"I'D LOVE TO BE THE ONLY FEMALE AT A "SUPER-BOWL" PARTY MADE UP OF 4-6 MEN! THE MEN TALK ME INTO MAKING A FRIENDLY WAGER ON THE 1/2 TIME SCORE. IF THE FELLAS TEAM IS LOSING AT 1/2 TIME...I GET $1000.00...IF THE TEAM THEY CHOSE IS WINNING...THEY ALL GET "ME" TO DO AS THEY PLEASE!!!"

assume the teams are 50/50.

Does she cost 1k and we are just kind of negative freerolling ourselves and not getting her half the time. Or does she cost 500 dollars? What's our net EV (im not sure if this is correct terminology). This prop is confusing me.

This is a clear -EV bet. Why on earth would you want to bang a chick with 4-6 other dudes? Well...scratch that. I'm sure some people think that is +EV (and that is fine by me), but I certainly think this is -EV no matter how you slice it up.

But to answer your question serioulsy (not that anyone should), if it is 50/50 on the teams and you think her prop is worth $1K, then the bet is break even. Basically you need to figure out how much you think the prop is worth (I think it is worth $0) before you can figure out what kind of odds you would need to take the bet.

Sherman

The part of the question im interested in is the pay 0 dollars when u win and get her and the pay 1k when u lose and not get her. I am wondering if we are paying the same amount as if we just payed 1k for her everytime and got her everytime. Are we giving up the same amount of money longterm here as if we payed her everytime and just having increased negative variance (not getting her half the time) or are we getting a better deal moneywise when playing the half time game.

The EV is exactly half of the difference between having her & -$1000, because both happen half the time.
E.g. If she was worth $0 then the EV would clearly be -$500.

Look, the problem you posted in the OP is a break-even bet if you think "having your way with her" is worth $1K.

You win her 1/2 time. That is, you get .50 * X, where X is the value of winning her. If X is exactly $1000 you get .50 * $1000 which is $500.

You lose $1000 1/2 the time. That is, you get .50 * -$1000, or -$500 (as a group of course, not you individually, I am assuming you pool your money).

So if you played this "game" twice and achieved each outcome once based on its probability of happening (50/50) you would win $500 once (.5 * X) and lose $500 once (as a group). Therefore the game is break even.

You are never risking $0 to win her, or risking $1000 to win nothing. You are risking $1000 to win X (her). If X = $1000 the bet is breakeven. If X is less than $1000 the bet is -EV. If X is greater than $1000 the bet is +EV.

So as I said before, you have to decide how much banging a chick with 5 other dudes (X) is worth.

Sherman

FWIW - If you think she is worth $1K, I suggest considering the following:

1) You could probably get a decent hooker for that price.
2) Will there be any reputational consequences if people find out about this incident (for you)? Your reputation is more valuable than you think.
3) Is there any probability she ends up accusing you all of rape?

What a waste of time, missing the second half the Superbowl...

Aren't you paying $200-$250 or whatever? I'm assuming the $1000 is split evenly among the guys.

Unfair assumption. Team Versus 1. We shouldn't pretend to know the intricacies of the team.

Team 1 could split the cost of an awesome hooker for 1k.

How does it affect EV if she's missing a front tooth?

No, only if all teeth are missing.

+ or - EV is subjective in this case.