Rules

1 - Infinities are not allowed. All numbers must be finite and real.

2 - Subsequent entries cannot rely on previous entries (e.g G(64) could not be beaten by G(64) + 1, but it could by TREE(3)).

3 - Semantic wordplay is not allowed (e.g 'the biggest number anyone could think of' or 'the largest number you could possibly describe' are illegal).

Vanessa 02/02/2026

Introduction

Vanessa if my first attempt at an incredibly large number, in the following section I will describe her and her failure.

To begin, it is necessary to understand what I mean by 'lottery'. A lottery size n order m is a game wherein a player picks n truly random numbers between one (1) and m inclusive.

There is then another random draw of exactly n random numbers between 1 and 2m. The player wins if all of his numbers are chosen during this draw (regardless of order).

In general, the chance of winning a lottery size n order m (or lottery(n,m) from hereout) is:

(1/2m)^(n)

Description

1) To begin, play a lottery(85,(10^100)). When drawing numbers in the lottery, do so only once every googolplex years (googolplex = 10^(10^100))

2) If you fail this lottery, adapt it so that it is now lottery(n^(10^(10^100)),m^(10^(10^100))), and raise draw time to (10^(10^100)) So for iteration 2, the lottery should be lottery(85^(10^(10^100)),(10^100)^(10^(10^100))) draw time (10^(10^100))^(10^(10^100)) years.

3) If the lottery is failed a subsequent time, the size, order and draw time shall be raised to themselves. This is to be followed for all failures henceforth.

4) Vanessa is defined as the average number of femtoseconds one is expected to play this game.

Gauging the size

Vanessa seems a strange size at first glance - she could range from reasonably large (smaller than G(64)) to unbelievably huge. I thought it could not be larger than Rayo(10^100) by its own nature (describable w/ far less than 10^100 symbols), but I considered it likely larger than TREE(3).

The probabiity of winning at even the first step is ludicrous:

P(Win first it) =

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ =

It goes without saying that (2^85)(10^8500) is a large number, but this is the very best chance you will get. All subsequent chances will be much lower, and will get smaller and smaller at a vanishing rate.

The number seems like a runaway train, the point in the distant future that a linear line meets an exponential. But they must meet right? Surely you can't play Vanessa's lottery forever?

Let us attempt an easier problem, calculating a (seemingly) comparitively much smaller number Vanessita:

Vanessita

1) Instead of playing lottery (85, 10^100), play lottery(1,2) (50% chance of winning). Draw once every one (1) second.

2) If you fail, raise the odds to lottery(2,4) and draw time to 2 seconds. From here on, raise all variables to themselves upon failure (e.g it 3 => lottery(4,256), draw time 4 seconds)

3) Vanessita is the number of years one is expected to play this game

But when are we expected to win? Again we can simplify to a coin flip (1/2), where it is clear we are 'expected' to win after 2 iterations (or the denominator of the fraction). Therefore, it follows that where i is equal to 1/(the probability of success), the player is expected to win by i iterations.

But of course, this will NEVER HAPPEN! In fact, if we calculate the cumulative probability of winning over an infinite amount of time, this does NOT resolve to one!

time on iteration i is 2^^2^^2^^.., with i - 1 tetrations. This grows faster than exponentially.

P(win on it i) =

So is Vanessa infinity?? Is Vanessita infinity??

The answer is yes! Over an infinite period of time, winning Vanessa's lottery or even Vanessita's Lottery is not guaranteed. Busted.