Tuesday, 12 May 2015 09:50

## Conditional probabilities of parity – Finnish lotto and Eurojackpot (reposted)

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Hello guys! I am pleased with your comments about my site and articles just have no physical recourse to answer to all of you, so in brief. Yes the idea of newfinns.com is mine and all the articles on it are written by me Rosti, but I am only a small person striving for better place under the Sun, for as many of us, as possible. So yes you are welcomed and allowed to share and quote my articles. That way the power of our shared thoughts grows and slowly but surely moves ahead, closing the gap between some of the realities around us and our better tomorrow.

Those of you, who have watched through the site have probably noticed that the idea of a New Finnish and European generation was also my candidacy for the Finnish Parliamentary Elections, but instead of New Finns a vast majority of the local voters chose “True” Finns .....

Well someone has said that “life is what happens with us, while we are making plans about it”.... So through my articles I try to continue my renewal aspiration and since the mobile version of the site focuses on my comments, here I will repost the “conditional probabilities of parity” of the Finnish lotto and the Eurojackpot, as the given algorithm can be developed for any limited distribution of type “N” out of which are chosen “n” numbers without repetition.

In Finnish lotto for the prime reward you select 7 main numbers out of 39. Did you know that if we speak about the top prize, the probability that within the selected 7 prime numbers there will be at least 2 consecutive numbers is as high as 72.23%? Or for example, that the probability that within the 7 selected numbers will be, let’s say, exactly one pair is equal precisely to 43.21%. Well here I have developed for you a complete table-calculation, in which I have presented all possible probability results, regarding the sample distribution of the 7 selected numbers:

 N Distribution g (=n. elem.) r (=n. elem. perm.) Combin. ((N-n)+1)! g! (33-g)! Probab.% 1 1+1+1+1+1+1+1 7 1 4272048 8.683E+36 5040 4.03E+26 27.77 2 2+1+1+1+1+1 6 6 6645408 8.683E+36 720 1.09E+28 43.21 3 2+2+1+1+1 5 10 2373360 8.683E+36 120 3.05E+29 15.43 4 3+1+1+1+1 5 5 1186680 8.683E+36 120 3.05E+29 7.72 5 2+2+2+1 4 4 163680 8.683E+36 24 8.84E+30 1.06 6 3+2+1+1 4 12 491040 8.683E+36 24 8.84E+30 3.19 7 4+1+1+1 4 4 163680 8.683E+36 24 8.84E+30 1.06 8 3+2+2 3 3 16368 8.683E+36 6 2.65E+32 0.11 9 3+3+1 3 3 16368 8.683E+36 6 2.65E+32 0.11 10 5+1+1 3 3 16368 8.683E+36 6 2.65E+32 0.11 11 4+2+1 3 6 32736 8.683E+36 6 2.65E+32 0.21 12 5+2 2 2 1056 8.683E+36 2 8.22E+33 0.01 13 6+1 2 2 1056 8.683E+36 2 8.22E+33 0.01 14 3+4 2 2 1056 8.683E+36 2 8.22E+33 0.01 15 7 1 1 33 8.683E+36 1 2.63E+35 0.00 Sum 15380937 Sum 100.00

Thus for example, the distribution of type 5+2 is one which has five consecutive numbers (ex. 23,24,25,26,27) and one pair (ex. 36,37) and out of the all 15 380 937 possible combinations there are exactly 1 056 pieces (0,01%) 5+2 ones. Then if you want to see the number of the possible combinations in which there is NO consecutive numbers, the type 1+1+1+1+1+1+1, (ex. 2,7,9,14,19,23,33), there are exactly 4 272 048 such combinations (27,77%) ! Or...there are exactly 33 combinations (0,0002%) where there are 7 subsequent numbers , the first is 1,2,3,4,5,6,7 and the last is 33,34,35,36,37,38,39, the rest 31 are between them...this way you could check if my calculations work, but If you are interested in a more gentle way of how I calculated the probabilities send me a message. The solution is not directly in the textbooks !

### Eurojackpot

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:

 N Distribution g (=n. elem.) r (=n. elem. perm.) Combin. ((N-n)+1)! g! (46-g)! Probab% 1 1+1+1+1+1 5 1 1370754 5.5E+57 120 3.34525E+49 64.70 2 2+1+1+1 4 4 652740 5.5E+57 24 1.40501E+51 30.81 3 2+2+1 3 3 45540 5.5E+57 6 6.04153E+52 2.15 4 3+1+1 3 3 45540 5.5E+57 6 6.04153E+52 2.15 5 3+2 2 2 2070 5.5E+57 2 2.65827E+54 0.10 6 4+1 2 2 2070 5.5E+57 2 2.65827E+54 0.10 7 5 1 1 46 5.5E+57 1 1.19622E+56 0.00 Sum 2118760 Sum 100.00

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.

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.

 A B C D E F G H I N Winclass Com(n;r) Com (N-n;n-r) Com(a;o) Com (A-a;a-o) Com(N;n)*Com(A;a) Probab. 1/Probab Com spread 1 5+2 1 1 1 1 95344200 1,04883E-08 95 344 200,0 1 2 5+1 1 1 2 8 95344200 1,67813E-07 5 959 012,5 16 3 5 1 1 1 28 95344200 2,93673E-07 3 405 150,0 28 4 4+2 5 45 1 1 95344200 2,35987E-06 423 752,0 225 5 4+1 5 45 2 8 95344200 3,77579E-05 26 484,5 3 600 6 4 5 45 1 28 95344200 6,60764E-05 15 134,0 6 300 7 3+2 10 990 1 1 95344200 0,000103834 9 630,7 9 900 8 2+2 10 14190 1 1 95344200 0,001488292 671,9 141 900 9 3+1 10 990 2 8 95344200 0,001661349 601,9 158 400 10 3 10 990 1 28 95344200 0,002907361 344,0 277 200 11 1+2 5 148995 1 1 95344200 0,007813532 128,0 744 975 12 2+1 10 14190 2 8 95344200 0,02381267 42,0 2 270 400 13 2 10 14190 1 28 95344200 0,041672173 24,0 3 973 200 14 1+1 5 148995 2 8 95344200 0,125016519 8,0 11 919 600 15 1 5 148995 1 28 95344200 0,218778908 4,6 20 859 300 16 0+2 1 1221759 1 1 95344200 0,012814193 78,0 1 221 759 17 0+1 1 1221759 2 8 95344200 0,205027091 4,9 19 548 144 18 0 1 1221759 1 28 95344200 0,35879741 2,8 34 209 252 1 95 344 200

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