Okay, let's say I am that crazy. I roll into Atlanta with a Multiple Man deck that wins two out of ten against the top decks. How much luck will I need to make a decent showing?

In case you have not played in one of these things, you must take into account the dynamic of the "top table effect".

If you win your first three matches, you will suddenly be paired against the top pros almost exclusively. If you lose early, you will be paired against some wacky decks from freaky fun-guys and a few angry top pros who got character screwed.

So, with Avengers and Faces and Squadron and those control decks rising to the top of the pro guantlet as we reach the short strokes, if I have enough brain damage to actually play Multiple Man...

Exactly how much luck would it take?

Hmmm...about this much:
<<<<<<<<<<<<<<<<<<<<<>>>>>>>>>>>>>>>>>>>>>>

The best way I can think to do this would be to estimate how much your chances increase or decrease with each win or loss (those are actually inversely respective, with decrease associated with win, but meh).

The first calculation will then be easy, to get to 10-0 there is exactly one path (win every round). However, for EVERY OTHER CALCULATION it gets REALLY tricky. For 9-1 alone, you have 10 iterations, as you can lose in the 1st round, the 10th round, or anyone inbetween. Finally, it looks like the number of iterations JUST FOR A SPECIFIC RECORD will be 10^n, where n is your number of losses.

In other words, I do NOT suggest calculating this unless you want to find out your chances of going 10-0, or MAYBE 9-1. Anything else will hurt your head. It's hurting mine just explaining it.

EDIT: I think I'm actually wrong on this. Since you can't lose twice in the same round, the number of ways to get 8 wins are 10*9 (or 90), not just 10^2 (or 100). Similarly, 7 wins would be 10*9*8 (or 720), not just 10^3 (or 1000). As you can tell, this mistake would've cost you quite a bit of time, assuming you weren't just using a computer program to do this for you. However, setting up that program in, say, Excel, using VBA, where you have a cell for your odds between 0 and 1 going into round 1, and how your odds change with every win or loss, could probably be done. Therefore, I'll leave it to Gator to do.

The answer is 42.

Obviously.

Define "decent". 6-4? About as much chance as I have of having a 3some w/ Jessica Alba and Scarlett Johansson.

Oh man Mike, you are crashing my dreams hardcore. And you know your math.

Nah, its 69. The answer is always 69.

Do the odds increase or decrease if you replace one of them with Sara Michelle Gellar?

Since it's me we are talking about, and the only other woman that Nina would ever let me do is Kate Winslet, it's a wash.

Clearly increase, what has she done since Buffy? Frankly I'm not entirely sure she has anything better to do than to have a 3some w/ me (I should let her know I'm open to the idea).

#### you Dalton, this is my gig. You would have never even asked about playing Multiple Man... let alone actually consider it.

Yeah but Sara Michelle Gellar has been married for a couple years so she might be more inclined to say no, whereas Jessica Alba and Scarlet Johansson are still in the random skankiness phase of their careers so the odds might actually be higher for them.

I heard a rumor on another site that Scarlett is actually a butter face. Confirmation?

(Translation: Her body is perfect, but her face...)

Okay then, back on task.

I tend to believe at least half of what Dalton says. When it comes to math, he can't be beaten... even in the finals.

But I refuse to give up hope. I mean, with his PC cash he could probably buy himself a couple lookalike strippers and take some blurry pictures or something.

Do I really have no shot with Multiple Man?