Eurojakpot (and Euromillions) probabilities
P.S.-5 (6.12.2022) This article was originally published at www.newfinns.com back in 2015 (here is the original article) and since then some of the rules of Eurojackpot have changed and now they mathematically (5 out of 50 and 2 out of 12)equal those of Euromillions.
I got a feedback from a member of the New Finnish and European generation that he wants to see the conditional probabilities of sequence of the Eurojackpot regarding the top prize and here I give you the solution:
For the main prize in Eurojackpot first we select 5 main numbers out of 50 and then 2 additional numbers out of 10. Thus, the 5 main numbers can be selected in 2 118 760 different ways and the two additional numbers in 45 ways. Then for the jackpot (5 and 2) the number of all possible combinations is 95 344 200 (= 2 118 760 * 45). Hence speaking about the probability of the top prize we have 1/95 344 200 = 0.00001 per thousand (if you play with one combination☺). On the other hand, for example, the probability that the selected 5 prime numbers could contain a pair (ex. 7,8 (or another one out of the 49☺), distribution of type 2+1+1+1) is equal to 30.81% and etc. for the other alternatives.
Probably you have also noticed that compared to the Finnish lotto here the probability that within the 5 prime numbers there will be NO consecutive ones is as high as 64,70% (vs 27,77%), which said in simple language is due to the fact that we select fewer numbers (5) out of a larger initial set (50).
Since in the coming week the top prize in Eurojackpot is close to 90 000 000 € one could ask how the above calculations could help, to which I would answer — by giving you a better understanding of what could happen, especially in a long run. In general in a long run you are supposed to be a more successful player if you play Eurojackpot with unpaired numbers (ex. 2,8,11,25,43) for the main draw. For the additional 10 numbers draw you have (n-1)=9 possible pairs ex. 1:2,2:3…9:10 (here that’s the only parity possibility), hence a probability of (9/45)*100=20% to occur. (6.12.2022) So with 12 additional numbers the probability for a pair is (11/66)*100=16,67%
P.S. Because some of my comments were spammed I asked my administrator to activate a registration system, which shall be on place in the next few days. I hope that that way I will be able to better follow and hopefully answer your comments.
P.S.2 Since recently I’ve updated the probability calculations for the Finnish Lotto, which rules are changing (for those who want to see it it’s here (In Finnish:))), I’ve decided to add to this article the general probability distribution table for the Eurojackpot as well. The results of the latter are of course the same as these in Wikipedia and other sites , but I also want to show how they can be calculated.
So given the current rules where we chose 7 numbers in 2 separate draws, 5 out of 50 and 2 out of 10, we could present a table with all possible winning (column A, rows 1:12) and not winning classes (column A; rows 13:18) like the one above.
In addition I presented the classes of all possible combinations and their corresponding probability as decimal number (column G) and 1/probability ratio (column H). The latter might be an easier way to grasp your winning chance. For instance your chance to get 5+1 correct is 0,0000001678(raw 2, column G) which is small, but 1 out of 5 959 012,5 (which is the same) could give you a better comparison of how small. Reciprocally you could calculate the probability per 1, by dividing all possible combinations 95 344 200 (column F) by the “combination spread” per class (column I), as I call it. So again for 5+1 we will have 95 344 200/16=5 959 012,5. Said in other words, out of the all 95 344 200 possible combinations we have exactly 16 of the class 5+1, exactly 9 900 of the class 3+2 and so on, and the corresponding chance to win can be calculated by dividing all possible combinations by those possible per class, which gives you winning chance per 1.
For those of you who might want to make the calculations themselves I have presented the pieces of the equation in the heading of the above Excell or any other spreadsheet programme that has combinations, but let’s put them together (Com=combination):
(Com(n,r)*Com(N-n;n-r))*(Com(a;o)*Com(A-a;a-o))/((Combination(N;n)*Combination(A;a))=Probability(in decimals) where “n” is the chosen numbers of the prime draw (5;) “r” is the number of the correct numbers in your coupon (anything from 5 to 0); “N” is the numbers of the first prime draw (50); “a” is the chosen numbers of the second draw (2); “o” is the correctly guessed numbers out of it (2,1 or 0) and “A” is the numbers in the additional 10 numbers draw huh… So substituted with numbers for let’s say 5+1 it would be:
(Com(5;5)*Com(45;0))*Com(2;1)*Com(8;1))/(Com(50;5)*Com(10;2))=0,0000001678. Or for exactly 3 (3 from the prime draw and 0 from the additional draw) correctly guessed numbers and using any Calculator which has combinations, normally marked by nCr, we have: (5C3*45C2)*(2C0*8C2)/(50C5*10C2)=0,002907. If you want to read more for instance here and here are given well explained similar examples of the mathematical principles applied.
So once again good luck, knowledge or both! (Helsinki 4 Oct 2016)
Kind regards, Rostislav Georgiev Dinkov
P.S.3 (13.3.2022) Read from an Eurojackpot cupon — starting 19.3.2022 there will be a second draw per week as the number of the second probability draw grows to 12 (from 10) which makes 66 additional combinations and the number of all combinations grows to 2 118 760*66 = 139 838 160
P.S.4 (15.11.2022) Here just the new general probability table of Eurojackpot (and Euromillions), which takes into account to two additional numbers in the second draw (look P.S.3)
