justlookin'
Registered User
Re: ECAC Pick the Playoffs 2010
Clarkson 2-1
RPI 2-1
Dartmouth 2-1
Princeton 2-1
The North Country Teams. OK SLU
Clarkson 2-1
RPI 2-1
Dartmouth 2-1
Princeton 2-1
The North Country Teams. OK SLU
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ECAC Pick the Playoffs 2010
- Thread starter vicb
- Start date
Re: ECAC Pick the Playoffs 2010
Heh. I never really paid attention to the scoring system before. I'm not sure your hedging strategy really works, though. Given the historical unlikelihood of game 3s, you're going to end up losing more points by turning potential 4-pointers into 3s than you're going to gain by turning 0s into 1s.
FlagDUDE08
Banned
Re: ECAC Pick the Playoffs 2010
There are 4 possible outcomes to every series, and each is equally likely to happen, assuming random outcomes. The long run situation will be that you pick the same option for all four games (home in 2, visitor in 3, etc.), and all 4 possibilities happen (home wins in 2, home wins in 3, visitor wins in 2, visitor wins in 3). If you pick a sweep in the long run, you will receive 8 points, or an average of 2 points per series (you'd get 4, 3, 1, and 0). If you pick a 3-game series in the long run, you will receive 10 points, or an average of 2.5 points per series (you'd get 4, 3, 2, and 1). Therefore statistically, burgie12's hedging strategy is more profittable. Now obviously the beauty of betting is we are playing the short run, and anything can happen. If this were a 5,000 series tournament, I'd take burgie's strategy any day. But with only 8 series, play your heart.
There are 4 possible outcomes to every series, and each is equally likely to happen, assuming random outcomes. The long run situation will be that you pick the same option for all four games (home in 2, visitor in 3, etc.), and all 4 possibilities happen (home wins in 2, home wins in 3, visitor wins in 2, visitor wins in 3). If you pick a sweep in the long run, you will receive 8 points, or an average of 2 points per series (you'd get 4, 3, 1, and 0). If you pick a 3-game series in the long run, you will receive 10 points, or an average of 2.5 points per series (you'd get 4, 3, 2, and 1). Therefore statistically, burgie12's hedging strategy is more profittable. Now obviously the beauty of betting is we are playing the short run, and anything can happen. If this were a 5,000 series tournament, I'd take burgie's strategy any day. But with only 8 series, play your heart.
engineerhockeyfan
PUCKMAN RULES!!!
Re: ECAC Pick the Playoffs 2010
Clarkson 2-1
RPI 2-0
Quinnipiac 2-1
Princeton 2-1
RPI
Clarkson 2-1
RPI 2-0
Quinnipiac 2-1
Princeton 2-1
RPI
Re: ECAC Pick the Playoffs 2010
I pretty much stopped reading right there. I just went through the trouble of showing that each outcome IS NOT equally likely to happen, so your whole analysis is based on a faulty assumption. Sweeps outnumber game 3s by 1.6:1, so if you always pick 2-1s, you will definitely lose in the long run.
FlagDUDE08
Banned
Re: ECAC Pick the Playoffs 2010
Regardless of who wins game 1, it is 50% likely that the same team will win game 2, assuming random outcomes. Either the home team wins, or the visiting team wins.
Your theory is based upon ECAC history. My theory is based upon a team's probability of winning a game, without any regard to what has happened in the past, or what any current record or skill level of a team is. I am assuming that if the home team swept the past 100 played series, that the probability of the outcome of the 101st series is exactly the same as the 1st. It's just like if you're shooting craps, and you roll snake eyes (1-1) 50 times in a row. The probability of you rolling a 7 on the next roll is STILL 1 in 6.
Now, since we're talking about sports, and one team generally has more skill than another, is it truly random? Probably not. But, as we've said MANY times on this forum in various threads, any team can win on any given night in the ECAC. Therefore, I have to assume random outcomes.
Regardless of who wins game 1, it is 50% likely that the same team will win game 2, assuming random outcomes. Either the home team wins, or the visiting team wins.
