Probability winning a lottery with supplementary numbers.

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For a particular lottery a player picks a selection of 6 numbers from the numbers 1 to 45. To determine the winners 8 numbers are chosen at random. The first 6 are designated as the winning numbers and the other 2 as supplementary numbers. Prizes are determined as follows;



Div 1: 6 winning numbers



Div 2: 5 winning numbers + 1 suplementary



Div 3: 5 winning numbers



Div 4: 4 winning numbers



Div 1 is simple enough just the number of combinations and I understand that Div 3 is actually 5 winning numbers and 0 supplementary. Div 4 is also 4 winning numbers plus any amount of supplementary. But I'm unsure the methodology to calculate Div 2.



Any explanation would be greatly appreciated.




probability




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    For a particular lottery a player picks a selection of 6 numbers from the numbers 1 to 45. To determine the winners 8 numbers are chosen at random. The first 6 are designated as the winning numbers and the other 2 as supplementary numbers. Prizes are determined as follows;



    Div 1: 6 winning numbers



    Div 2: 5 winning numbers + 1 suplementary



    Div 3: 5 winning numbers



    Div 4: 4 winning numbers



    Div 1 is simple enough just the number of combinations and I understand that Div 3 is actually 5 winning numbers and 0 supplementary. Div 4 is also 4 winning numbers plus any amount of supplementary. But I'm unsure the methodology to calculate Div 2.



    Any explanation would be greatly appreciated.




    probability




    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      For a particular lottery a player picks a selection of 6 numbers from the numbers 1 to 45. To determine the winners 8 numbers are chosen at random. The first 6 are designated as the winning numbers and the other 2 as supplementary numbers. Prizes are determined as follows;



      Div 1: 6 winning numbers



      Div 2: 5 winning numbers + 1 suplementary



      Div 3: 5 winning numbers



      Div 4: 4 winning numbers



      Div 1 is simple enough just the number of combinations and I understand that Div 3 is actually 5 winning numbers and 0 supplementary. Div 4 is also 4 winning numbers plus any amount of supplementary. But I'm unsure the methodology to calculate Div 2.



      Any explanation would be greatly appreciated.




      probability




      share|cite|improve this question













      For a particular lottery a player picks a selection of 6 numbers from the numbers 1 to 45. To determine the winners 8 numbers are chosen at random. The first 6 are designated as the winning numbers and the other 2 as supplementary numbers. Prizes are determined as follows;



      Div 1: 6 winning numbers



      Div 2: 5 winning numbers + 1 suplementary



      Div 3: 5 winning numbers



      Div 4: 4 winning numbers



      Div 1 is simple enough just the number of combinations and I understand that Div 3 is actually 5 winning numbers and 0 supplementary. Div 4 is also 4 winning numbers plus any amount of supplementary. But I'm unsure the methodology to calculate Div 2.



      Any explanation would be greatly appreciated.





      probability





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited yesterday









      David G. Stork

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      Bodmas12

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          There are $6$ ways to choose the five of your numbers that will be winning and $39$ ways to choose a winning number that you didn't pick. Then the supplementary number you picked is determined and there are $38$ ways to choose the supplementary number you didn't pick. Overall the chance is then $$frac6 cdot 39 cdot 3845 choose 639 choose 2=frac 1678755$$






          share|cite|improve this answer























          • this assumes sequential picking? I would think they picked the 8 out of 45 in $binom458$ ways?
            – sku
            yesterday










          • @sku: It was defined that there were six picked as winning and two as supplementary numbers, so we have to pick one batch and then the other from what is left.
            – Ross Millikan
            yesterday










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          There are $6$ ways to choose the five of your numbers that will be winning and $39$ ways to choose a winning number that you didn't pick. Then the supplementary number you picked is determined and there are $38$ ways to choose the supplementary number you didn't pick. Overall the chance is then $$frac6 cdot 39 cdot 3845 choose 639 choose 2=frac 1678755$$






          share|cite|improve this answer























          • this assumes sequential picking? I would think they picked the 8 out of 45 in $binom458$ ways?
            – sku
            yesterday










          • @sku: It was defined that there were six picked as winning and two as supplementary numbers, so we have to pick one batch and then the other from what is left.
            – Ross Millikan
            yesterday














          up vote
          0
          down vote













          There are $6$ ways to choose the five of your numbers that will be winning and $39$ ways to choose a winning number that you didn't pick. Then the supplementary number you picked is determined and there are $38$ ways to choose the supplementary number you didn't pick. Overall the chance is then $$frac6 cdot 39 cdot 3845 choose 639 choose 2=frac 1678755$$






          share|cite|improve this answer























          • this assumes sequential picking? I would think they picked the 8 out of 45 in $binom458$ ways?
            – sku
            yesterday










          • @sku: It was defined that there were six picked as winning and two as supplementary numbers, so we have to pick one batch and then the other from what is left.
            – Ross Millikan
            yesterday












          up vote
          0
          down vote










          up vote
          0
          down vote









          There are $6$ ways to choose the five of your numbers that will be winning and $39$ ways to choose a winning number that you didn't pick. Then the supplementary number you picked is determined and there are $38$ ways to choose the supplementary number you didn't pick. Overall the chance is then $$frac6 cdot 39 cdot 3845 choose 639 choose 2=frac 1678755$$






          share|cite|improve this answer















          There are $6$ ways to choose the five of your numbers that will be winning and $39$ ways to choose a winning number that you didn't pick. Then the supplementary number you picked is determined and there are $38$ ways to choose the supplementary number you didn't pick. Overall the chance is then $$frac6 cdot 39 cdot 3845 choose 639 choose 2=frac 1678755$$







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday


























          answered yesterday









          Ross Millikan

          275k21183348




          275k21183348











          • this assumes sequential picking? I would think they picked the 8 out of 45 in $binom458$ ways?
            – sku
            yesterday










          • @sku: It was defined that there were six picked as winning and two as supplementary numbers, so we have to pick one batch and then the other from what is left.
            – Ross Millikan
            yesterday
















          • this assumes sequential picking? I would think they picked the 8 out of 45 in $binom458$ ways?
            – sku
            yesterday










          • @sku: It was defined that there were six picked as winning and two as supplementary numbers, so we have to pick one batch and then the other from what is left.
            – Ross Millikan
            yesterday















          this assumes sequential picking? I would think they picked the 8 out of 45 in $binom458$ ways?
          – sku
          yesterday




          this assumes sequential picking? I would think they picked the 8 out of 45 in $binom458$ ways?
          – sku
          yesterday












          @sku: It was defined that there were six picked as winning and two as supplementary numbers, so we have to pick one batch and then the other from what is left.
          – Ross Millikan
          yesterday




          @sku: It was defined that there were six picked as winning and two as supplementary numbers, so we have to pick one batch and then the other from what is left.
          – Ross Millikan
          yesterday












           

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