[Pancho sanza]

100000 players each put $5 into an nfl pool, the objective is to pick the most winners straight up.

40 % of the pool is taken out as an admin fee, leaving 60 % for the winner(s)

There are 13 games in the pool, the first number represents the favored teams win probability, the second number represents the ratio at which the players have picked the favored team:

.59 .68
.65 .8
.556 .62
.9 .99
.79 .92
.55 .56
.64 .72
.67 .77
.65 .82
.88 .96
.81 .96
.54 .5
.57 .58

You have the option to box 10 games for a cost of $5120 and have to take a side on the other 3.

This will now put $505,120 into the pool before the admin fee takeout.

If your 3 win, you have a perfect ticket and are guaranteed to be one of the winners, perhaps the only winner, which would pay you $303,072.

Is there a way to structure this to make it + ev?

500 points to the one with the best answer.

I've come up with a solution but not sure its correct.

[RickySteve]

Sexy Q.

[RickySteve]

Aren't there 1024 tickets, $5 each, so $5120 contributed? 2^10 = 1024, what am I missing?

[Pancho sanza]

Aren't there 1024 tickets, $5 each, so $5120 contributed? 2^10 = 1024, what am I missing?

Yes, my bad, fixed.

[RickySteve]

To make the problem solvable without knowing the selections of each ticket, one must assume that there is no cross-correlation between picks and each ticket is randomly distributed based on the overall pick %.

[RickySteve]

Fade both .65 teams, back the .54.

[RickySteve]

I could be convinced that backing the .88 is better than the .54.

[pokernut9999]

Best option is to run the pool and take the 40%

[RickySteve]

Best option is to run the pool and take the 40%

Edge for optimal play is >40%. Of course you could enter tickets AND run the pool.

[Pancho sanza]

Fade both .65 teams, back the .54.

I get about +150 % roi using those 3, is that what you get?

I understand why you picked them, but I don't think that's the correct set.

Reason being you only get a perfect ticket about 6.6 % of the time, your ticket is somewhat of a longshot and a lot of + ev pots go uncollected.

If you take some stronger chalk, you participate in more pots, try plugging in the .54 .55 .57 teams, the ROI is closer to 200 %.

[RickySteve]

Unless I'm making a big error somewhere, taking both .35 teams is required.

I would change my answer to .88 rather than .54 though.

[RickySteve]

Wait, I change my mind again...

[Pancho sanza]

Unless I'm making a big error somewhere, taking both .35 teams is required.

I would change my answer to .88 rather than .54 though.

Why required?

You still get the benefit of knocking out a disproportionate # of players if the .35 side wins and its not 1 of your 3 sides.

[RickySteve]

.90 + .19 + .35 (.18) is the best I've seen so far.

You are adjusting for the chopped pots right?

[Pancho sanza]

.90 + .19 + .35 (.18) is the best I've seen so far.

You are adjusting for the chopped pots right?

Yes.

I don't think you would ever want to pick the .9 side outright, you want to leave it in there in case the dog wins since you knock out 99 % of the players.

if it were a 1 game pool, you essentially get about 60-1 where it should be about 10-1

If you take it and the upset happens, your out of the running for a huge pot.

[RickySteve]

My b, must be a flaw somewhere in my model.

[That Foreign Guy]

I'd approach this by picking the three sides with the highest (1-%picked) * (% of win)

This seems like it would balance risk and reward.

The other approach is to assume that because there are so many people (2^13 is only 8192) you will need a perfect ticket to win, and therefore you take sides on the three highest chance teams to maximise the chance of a perfect ticket but this minimises your return.

Hmmm, good puzzle.

[Pancho sanza]

I'd approach this by picking the three sides with the highest (1-%picked) * (% of win)

This seems like it would balance risk and reward.

The other approach is to assume that because there are so many people (2^13 is only 8192) you will need a perfect ticket to win, and therefore you take sides on the three highest chance teams to maximise the chance of a perfect ticket but this minimises your return.

Hmmm, good puzzle.

Taking the 3 highest sides guarantees you'll be winning a lot of small pots and missing out on the big ones.

Remember everyone is taking the sides that have a high win %.

[That Foreign Guy]

That's why I think multipling by the number of people who a correct pick would knock out to maximise winning EV not just winning %

The conservative approach may have better EG though

[Pancho sanza]

That's why I think multipling by the number of people who a correct pick would knock out to maximise winning EV not just winning %

The conservative approach may have better EG though

you may be right, frankly im not sure my solution is correct.

[laxbrah420]

this can't be +ev. even if you're on the longest shots, you're with .01*.04*.04 *100000= 1.6 people =->2.
your 10%, 12%, and 19% parlay should also pay 437-1.
you're getting either 30-1 or 60-1. awful
you're also assuming that only you are playing optimally which I feel is quite a fallacy.

