Conclusions About The SBR Best Bet Thread

[VideoReview]

Now, I realize there are numerous problems with the general population sample of the SBR Best Bet thread which include things like:

1) Juice is inconsistent among participants and rules on juice had varied from the start of the thread
2) A bias towards participants who start off poorly to drop out
3) Recording net profit/loss as amount won per unit bet instead of amount to win unit

However, now that it has been running for a significant time and the sample size has grown, I am wondering if we can start drawing some conclusions about the aggregate results yet?

As of the end of betting for Wednesday, May 21st, 2008, the total posted results for the entire population were:

Wins 1359
Losses 1367
Ties 38
Total bets (Not Including Ties) 2726
Net Profit -29.09
ROI Betting 1 Unit -1.06713%

Since the profit is negative at this point, I would like to explore the negative side of the results (a discussion on positive results might be appropriate if they ever occur). Notwithstanding the problems I mentioned above (and others I missed), assuming the trend of -29.09/2726= -1.06713% ROI continues, how large does the sample size have to be to say with 95% certainty that the group, as a whole, has:

1) No statistical advantage when juice is a factor
2) No statistical advantage if the lines were fair odds (i.e. no juice)

Anyone have any ideas on how to calculate this?

Off topic from this thread: FWIW although I have not participated in a tournament or best bet thread myself, I really like the idea. I think the posters who have volunteered to keep track and grade bets are doing a great service at SBR. I wonder, because of the fairly widespread popularity, if SBR at some point would not want to consider automating the entire process (time stamping, grading, consistent juice, etc.) to relieve the admins of these popular threads of any burden or at least assist them to make the job less onerous. Just a thought.

[Ganchrow]

We'd have to know the odds at which the bets were placed.

[VideoReview]

Yeah, I kind of figured that may have to be done. Since I don't really feel like going through all of the posts to get that information but still would like an idea, how about if we assume that all of the bets were at -107 (a compromise between some -110 books and the top -104 books).

What would the calculation and result look like?

Without doing the calculations myself it does appear that the "group" may be able to reject the null hypothesis of a fair bet at some point based on the above assumptions at some point.

[Ganchrow]

The average line based on the data (and assuming 1 unit risked per bet) is roughly -101.58, so perhaps you might just want to use that value in lieu of -107? Also you'd need to actually specify the juice in question #1. Remember that for the problem to have a solution the juice specified must be strictly less than 1.06713%.

Anyway with the assumptions specified we have what amounts to two straightforward hypothesis tests.

How would you proceed, VR?

[LT Profits]

VR,

You can't assume -107 when a majority of the bets are on Moneylines. I don't think I have submitted more than three NBA plays in the Best Bet thread since baseball season started.

[BuddyBear]

The main conclusion is that forums are better for finding fades than finding winners....

[Ganchrow]

You can't assume -107 when a majority of the bets are on Moneylines. I don't think I have submitted more than three NBA plays in the Best Bet thread since baseball season started.

I think VR is just looking for a first-order approximation.

Were we to break the results down by the actual odds received what we'd find would be that in general the more the lines deviated from the average value the less significant the observed aggregate results would be.

[Ganchrow]

The main conclusion is that forums are better for finding fades than finding winners....

I think the conclusion is that generally speaking these contests are, when taken in a vacuum, too small to tell us much of anything.

[VideoReview]

The average line based on the data (and assuming 1 unit risked per bet) is roughly -101.58, so perhaps you might just want to use that value in lieu of -107? Also you'd need to actually specify the juice in question #1. Remember that for the problem to have a solution the juice specified must be strictly less than 1.06713%.

Anyway with the assumptions specified we have what amounts to two straightforward hypothesis tests.

How would you proceed, VR?

I'll think about it and get back to you tomorrow.

[VideoReview]

I think VR is just looking for a first-order approximation.

Were we to break the results down by the actual odds received what we'd find would be that in general the more the lines deviated from the average value the less significant the observed aggregate results would be.

That is exactly what I was thinking. Assuming that the bets were all even money (+vig) would give the best possible chance for the results to be positive. The true answer would have to be worse than that figure since all of the odds are obviously not even money bets. Depending on the result, I could get an idea if it was worth pursuing all of the data in detail.