Your theory is based upon ECAC history. My theory is based upon a team's probability of winning a game, without any regard to what has happened in the past, or what any current record or skill level of a team is. I am assuming that if the home team swept the past 100 played series, that the probability of the outcome of the 101st series is exactly the same as the 1st. It's just like if you're shooting craps, and you roll snake eyes (1-1) 50 times in a row. The probability of you rolling a 7 on the next roll is STILL 1 in 6.
Now, since we're talking about sports, and one team generally has more skill than another, is it truly random? Probably not. But, as we've said MANY times on this forum in various threads, any team can win on any given night in the ECAC. Therefore, I have to assume random outcomes.
FlagDUDE08
Banned
Re: ECAC Pick the Playoffs 2010
For the heck of it, let's try LF's theory of 8:5 probability that a series sweeps. The long run will consist of 26 series. The home team sweeps 8 times, the visiting team sweeps 8 times, the home team wins in three 5 times, and the visiting team wins in three 5 times.
Pick Home in 2: 3(5) + 1(5) + 4(8) + 0(8) = 52
Pick Away in 2: 1(5) + 3(5) + 0(8) + 4(8) = 52
Pick Home in 3: 4(5) + 2(5) + 3(8) + 1(8) = 62
Pick Away in 3: 2(5) + 4(5) + 1(8) + 3(8) = 62
It's still better to pick the 3 game series.
In fact, given vicb's scoring system, it is NEVER profittable in the long run to pick the sweep.
3+1+4x+0x < 4+2+3x+1x
For the heck of it, let's try LF's theory of 8:5 probability that a series sweeps. The long run will consist of 26 series. The home team sweeps 8 times, the visiting team sweeps 8 times, the home team wins in three 5 times, and the visiting team wins in three 5 times.
Pick Home in 2: 3(5) + 1(5) + 4(8) + 0(8) = 52
Pick Away in 2: 1(5) + 3(5) + 0(8) + 4(8) = 52
Pick Home in 3: 4(5) + 2(5) + 3(8) + 1(8) = 62
Pick Away in 3: 2(5) + 4(5) + 1(8) + 3(8) = 62
It's still better to pick the 3 game series.
In fact, given vicb's scoring system, it is NEVER profittable in the long run to pick the sweep.
3+1+4x+0x < 4+2+3x+1x
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Re: ECAC Pick the Playoffs 2010
You're misusing the word "random." If I buy a lottery ticket, I'll either win or I won't. Does that mean that I have a 50% chance of winning? No, of course not. The fact that there are two possible outcomes doesn't mean that the probability of each occurring is 50%. My analysis (that you're better off picking sweeps) is based on 62% chance that there will be a sweep in a given series - I'm still allowing for randomness.
You and I have the same basic model - it's just that you're using 50% as the probability that the game 1 winner wins game 2, whereas the historical data suggests that probability is actually closer to 62%.
Even if one team has significantly more skill than another, their winning can still be treated as a random event - a 90% chance of winning is still only a chance. So yes, by all means assume random events - but why ignore the information we do have about what the underlying probabilities of each outcome might be?
FlagDUDE08
Banned
Re: ECAC Pick the Playoffs 2010
OK, but even if it's closer to 62%, or even 90%, it's still better to pick 3-game series, based upon vicb's scoring system. Let's take the extra probabilities into account. Use x for the number of times a team will win via a sweep compared to 3 games, which we'll signify as 1. (With 62%, x = 100/62, or 1.613)
If you pick a sweep: 4 points if your team sweeps, 0 points if you get swept, f is your team wins in 3, 1 if your team loses in 3. Let's call this 4x + 0x + 3 + 1.
If you pick a 3-game series: 3 points if you team sweeps, 1 point if you get swept, 4 points if your team wins in 3, 2 points if your team loses in 3. Let's call this 3x + 1x + 4 + 2.
4x + 4 can NEVER be greater than 4x + 6 for all real values of x.