[Pancho sanza]

this can't be +ev. even if you're on the longest shots, you're with .01*.04*.04 *100000= 1.6 people =->2.
your 10%, 12%, and 19% parlay should also pay 437-1.
you're getting either 30-1 or 60-1. awful
you're also assuming that only you are playing optimally which I feel is quite a fallacy.

those other 1.6 people still have to go 10-0 on the other 13 games to share the pot.

Not sure how you got 437-1 either.

[laxbrah420]

what do you calculate that parlay at?
and sorry i figured every player had the same condition win 3 thing

[Pancho sanza]

what do you calculate that parlay at?
and sorry i figured every player had the same condition win 3 thing

Should be about 252-1

[laxbrah420]

you calculate 10%, 12%, 19% to be 252? maybe I don't know how to calculate parlays? figured it was a series of all in bets --.10*.12*.19=0.00228
1/.00228=438.6 actually.

[Pancho sanza]

you calculate 10%, 12%, 19% to be 252? maybe I don't know how to calculate parlays? figured it was a series of all in bets --.10*.12*.19=0.00228
1/.00228=438.6 actually.

odds would be roughly +875 +715 and +405

I think the correct strategy might be to actually take the 3 biggest fav's here, the idea being its more important to share in a lot of little pots, and the occasional big one, than to share in less pots, even though most of those would be big.

[laxbrah420]

odds would be roughly +875 +715 and +405

I think the correct strategy might be to actually take the 3 biggest fav's here, the idea being its more important to share in a lot of little pots, and the occasional big one, than to share in less pots, even though most of those would be big.

I mean I think it's poor form to round odds mid conversion, but even with those odds, I have a tough time figuring out how you're coming up with 252-1

[Pancho sanza]

i mean i think it's poor form to round odds mid conversion, but even with those odds, i have a tough time figuring out how you're coming up with 252-1

(8.75*7.15*4.05)-1

[TomG]

there's no way to know if this pool is +ev. in fact, it's very likely -ev. you don't have each players selections but rather the aggregate ratios for all 100,000 participants. all it would take is for enough participants out of 100,000 to know nothing about the teams and pick randomly. or worse, another person employing a strategy similar to the fade the big winners strategy discussed above. that will quickly create duplicate entries. with a 40% takeout, it doesn't take much overlap to make this pool a huge -ev play.

i will accept 250 points (half points) as a consolation prize since there is no known +ev solution as requested.

[Pancho sanza]

there's no way to know if this pool is +ev. in fact, it's very likely -ev. you don't have each players selections but rather the aggregate ratios for all 100,000 participants. all it would take is for enough participants out of 100,000 to know nothing about the teams and pick randomly. or worse, another person employing a strategy similar to the fade the big winners strategy discussed above. that will quickly create duplicate entries. with a 40% takeout, it doesn't take much overlap to make this pool a huge -ev play.

i will accept 250 points (half points) as a consolation prize since there is no known +ev solution as requested.

This is an actual pool that is run weekly by the Ontario government.

You can't say its negative ev because as you said, you don't know what strategies folks are employing.

I'm sure there are people who don't know squat and just pick at random, im guessing they are a small minority.

And yes there are groups who box a lot of games, im aware of a few of them.

The pool has been very much + ev since its inception if you pick your spots, ie, play the weeks where there's a lot of big dogs.

[TomG]

it's a parimutuel pool with a massive (40%) takeout.

if just a handful of people employ basic strategy it becomes hugely -ev.

good luck.

[Pancho sanza]

As far as people just picking at random, if everyone were to do that, you would have a huge +ev by boxing the 3 biggest chalks.

[RickySteve]

there's no way to know if this pool is +ev. in fact, it's very likely -ev. you don't have each players selections but rather the aggregate ratios for all 100,000 participants. all it would take is for enough participants out of 100,000 to know nothing about the teams and pick randomly. or worse, another person employing a strategy similar to the fade the big winners strategy discussed above. that will quickly create duplicate entries. with a 40% takeout, it doesn't take much overlap to make this pool a huge -ev play.

i will accept 250 points (half points) as a consolation prize since there is no known +ev solution as requested.

In other words, post #5.

[laxbrah420]

(8.75*7.15*4.05)-1

Using your formula, I calculate that a parlay of three even money teams (+100) is 1*1*1-1=0.

You forgot that they get their money back on the win. So it's (8.75+1)*(7.15+1)*(4.05+1)-1 =400-1
Again, that's not very accurate because you rounded the decimals to american odds to suit your formula.
Forgive me if I'm wrong --I really just don't see it though.

[laxbrah420]

My revised understanding is that every other player is trying to max 13 and you only have to get 3.