[rsk111]

[ QUOTE ]
I know there are more qualified people here than I to do the math, as I've never had any formal stats/probabillity classes other than sections/blurbs in my math studies almost 2 decades ago (that never went beyond diff eq)...but this is how I see it:

This pseudoanalysys is based on only the weekly results at $10 (OP said $10 went to the weekly pool and $2 to the season) to win $200 (20 teams w/ "Mr Questionable" only having one team in the first example.
Also assuming there are 2 coinflip games in the week.
Also assuming everyone has the same chance at winning the coinflip games.

So w/ Mr Questionable having one team, he is betting $10 to win $200 with the same 25% chance of winning as everyone else...and it would be expected that 5 teams would correctly pick the 2 coinflips (25% of the 20 teams). Net result, a 25% chance that Mr Q's $10 will win $40 ($200 split between 5 winners) from a $10 bet. That's a PK as expected since I assumed everyone had an equal chance of guessing the coinflips.

Now if Mr Q has 2 teams, and picks 2 different coinflip outcomes, he'd have a 50% chance of winning $42 (25% of $210 since there are 21 teams now) from a $20 bet. +104?

If Mr Q has 3 teams, he now has a 75% chance of winning $44 from a $30 bet. +110?

Looks like an advantage to me but perhaps my assumptions were too much.

Corrections/flames graciously accepted.

[/ QUOTE ]

I believe that you made a mistake here and it has nothing to do with whether there are coinflip games or not (which is an entirely different issue). If we simply use the assumptions that you have put forth, I believe that this analysis is incorrect.

Let's go through each scenario.

*Mr. Q has one 1 pick
25% of the time he will share a prize pool with 25% of the participants (5 people). So his share of the pot will be $200/5, which is $40. When this happens he will have a $30 profit.

75% of the time he will be wrong and lose $10.

EV = 0.25*(+30) + 0.75*(-10) = 0

*Mr. Q has two picks.
Now here is where the mistake is made.
With two picks Mr. Q will have a 50% chance of winning the the prize pool and sharing it with 25% of the contestants (5.25 people). IT IS NOT 5 people as you presented. It is 25% of the contestants. If we ran this a million times, sometimes there would be 3 winners, sometimes 4, sometimes 5, sometimes 6, etc. and the average number of winners would be 5.25. So, when he wins, his share of the pot will be $210/5.25 = $40, which will give him a net profit of $20.

50% of the time Mr Q will lose $20.

Therefore

EV = 0.5*(+20) + 0.5*(-20)= 0

*Mr. Q has three picks
75% of the time he will share the pot with 25% of the participants (5.5 people). If he wins, his share will be $220/5.5 = $40, giving him a profit of $10.

25% of the time he will lose $30.

Therefore,

EV = 0.75*(+10) + 0.25*(-30) = 0


So, in your example, one, two, or three picks gives no advantage. EV is zero for all scenarios.

I too admit that I could be wrong and graciously accept any flames/corrections.