Ask The Wizard #350

There does appear to be a reason to strive for overtime in hockey, as you've mentioned. Let's delve into some statistics to clarify your question. The data being referenced comes from four hockey seasons, starting with the 2017/2018 season.

The table below provides a detailed overview of the 7,846 hockey games analyzed over these four seasons, categorizing them by regular season and playoff matches and indicating how many extended into overtime. It reveals that 11.27% of regular season games went to overtime, in contrast to only 9.03% of playoff games.

The critical question to consider is whether the disparity of 11.27% against 9.03% is significant statistically or if it might simply be due to natural variance. To assess the means of the two samples, I will utilize a chi-squared test similar to the one available at MedCalc.org. Comparison of proportions calculator Across the total of 7,846 games, there were 871 instances of overtime, translating to a probability of 11.10% for such occurrences. This indicates an 88.90% chance that a game would not go into overtime. Assuming there is no statistical disparity between regular and playoff games, it can be expected that approximately 804.6 regular season games and 66.4 playoff games would go to overtime.

The table provided compares the observed outcomes against these expectations, under the premise that the actual probability of overtime remains consistent for regular and playoff games. The column on the right displays the chi-squared statistic, which is derived from squaring the difference between actual and expected values and then dividing by the expected total.

According to the chi-squared statistic displayed in the table above, we find a value of 2.813628. With a single degree of freedom, the likelihood of observing results this skewed, or even more extreme, stands at 9.347%. In simpler terms, if there were no behavioral differences between regular and playoff games, and the probability of overtime was genuinely equal, we would see such a 2.24% variance in games extending into overtime, or higher, with a 9.347% chance. This suggests a statistically notable difference in overtime rates between the two game types, although there remains a 9.35% possibility that it could be attributed to normal random variance.

It's worth mentioning that the MedCalc calculator and other analytical resources implement an 'N-1' adjustment to the chi-squared statistic. Specifically, this involves multiplying the chi-squared value by (N-1)/N, where N represents the total observations. In this context, the adjusted chi-squared figure would amount to 2.813628 * (7845/7846) = 2.813270. This yields a p-value of 9.349% for this chi-squared statistic with one degree of freedom. Though it may seem like an unnecessary complication, I felt it important to address it to avoid leaving my readers with questions about its absence.

From my perspective, it appears that teams are more inclined to aim for overtime in the regular season compared to the playoffs, and the data does lend some support to this view, albeit not conclusively.

External Links

• Use of the Chi-Square Statistic at the Johns Hopkins Bloomberg School of Public Health.

Estimating Verlander's performance poses a challenge, so let's tackle his statistics last.

During the 2019/2020 season, Steph Curry achieved an impressive free throw success rate of 93.10% (source: Basketball Reference ).

The average success rate for an NFL 40-yard field goal sits at 85.83%. Nevertheless, I would contend that Justin Tucker performs better than the average. For field goals placed between 30 and 39 yards, the NFL average is 89.32%, but Tucker boasts an impressive 96.63%. By applying Tucker's percentage in relation to the NFL average, I calculate his probability of converting a 40-yard field goal attempt as 85.85% × (96.63%/89.32%) = 92.86%.

Baseball scenarios introduce complexity. We must clarify whether we're analyzing actual game pitches or controlled practice sessions. This distinction is crucial because, in real games, pitchers do not aim to throw strikes each time; instead, they often target the edges of the strike zone to make clean hits more difficult for batters.

While I lack concrete statistics to support this, I have observed pitchers in minor league bullpens consistently hitting their catcher's mitt directly without requiring movement time after time. I would estimate that someone like Verlander could achieve a strike with 95% accuracy in a controlled environment, but in actual game situations, his strike rate is only 68.50%.

To determine the likelihood of achieving 100 consecutive successful attempts without considering fatigue, you only need to raise the probability of one successful attempt to the power of 100.

