arriba
arriba
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September 24th, 2014 at 9:01:29 AM permalink
What is the statistical benefit of having multiple entries in suicide pools?

For purposes of this question, assume that the odds of one (1) entry winning* is .5%.

* "winning", means getting some money either by winning the pool outright or dividing the pool money with other entrants.

If the odds of one entry winning are .5%, what would be the odds of winning if a player had three (3) entries? Would the odds of three be 1.5% (3 x .5 = 1.5)?

In other words, are the odds of winning with multiple entries greater/or the same as the sum of the odds for individual entries?

My sense, which is intuitive and without any statistical basis, is that multiple entries have a greater advantage that the sum of the individual entries. Thus, in my example, the odds of winning for the multiple entries would be greater than 1.5%.

Hoping there's a mathematician or statistician out there who has an answer.
GWAE
GWAE
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September 24th, 2014 at 9:21:53 AM permalink
Quote: arriba

What is the statistical benefit of having multiple entries in suicide pools?

For purposes of this question, assume that the odds of one (1) entry winning* is .5%.

* "winning", means getting some money either by winning the pool outright or dividing the pool money with other entrants.

If the odds of one entry winning are .5%, what would be the odds of winning if a player had three (3) entries? Would the odds of three be 1.5% (3 x .5 = 1.5)?

In other words, are the odds of winning with multiple entries greater/or the same as the sum of the odds for individual entries?

My sense, which is intuitive and without any statistical basis, is that multiple entries have a greater advantage that the sum of the individual entries. Thus, in my example, the odds of winning for the multiple entries would be greater than 1.5%.

Hoping there's a mathematician or statistician out there who has an answer.



sorry can not provide any numbers but the obvious answer is that you gain a huge advantage as the season goes along IF all of your teams are still alive. If you get to the final week and you have 2 entries still alive against 1 or 2 other people then you obviously can take both sides of a game and guarantee yourself a win.
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dwheatley
dwheatley
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September 24th, 2014 at 9:33:35 AM permalink
odds are 1 - ( 1 - 0.5% ) ^ 3 = 1.4925%
the difference between your way (the wrong way) and the correct way becomes more significant when the odds of winning are higher or there are more entries.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
dwheatley
dwheatley
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September 24th, 2014 at 9:36:26 AM permalink
I'm assuming there are no correlations between the entries, which is probably reasonable considering how many possible entries there are.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
arriba
arriba
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September 24th, 2014 at 10:14:05 AM permalink
The 0.5% value was selected randomly. Obviously, this value would increase or decrease depending upon the total number of entrants. For example, the odds of winning a pool having 20 entrants is substantial greater than it would be if the pool had 300 entrants.

Whatever the true odds might be, 0.5% or 10% or something else, my question is whether multiple entries increase the odds of winning by a factor greater than the total of the individual odds. I can't understand how the odds of multiple entries is LESS than the total of all the individual entries odds. That seems to be what dwheatley's equation shows... 1.4xxx vs 1.5, or am I missing something here?

I'm not sure what you mean "no correlations between the entries".

If a player has just 1 entry, he is in a "one and done" situation (the team loses, the player is out); if the player has three (3) entries, he is not in a "one and done" situation. In my mind, this creates a statistical advantage greater than 3 x 0.5% (or whatever the odds are for 1 entry). Or, maybe it doesn't... .
dwheatley
dwheatley
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September 24th, 2014 at 10:33:35 AM permalink
Extend the argument to a large number of entries. If you have 200 entries, your simple formula says you have a .5% *200 = 100% chance of winning. Clearly you don't, since there are many many possible entries and other players.

Your true chance is less than the sum of the odds of the entries, and this holds for any number of entries. Odds of winning go up, but not linearly.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
arriba
arriba
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September 24th, 2014 at 11:32:45 AM permalink
The odds of winning change depending on the total number of entrants in the pool. If the odds of winning were .5% (as in my example) , the poll would have to have a total of 200 entrants ( 0.5% x 200 = 100%) ... In this senario, I would have all the entrants and a guaranteed win.

I am currently in a winner take all pool... started with 325 , now at 180, and have only 1 entry. I have been offered to partner with 2 other entrants. I would be giving up 2/3 of my entry, and getting 1/3 in each entry of the two other guys. Better odds, smaller payout...


