Interesting Situation in my NFL Picks League

[paperboyNC]

[ QUOTE ]
Please explain. I can't see how it is anything other than the same EV as one team.

[/ QUOTE ]

He isn't picking against the spread which means there are no "coin-flips" in a typical week.

Thus after he calculates the optimals picks, he is forced to pick suboptimal picks for his other 2 entires since they have to be different. If his optimal picks are breakeven, then he is losing money on his suboptimal picks.

If 20 people play the pool and he has three entries, He breaks even at a 15% chance to win. If the game were complete luck, each of his $10 entries would have an EV of $10 and his total EV would be $30.

~PBNC

[BiPolar_Nut]

[ QUOTE ]
He isn't picking against the spread which means there are no "coin-flips" in a typical week.

[/ QUOTE ]

Dunno why I didn't take that into account.

Essentially eliminating coinflip situations effectively negates any advantage multi-teaming would have...so basically anyone multi-teaming is just pumping the pool, right?

OP: Suggest to whomever runs the league to make it clear that running multiple teams IS allowed, then use my pseudoanalysis a few posts up to convince everyone else that they'd have an advantage running multiple teams, then just run one team yourself [img]/images/graemlins/wink.gif[/img]

[sp3ctr3]

If the lines were 100% accurate this forum probably would not exist. For the sake of this, lets say that there are two games each week that are essentially coinflips (spread being 1-2 points). There is one +-1 pt game this week and there were multiple games last week.

It is looking like there is no advantage, but the person doing this sports bets often and has done well at the picks league in the past. I see the argument for having multiple teams creating teams with suboptimal picks, but if the game is a true coinflip, wouldn't this create an advantage?

[rsk111]

[ QUOTE ]
[ QUOTE ]

I thought about it in exactly this way and I don't think it is right. Instead of the classic "do you see why?" I'll say that upon request I think I can provide a specific counterexample to refute the lottery example. While my counterexample can refute the lottery example I am *not* sure that it generalizes to apply to the football pool example (I am still thinking about how to do that).

[/ QUOTE ]

Noobie:

After hanging around here for a while, you'll realize that I'm wrong most of the time. Therefore, do not wait for me to request a counterexample to prove me wrong, just do it like everyone else does.

[/ QUOTE ]
I just didn't want to type it all out, unless someone really cared or if you figured it out. Anyway, here it is.

Let's say you have a lottery with one number (1-10). You can buy a ticket with any number from 1-10 printed on it and each week one number from 1-10 will be pulled out of a hat. Each ticket costs $1 and you can buy as many tickets as you want (and for each ticket you can choose whatever number you want). The prize pool is the total receipts collected from ticket sales. Any player who has a ticket with the winning number is entitled to a share of the prize pool (i.e., if there is only one winning ticket, that person gets all of the prize money, if there are two winning tickets then each person gets 1/2 of the prize pool, etc).

Now let's say three players decide to play this lottery. Player A decides to buy 10 tickets with so that he has one ticket with each number. Player B and Player C decided that they will just buy one ticket and they randomly choose which number to play (it really doesn't matter what they choose, all that is important is that they just pick one number).

In this scenario, total prize pool = $12 ($10 from Player A, $1 from Player B, and $1 from Player C)

According to your reasoning, Player A should have no advantage, because even though he has 10x the chance of winning, he paid 10 times as much, so it should even out. Let's see if this is the case with some simple EV calculations.

EV for Player A

*81% of the time Player A will have the only winning ticket and win 100% of the prize pool for a net win of $2 (i.e., no tie)
*18% of the time Player A and only one of player B or player C will have the same winning ticket so Player A will win 50% of the prize pool for a net loss of $4 (i.e., two way tie)
*1% of the time Player A, B, an C will have a winning ticket and each get 1/3 (i.e., three way tie), so player A will have a net loss of $6

So, therefore the EV equals

0.81*2 + 0.18*(-4) + 0.01*(-6) = 0.84,

So Player A has a EV of $0.84. Similar calculations will yield an EV of -$0.42 for the other players.

So in this instance, by buying multiple entries, player A has managed to gain an advantage in the lottery.

However, for this same lottery game, it is easy to also come up with examples where it is -EV to buy multiple tickets. So, in the end, we can show that buying multiple tickets can be either +EV or -EV, depending on the strategy employed.

I may have made a mistake, somewhere. If so, let me know. Furthermore, as I said previously, I am not sure how to apply this to the football pool case.

[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.

[Lori]

[ QUOTE ]
If he has an advantage by having three teams, then the other twenty players should get together and play as one unit and have an incredible super advantage. [img]/images/graemlins/tongue.gif[/img]

Lori

[/ QUOTE ]

I can't believe I failed to end the thread with this [img]/images/graemlins/mad.gif[/img]

Of course, if the guy is a winning capper, then his +EV is multiplied ZeeJustin style and that would be a good reason to not allow multiple entries in the future.

Lori

[BiPolar_Nut]

[ QUOTE ]

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

[/ QUOTE ]

Thanks, rsk111. Makes sense to me...I tried to think deeper on it but my brain started to hurt. All I know is it's -EV for me to devote any more time on this one [img]/images/graemlins/wink.gif[/img]

[linesacker]

Many years ago, I was in a pick 'em pool and someone purchased two entries each week. One with his picks and one with the exact opposite picks.

I guess he was buying insurance. He won more often than anyone else that year.