Ask the Wizard #237
I heard that the same numbers were chosen in the German 6/49 lottery on different days. Something seems fishy with that. What are the odds?
Lotte from Hamburg
That is true, but it is not as fishy as you think. According to Understanding Probability: Chance Rules in Everyday Life by H. C. Tijms, the same set of numbers were drawn on June 21, 1995, and December 20, 1986, in bi-weekly drawings. The drawing of December 20, 1986 was the 3,016th drawing. The number of combinations in a 6/49 lottery is combin(49,6) = 13,983,816. The probability the numbers in the second drawing will not match those of the first is (c-1)/c, where c is the number of combinations, or 13,983,816. The probability the third drawing will produce a unique set of numbers is (c-2)/c. So, the probability that every drawing from the 2nd to the 3,016th will produce unique numbers is (c-1)/c × (c-2)/c × ... (c-3015)/c = 0.722413. So, the probability of at least one set of common numbers is 1- 0.722413 = 0.277587, or 27.8%. The following table shows the probability of at least one pair of matching numbers drawn by the number of years, assuming two draws per week.
Probability of MatchingNumbers in 6/49 Lottery
In case you were wondering, the number of draws for the probability of a matching draw to first exceed 50% is 4,404.
I have a puzzle that I’ve been trying to solve for a few months, with absolutely no progress. Time permitting, I’m hoping you can indulge me, as it’s been keeping me up at night :-). Anyway, in the glossary of Beyond Counting -- Exhibit CAA, three sequences of numbers and letters are given as the glossary entry for "Magic Numbers." One of these numbers even graces the book’s cover, so I assume they’re of some importance. Do you have any thoughts?
It isn’t often I say this, but I have no idea. As you noted in another e-mail, they take the format of serial number on US currency, two letters, with a ten-digit number in between. Out of respect for copyright, I won’t indicate what the numbers are here.
Say you are playing bonus deuces wild (or any other game where the correct strategy is to hold and draw to only one pair out of two dealt pairs), when playing this game on Spin Poker with 9 patterns, for equal value pairs, does the position of each pair make any difference in terms of which pair should be held and if so, which are the best and worst positions to hold?
Joe from Denver
For the benefit of other readers, sometimes in deuces wild games the odds favor holding a single pair over a two pair. This is true in full pay deuces wild (100.76%) and any common version of bonus deuces where a full house pays 3. An exact calculation of this would be very tedious and time consuming. However, it is easy to see that on reels 1, 2, 4, and 5, the nine pay-lines run through each position three times. Yet on reel 3, the top and bottom positions are crossed only two times each and the middle position 5 times. It will lower your volatility to hold a pair that includes the middle column. In the 20% of cases where the middle column is the singleton, I would hold a pair if it consists of columns 1 and 5, or 2 and 4, if you can. If that is not possible, then hold a pair in columns 1 and 2, or 4 and 5, if you can. Otherwise, it doesn’t make any difference which pair you hold.
I’m one of six people in an NFL handicapping contest. Each of us has to pick 70 games over the season against the line at a major Internet sports book, which has a 20-cent line. One of the other participants offered an over/under prop on the highest score at the end of the season. His line was a profit of 8.5 units. Let’s assume the season has not started yet, and that the participants are experienced sports bettors. What do you think of that line, and how would you analyze it?
Rob from Las Vegas, NV
The big question to ask yourself with a prop like this is what is the probability that a given pick will end in a win, loss, or push. From my section on betting the NFL, we can see that 2.8% of games fall right on the line. Let’s just say 3%, to keep it simple. Let’s call p the probability of a win, given that the bet was resolved. For a purely random picker, p would obviously be 50%. It is easy to improve upon that by only picking underdogs. As my previously mentioned page shows, over 25 seasons flat betting underdogs would have resulted in a win rate of 51.5%. It is also easy to improve upon that a bit more by cherry picking the softest lines against the market in general. Between those two, I think it isn’t hard to get to 52%. So I’m going to take it on faith that these guys can at least get as far as 52%.
So, assuming 52% of resolved bets win, the overall probabilities are:
Using basic statistics, it is easy to see that the expected win per pick, laying -110, is -0.0078. The standard deviation per pick is 1.0333. The expected win over 70 picks is -0.5432, and the standard deviation is 701/2×1.0333 = 8.6452. A win of 8.5 units is 9.0432 units above expectations, or 9.0432/8.6452=1.0460 standard deviations to the right of expectations on the Gaussian Curve. I think we can ignore the adjustment for a discrete distribution because of the pushes, and some games not being -110/-110, will result in a fairly smooth curve down the a factor of 0.05 units.
So, the probability of any one player finishing more than 1.046 standard deviations above expectations is 14.77%. That figure can be found in any table of the Gaussian curve, or with the formula =1-normsdist(1.046) in Excel. The probability of all six players finishing under 1.046 is (1-0.1477)6=38.31%. Thus, the probability of at least one player finishing above 1.046 standard deviations up is 61.69%. That makes the over look like a solid bet laying -110. I show it is fair at -161.
The following table shows the probability of the over 8.5 winning given various values of p. Perhaps the person setting the prop was assuming a value closer to 51% for p.
NFL Handicapping Prop
|Prob. Correct Pick||Prob. Over Wins|