What I am trying to determine is whether there is a statistical advantage (greater than 3 times) to partnering. My current odds, based on the 180 remaining, is 0.55%. Is the advantage of a partnership greater than 0.55% x 3? If not, I'll decide based on whether to go for a big payout or try to book a winner.

dwheatly, could you explain your equation. Did it show that the 3 entries would yield LESS odds than 3 x the individual odds? How is that possible?
dwheatley
dwheatley
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September 25th, 2014 at 6:51:24 AM permalink
If the number of entries is fixed now, then there is correlation between the entries. Assuming you can't judge which entries are better, in a last entry standing wins, your original equation is correct. Having three entries triples your chance of winning.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
FinsRule
FinsRule
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September 25th, 2014 at 8:26:03 AM permalink
In most cases I give practical answers, not mathematical ones. This is one of those cases. Suicide pools aren't really mathematical at all.

1) You're too early to partner.
2) You should only partner if you have selected different teams than your partners have.

1 - If you partner now, are you going to keep picking the best team (SD this week) or are you each going to have a different team? If you each pick a different team at this point, your partnership might last a bit longer, but you'll probably have fewer entries. If you make it to a point where there's like 50 entries, then it might make sense.

2 - And it makes sense if the partnership gives you a wider variety of teams to pick. If you have made the same picks through 8 weeks, it's a lot less helpful.

My friend is in a suicide pool with 8000. That's pretty much a lotto ticket, and math won't help you.

I'm in one with 20. There's 7 left. Typically I'm picking the 2nd biggest favorite and hoping the other 6 just pick the top choice.

Now I'll start talking math/statistics, although you've probably stopped reading.

With 200 left, you can look at the Yahoo overall pick distribution and assume that your pool will follow that. You would also use the Wizard's NFL point spread / win % expectancy chart.

So let's say Yahoo says 60% of people have picked the Chargers, and the Wizard's point spread win % for the Chargers is about 92%. I would say that means the Chargers are the right pick. Of course, that works earlier in the season when more teams are available. As it gets later, and less teams are available, it's just survive and advance.

The final issue that makes it tricky is that you are using game theory (I think) to predict what people will do. Example: There's 4 people left in suicide. Team A has a 75% chance of winning. Team B has a 65% chance of winning. (Both teams are available and the only real choices this week) Logic says everyone just picks team A. But, then team B has the value. Because if one person picks Team B, then they now have the best chance to win. Well, you're just trying to predict human behavior now (Which I have to do on the last race of horse racing contests and I hate it).

So, in conclusion - Just pick who you think will win, and only partner if there's less than 50 left and it's like Week 8 or later and you've picked different teams. But I don't think math can help you solve this one.
arriba
arriba
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September 25th, 2014 at 9:13:56 AM permalink
No, FinsRule, I didn't stop reading.

Thus far, we have picked different teams, with one exception where two of the partners picked the same team one week. It concerns me that if we wait to partner, the three of us may pick the same teams in the upcoming weeks... and this would substantially reduce the benefits of a partnership later on. If we start the partnership now, we can control the picks for each of us and avoid duplicate picks.

The proposed partners have discussed how we will proceed if we partner up. Because everyone in the pool picks favorites, we know that eliminations are caused by upsets. It is our intention to pick 3 different teams each week... 1st favorite, 2nd favorite and 3rd favorite (those teams should be available to one us, at least, until later in the season). I have been using the money line, rather than the point spread, to assist in determining the favorites (but, also, consider other factors). Actually, we look for the team that is least likely to be upset... more times than not, this is a favorite but not necessarily. For example, last week NO was a big favorite on the $ line and pt spread... but I couldn't pull the trigger on a team that was 0-2... Cinn was my #1 pick last week.

The biggest advantage of a partnership, imo, is that we are no longer 'one and done'. Each week, we will pick the #1, #2 and #3 favorite teams. Presumably, others in the pool will make similar picks, and when one of these teams gets upset, we lose an entry but are still in the pool with 2 remaining entries. And, hopefully, when one of these favorites gets upset, we will lose a large number of entrants.

Thanks for your analysis... it was helpful.

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