Ask The Wizard #237

Yes, that’s correct, but it’s not as suspicious as it seems. As per Grasping Probability: How Chance Influences Daily Life by H. C. Tijms, the identical numbers were selected on June 21, 1995, and December 20, 1986, during bi-weekly drawings. The drawing held on December 20, 1986, was the 3,016th in this series. With a 6/49 lottery, the total combinations are given by combin(49,6) = 13,983,816. The probability that the numbers drawn in the second event don’t align with the first is calculated as (c-1)/c, where c equals the total combinations, or 13,983,816. The chance of the third event yielding a unique set of numbers is (c-2)/c. Therefore, the likelihood that every subsequent drawing from the second up to the 3,016th produces distinct numbers is (c-1)/c × (c-2)/c × ... × (c-3015)/c = 0.722413. This suggests that the probability of encountering at least one set of matching numbers is 1 - 0.722413 = 0.277587, or 27.8%. The table below presents the likelihood of at least one pair of matching numbers according to the number of years, assuming two drawings occur each week.

If you're curious, the number of draws required for the chance of a matching selection to surpass 50% is exactly 4,404.

It’s rare for me to admit this, but honestly, I’m stumped. As you pointed out in another message, they appear similar to serial numbers found on US bills, consisting of two letters flanking a ten-digit number. To respect copyright, I won't disclose the specific numbers here.

For clarification for other readers, in certain deuces wild games, the odds can favor holding a single pair instead of two. This is especially true in full pay deuces wild (100.76%) and other popular variations where a full house awards 3 times the bet. Calculating the exact odds can be labor-intensive. However, it's straightforward to see that on reels 1, 2, 4, and 5, each of the nine pay-lines intersects every position three times. In contrast, reel 3 only intersects the top and bottom positions twice each, but crosses the middle position five times. Opting to hold a pair that includes the middle column can decrease your volatility. If the middle column happens to be the lone singleton in 20% of cases, I recommend holding a pair that consists of columns 1 and 5 or 2 and 4, if possible. If that isn’t an option, then consider holding a pair from columns 1 and 2 or 4 and 5, if you can. Otherwise, the order of the pairs won’t significantly impact your strategy.

The key question to consider with a proposition like this is the probability that any specific pick results in a win, loss, or push. Referring back to my prior section on betting the NFL , we’ve found that 2.8% of games end up right on the line, so let’s simplify it and consider it to be 3%. We’ll denote p as the win probability, given that the bet has been resolved. For someone picking at random, p would naturally be 50%. However, simply selecting underdogs can easily enhance that probability. As I previously illustrated, over a span of 25 seasons betting only on underdogs led to a win rate of 51.5%. By further refining selections to focus on softer lines relative to the overall market, I believe reaching a 52% win rate isn’t overly challenging. Therefore, I'm inclined to believe these players could achieve at least that threshold.

Assuming that 52% of the resolved bets end up winning, we can derive the overall probabilities:

Win: 50.44%
Draw: 3.00%
Loss: 46.56%

Using fundamental statistics, it becomes clear that the expected win for each selection, when laying -110, is -0.0078. The standard deviation for each selection is calculated as 1.0333. Over the span of 70 selections, the expected win totals to -0.5432, while the standard deviation becomes 70 × 1.0333 = 8.6452. A profit of 8.5 units stands 9.0432 units above the anticipated outcome, translating to 9.0432/8.6452 = 1.0460 standard deviations to the right of the expected mean on a Gaussian distribution curve. I believe we can disregard adjustments for a discrete distribution due to the presence of pushes and variations in line adjustments, which would result in a relatively smooth curve with an adjustment factor of 0.05 units.^1/2Thus, the likelihood that any single player finishes more than 1.046 standard deviations above forecasted results is 14.77%. This figure can be easily referenced from any Gaussian curve table or calculated with the formula =1-normsdist(1.046) in Excel. The chance of all six participants finishing below 1.046 stands at (1-0.1477) = 38.31%. Consequently, the probability that at least one player exceeds this benchmark is 61.69%. Therefore, betting on the over seems to be a promising option with a -110 spread. I calculate its fair value around -161.

The subsequent table illustrates the probability of the over 8.5 bet winning based on varying values of p. It’s possible that the individual who set this prop was estimating a value closer to 51% for p.⁶Correct mathematical strategies and insights for various casino games, including blackjack, craps, roulette, and many others you may wish to explore.

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