I shall humbly admit that my earlier claims of randomness were misused. However, based upon the above calculations, I still believe it is better to choose a 3-game series over a sweep in the long run. But as I said earlier, this is the short run, so feel free to bet to your heart's content, because anything can happen.
OK, but even if it's closer to 62%, or even 90%, it's still better to pick 3-game series, based upon vicb's scoring system. Let's take the extra probabilities into account. Use x for the number of times a team will win via a sweep compared to 3 games, which we'll signify as 1. (With 62%, x = 100/62, or 1.613)
If you pick a sweep: 4 points if your team sweeps, 0 points if you get swept, f is your team wins in 3, 1 if your team loses in 3. Let's call this 4x + 0x + 3 + 1.
If you pick a 3-game series: 3 points if you team sweeps, 1 point if you get swept, 4 points if your team wins in 3, 2 points if your team loses in 3. Let's call this 3x + 1x + 4 + 2.
4x + 4 can NEVER be greater than 4x + 6 for all real values of x.
I shall humbly admit that my earlier claims of randomness were misused. However, based upon the above calculations, I still believe it is better to choose a 3-game series over a sweep in the long run. But as I said earlier, this is the short run, so feel free to bet to your heart's content, because anything can happen.
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Re: ECAC Pick the Playoffs 2010
Again, you're using equal probabilities for things that are not equally likely. The visiting team is never going to sweep 8 series out of 26. The right thought process is, given who I think is going to win the series, am I THEN better off picking them to sweep or go 3 games.
Re: ECAC Pick the Playoffs 2010
When you're doing this "expected value" math, each of the 4 outcomes must have its probability of occurring accounted for, so your 3rd and 4th terms should be multiplied by (1-x), not by 1.
Picking sweeps then gives an expected value of 4x+0x+3(1-x)+1(1-x) = 4
Picking splits is 3x + 1x + 4(1-x) + 2(1-x) = 6 - 2x
So you're only better off picking splits if 6 - 2x > 4, or if x < 1.
Now, in this formulation, we're "double counting" because the sum of the 4 probabilities must be 1, so we really have to divide by 2, so you're only better off picking splits if x (the probability of a sweep) is less than 50%. This makes complete sense - you're better of to hedge on whichever is more likely to occur, splits or sweeps. If there were more splits than sweeps, you're better off picking splits for everything, but since there are more sweeps than splits, it's the other way around, and you're better off picking sweeps.
Okay, last post, I promise...
When you're doing this "expected value" math, each of the 4 outcomes must have its probability of occurring accounted for, so your 3rd and 4th terms should be multiplied by (1-x), not by 1.
Picking sweeps then gives an expected value of 4x+0x+3(1-x)+1(1-x) = 4
Picking splits is 3x + 1x + 4(1-x) + 2(1-x) = 6 - 2x
So you're only better off picking splits if 6 - 2x > 4, or if x < 1.
Now, in this formulation, we're "double counting" because the sum of the 4 probabilities must be 1, so we really have to divide by 2, so you're only better off picking splits if x (the probability of a sweep) is less than 50%. This makes complete sense - you're better of to hedge on whichever is more likely to occur, splits or sweeps. If there were more splits than sweeps, you're better off picking splits for everything, but since there are more sweeps than splits, it's the other way around, and you're better off picking sweeps.
FlagDUDE08
Banned
Re: ECAC Pick the Playoffs 2010
And this is why I was a computer engineering major.
I should stick to the statistics of craps.
And this is why I was a computer engineering major.
I should stick to the statistics of craps.
allnightwong
Quinnipiac '09 Grad School '11
Re: ECAC Pick the Playoffs 2010
Good God fellas. Give it a rest!
Good God fellas. Give it a rest!
ZYanksRule
GO BOBCATS!!
Re: ECAC Pick the Playoffs 2010
Huh?
Good thing I was a journalism major. You guys can take care of the math.
Huh?
Good thing I was a journalism major. You guys can take care of the math.