Pocket AA - chances of winning vs. # of opponents?

[Karatitis]

Recently, a poker buddy (let's say his name is Oscar) was trying to calculate the odds of pocket AA winning versus different number of opponents, assuming you just "roll" it to the river and showdown.

Oscar calculated that AA wins something like ~ 40% versus 9 opponents (using the above assumption). I found this a bit startling and disagreed that it could be this high. Later in his analysis Oscar rightly pointed out that as each opponent is added the odds against AA winning increase, but the increments get much smaller as each opponent is added since effectively many hands are holding the "outs" for other contenders, and duplicate hands only beat the AA once (that is to say, the executed prisoner only dies once, but gets peppered with several bullets from different marksman?)

Might anyone know the formula or have a table that has the calculated percentages of AA winnings versus n number of opponents (assuming a fantasy maximum of opponents given a 52-card deck, but with a super hold'em table with seats to hold say 15-20 players).


Thanks.

[AaronBrown]

Here are some simulation results for all hands. It shows a 31.1% chance against 9 opponents.

Your logic is correct, however. Each opponent reduces the chance of winning by a smaller amount.