In summary, if the context is a controlled experiment, I would give the edge to Verlander; however, under actual game conditions, I would side with Curry.

This inquiry was initially posed at Barstool Sports . There is a wealth of discussion surrounding this topic, as well as on my forum at Wizard of Vegas .

Below are a few formulas that readers may find beneficial:

Continue scrolling for my response and explanation.

When I refer to 'efficient', I mean which arrangement minimizes empty space between the cannonballs.

To simplify things, let's consider the volume of either pyramid by assuming the centers of the cannonballs are positioned at the corners of the pyramid. Denote n as the number of cannonballs along one side of the base of any given pyramid.

First, let's examine the pyramid structured with a square base.

This pyramid contains a total of cannonballs equal to 1.² + 2² + 3² + 4² + 5² + 6²+ .. + n²= n*(n+1)*(2n+1)/6.

Now, let's calculate the height of this square pyramid where the base has a side length of n. As illustrated, the sides, aside from the square base, are equilateral triangles. Therefore, the slant height is also n. The distance from one base corner to the diagonally opposite corner measures n*sqrt(2). Consequently, the distance from a base corner to the midpoint of the base is n*sqrt(2)/2. We can call the height h. Consider the right triangle formed by the height, the base-to-center distance, and the slant height.

h²+ (n*sqrt(2)/2)²= n²
h = n*sqrt(2)/2.

Recall that the volume of a pyramid is given by base times height divided by three. Therefore, the volume of this pyramid is:

n²* n* sqrt(2)/2 * (1/3) = n³*sqrt(2)/6.

Thus, the ratio of cannonballs compared to volume will be [n*(n+1)*(2n+1)/6] / [n³*sqrt(2)/6] = sqrt(2)*n*(n+1)*(2n+1)/(2*n³) = sqrt(2)*(n+1)*(2n+1)/(2*n²)

Next, let's delve into the pyramid that has a triangular base.

The total amount of cannonballs within this pyramid is 1 + 3 + 6 + 10 + 15 + ... + n*(n+1)/2 = n*(n+1)*(n+2)/6.

Following that, let’s compute the base area. Remember, in a 30-60-90 triangle, the sides follow the proportions of 1/2, sqrt(3)/2, and 1. From this, it’s straightforward to calculate the height of an equilateral triangle with side length n, which is n*sqrt(3)/2. Hence, the area of the base becomes n', "The distance from a corner of the base to the center measures sqrt(3)/3. Using this alongside the pyramid's slant height of 1, we can apply the Pythagorean theorem to determine the height of the pyramid, which is sqrt(6)/3.²*sqrt(3)/4.

Now, we can determine the volume of the pyramid by using base times height divided by three = (n', "The ratio of cannonballs relative to volume is consequently [n*(n+1)*(n+2)/6] / [n', "Here’s a comparison chart of the ratios of balls to volume:

As n grows larger, the ratio of cannonballs to the volume will approach n for both pyramid configurations. Therefore, with substantial quantities of cannonballs, their efficiency becomes comparable.²*sqrt(3)/4) * (n*sqrt(6)/3) * (1/3) = n³*sqrt(18)/36 = n³*sqrt(2)/12.

This inquiry is also explored and discussed in my forum at³*sqrt(2)/12] = sqrt(2)*n*(n+1)*(2n+1)/(2*n³) = sqrt(2)*(n+1)*(n+2)/n²

Detailed strategies and mathematically accurate information regarding casino games such as blackjack, craps, roulette, and countless others.

• Square base: sqrt(2)*(n+1)*(2n+1)/(2*n²)
• Triangle base: sqrt(2)*(n+1)*(n+2)/n²

Let's divide both ratios by sqrt(2)*(n+1)/n²:

• Square base: (2n+1)/2 = n + 0.5
• Triangle base: n+2

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Given the volume of a cannonball, the efficiency in both pyramids, defined as the ratio of cannonball volume to total volume, approaches pi*sqrt(2)/6 = apx. 74.05%.

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