Math Question re:- the " Carib Football Poll "

[The Mayher]

The standings going into the final week are 113 pts.to 108 pts for the top 2 entrants.By the way it's a $ 55,000 1st place prize.......You make your top 5 picks against the spread.As you can see,the magic # is "one",but an unusal "one" because any pick that ends up being tha same eliminates the player in 2nd place.So what's the chance of the guy trailing,tying it up.It must be astranomical.Does anyone out there know the math.......Thanx....BB

[DeucesUp]

I think I understand what you're asking, you can make a simple estimate as:

Three things have to happen.
1. They must make 5 different picks.
2. Leader must miss all 5
3. Trailer must hit all 5.


#1 -- Assuming the two players have no knowledge about each other's picks and have no tendancies which would tend to make them more or less likely to pick the same teams, #1 can be estimated as: (26/32)*(25/32)*(24/32)*(23/32)*(22/32) or about 23.5%

#2 -- Assuming no ties and assuming the players don't beat the spread, the chance of leader losing all 5 is about (1/2)^5 or 3.1%

#3 -- Chance of trailer winning all 5 is about the same 3.1%

Chances of all 3 happening are then 0.235*.031*.031 = 0.023% about 1 in 4500.


It's not quite this bad, as the 3 conditions aren't completely mutually exclusive. That is, once condition 1 is met, it becomes much more likely the "trailer" picked the opposite team in the same game. Then when the leader looses that game (condition 2), the trailer has automatically won it.

On the other hand, all pushes are in the favor of the leader as a tie is as good as win for him and a tie is as bad as a loss for the trailer.

These two factors partially cancel each other out, I won't try to calculate the exact effect of these.

[MicroBob]

Wonder if the 2nd place guy can try to make picks that he thinks will be the opposite of the other guy.

Picking the games that he thinks are more likely to lose might actually give him the best shot since this gives him a better chance to stray from the other guy's picks.
to that end, he might even want to consider picking all of the playoff bound teams who are supposed to be resting their starters (PHI, PIT, IND, GB, etc) if he thinks the other guy will avoid these teams.


I played in a free weekly pool last year at a casino.
Top 5 winners of 12 random college games won money ($2.5k down to $100 I think). Tie-breaker was total points of one of the games. You were just picking winners...not picking ATS.
Normally 11-1 or 10-2 was good for first place (out of 600 entrants).
One week I went 12-0 and missed the tie-break total by only 3 points. Thought I had cashed-in (and it was on my birthday no less).
Learned that out of 600 entrants, 120 had a perfect 12-0 score and 10 players or so missed the tie-break by 0-2 points so I didn't cash-in.

From that point, I determined it was better EV to pick a team that I thought had only a 40-45% chance of winning IF I thought 70% or more of the public would be siding with the favorite.
In other words, picking an undervalued underdog had value to give me the best chance of winning in the pool. This is in spite of the fact that my chance of going 12-0 or 11-1 were decreased by choosing the underdog.

This would have been different, of course, if I was still in contention for the $10k or whatever they were giving for the best season-total. But I was already far enough behind in that race that it wasn't a concern.

[The Mayher]

[ QUOTE ]
I think I understand what you're asking, you can make a simple estimate as:

Three things have to happen.
1. They must make 5 different picks.
2. Leader must miss all 5
3. Trailer must hit all 5.


#1 -- Assuming the two players have no knowledge about each other's picks and have no tendancies which would tend to make them more or less likely to pick the same teams, #1 can be estimated as: (26/32)*(25/32)*(24/32)*(23/32)*(22/32) or about 23.5%

#2 -- Assuming no ties and assuming the players don't beat the spread, the chance of leader losing all 5 is about (1/2)^5 or 3.1%

#3 -- Chance of trailer winning all 5 is about the same 3.1%

Chances of all 3 happening are then 0.235*.031*.031 = 0.023% about 1 in 4500.


It's not quite this bad, as the 3 conditions aren't completely mutually exclusive. That is, once condition 1 is met, it becomes much more likely the "trailer" picked the opposite team in the same game. Then when the leader looses that game (condition 2), the trailer has automatically won it.

On the other hand, all pushes are in the favor of the leader as a tie is as good as win for him and a tie is as bad as a loss for the trailer.

These two factors partially cancel each other out, I won't try to calculate the exact effect of these.

[/ QUOTE ]

Interesting & gives me an idea.thanx for your effort.

[Michael Davis]

I play in a bowl pool where you have to pick winners of 22 games and you assign each number (1-22) once. If your team wins, you get that number of points. About 500 people enter every year.

I am fairly certain that the strategy that gives you the most chance to win is to pick the most absurd upset on the board. (Texas Tech?)

-Michael

[MicroBob]

I don't think you want to pick the most absurd upset on there....but you want to pick the game that you think EVERYONE ELSE thinks is the most absurd that you also think has a reasonable chance of coming through.


To that end, I think throwing a few points on T-Tech would have been a good move (I thought their chances were decent actually...see my previous posts where I mention my thoughts that the PAC-10 is overrated).


Since it's a confidence-points pool you can just bet 4 or 5 points on T-Tech...while everyone else is betting 15-20 points on California.

Certainly putting 20 points on T-Tech would have been risking too much...but taking a 5-point shot with them would have been a reasonable strategy I think.


I love confidence-points pools btw. I worked at a place that had 40-50 people on an NFL confidence pool each week and I took it down 2x in one season.
People got mad that I was "just picking the upsets" but I was actually picking situations where it was pretty obvious to me that the consensus-favorite was vastly over-rated (and my opinions were proven correct in the weeks that followed).
So there were a couple of teams where everyone thought the favorite was worth at least 8 to 10 confidence points...and I put anywhere from 2 to 5 on the underdog.
They lose their 10 point game and I'm winning 5. Not bad at all.