, or < 3% (the chance of double sixes)., your wife discovers you are in a group about to enter the room, and she is petrified, because she understands that inevitably ~90% of ALL players who participate will be shot before the game is over! Who is perceiving the odds correctly?BUT

*true and accurate, once the game*

**eventually***is over*, it doesn't change the real

*in-the-moment*probability of harm for any single individual, which is only 1 in 36 ...in-other-words, one cannot evoke reverse causation in adjudging the probability at a given point-in-time.

...I hope this is adequate explanation, but if someone feels they can state it more clearly or differently in the comments, please feel free to do so (I'm leaving out certain nuances and fine semantic points of the puzzle, which can be argued over, to try and convey the gist of it here).

## 1 comment:

I think the problem is in the varying number of people who are counted as participants as you go from round to round. In other words, the probabilities given do not account for the probability that you are in some future group and the game ends before you become a player.

The first player has a 1/36 chance of being unlucky. Given the conditional fact that they are called into the room, each member of the second group will also have a 1/36 chance of being shot; however before the first round there is a 1/36 chance that they will not be called into the room, so their true odds are 35/1296 or ~1/37.

Each group has better odds than the one in front of it because there is only a (35/36)^n chance that round n+1 will be called.

If the first player comes up unlucky, it is true that 100% of the players _who were ever called_ were shot, but this outcome saves hundreds or thousands from ever being subject to the game in the first place.

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