[louisvillekid]

i'm obviously not a math person, but for all the wins/losses/ties that VR posted above, wouldn't you have to take into consideration how many people were on the same team each day and then add them up. like if 6 people were on the Celtics one day, wouldn't that be more like -660.

[Ganchrow]

i'm obviously not a math person, but for all the wins/losses/ties that VR posted above, wouldn't you have to take into consideration how many people were on the same team each day

Yes, most definitely. (Although treating 6 picks on the same team as a -600 wouldn't be correct either -- you'd simply treat as 6 units on the same bet, this would imply standard deviation sqrt(6) = 2.45× as great as on a single unit bet).

Although not specified, the operating simplifying assumption would be zero correlation between picks.

[Ganchrow]

Depending on the result, I could get an idea if it was worth pursuing all of the data in detail.

Doing so would only likely to be useful as an instructive exercise in hypothesis testing.

Do it if you want to better understand the process -- don't waste your time if you're looking for significant results.

[VideoReview]

Doing so would only likely to be useful as an instructive exercise in hypothesis testing.

Do it if you want to better understand the process -- don't waste your time if you're looking for significant results.

I am only interested in it as a learning excerise and it is not a search for an angle. My "original" idea was to see just how close these types of best bet threads, when the total population is considered, mirrored the total market. My hunch at the beginning (before I did the totals) was that it would probably not be able to reject the fair bet hypothesis given a large enough sample size but because all "votes" are being considered equally, would never reject the profitable bet hypothesis. In otherwords, although I expect that there are a few people in the best bet thread that are advantage players, the proportion of people who are not is probably about the same as the whole market.

[Ganchrow]

We'll assume all bets were made at a line of -100*1359/(1367-29.09) ≈ and that there is no vig.

Approximation using the central limit theorem:

H₀: Contestants as a group are break-even
H_a: Contestants as a group are net losers

Let σ²_k = variance of the k^th bet
Let σ²_N = variance over N bets Σ^N_i=1 [ σ²_i ]
Let E(X_k) = expected unit return of the k^th bet
Let E(X_N) = expected unit return over N bets = Σ^N_i=1 [ E(X_i) ]

Because by assumption all bets are at the same odds () at and at the same vig (0%), σ²_N ≈ N * ( - 1)

Similarly, if we assume a constant realized pick rate of 1359/(1367+1359) ≈ 49.853%,
E(X_N) ≈ ( 49.853% * 1.98448 - 1) * N

So our text statistic over N bets (remember we're assuming no vig) would then be:

Z_N = E(X_N) / ( σ²_N ) ≈ ( 49.853% * 1.98448 - 1) * N (sqrt(N * 0.98448))

Our 1-tailed rejection region of the null hypothesis at α = 5% would be:

Z_N = ( 49.853% * 1.98448 - 1) * sqrt(N) sqrt(0.98448) < -1.644854

Solving for N yields:
N ≈ 23,390

This implies that we'd need to see a total of 23,390 picks (winning at rate of 49.853%) before we could conclude with 95% statistical confidence that the contestants were net losers versus lines at zero juice and at odds of about -101.58.

After only 2,726 trials, however, we'd be unable to reject the null hypothesis that the contestants as a group are at least break-even at zero vig.

Calculating N under the assumption of 0.5% juice (H₀ remaiing that contestants as a group are break-even) is left as an exercise for the reader.

[VideoReview]

This implies that we'd need to see a total of 23,390 picks (winning at rate of 49.853%) before we could conclude with 95% statistical confidence that the contestants were net losers versus lines at zero juice and at odds of about -101.58.

After only 2,726 trials, however, we'd be unable to reject the null hypothesis that the contestants as a group are at least break-even at zero vig.

Thanks for the detailed explanation. Considering how liberal the initial assumptions were and the fact that 23,390 picks at the current win rate would be needed, until we get to a much larger sample size, I am comfortable assuming that the group as a whole mirrors the entire market in terms of its advantage:recreation player ratio or possibly is better by only a very negligible amount. Any benefit would seem to be limited to following individual advantage players and not using any kind of consensus approach.