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Odds? 2 of 3, pick 3

Topic closed. 26 replies. Last post 11 months ago by Stack47.

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Burlington, VT
United States
Member #173303
February 25, 2016
17 Posts
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Posted: February 27, 2016, 12:14 pm - IP Logged

There are matching numbers missing from your table. Any with the 7 before 5 are missing, and so are combos like 574, 573 etc.

You're right, I messed up again! I don't know why I put criteria in my query that doesn't belong. Eek

It should be 54:

countn1n2n3
1057
2075
3157
4175
5257
6275
7357
8375
9457
10475
11507
12517
13527
14537
15547
16557
17567
18570
19571
20572
21573
22574
23575
24576
25577
26578
27579
28587
29597
30657
31675
32705
33715
34725
35735
36745
37750
38751
39752
40753
41754
42755
43756
44757
45758
46759
47765
48775
49785
50795
51857
52875
53957
54975
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    Madison, WI
    United States
    Member #172977
    February 11, 2016
    515 Posts
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    Posted: February 27, 2016, 12:25 pm - IP Logged

    I don't think you're seeing this one right, Tucker. The odds should be much better. I believe it looks like this:

    (10X10)/(3!).

     

    1 in 16.7

     

    Pretty sure anyway.

    These kinds of things are fun to think about. I now see what I have wrong here. This is the odds if you simply picked 2 distinct digits that they would both show up somewhere in the three picked. When you add in that you get to pick an additional digit and you are just looking for any of the two to hit tge odds should get better right?

    Pretty sure only still.

    Serge, can you explain your math on yours?

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      Burlington, VT
      United States
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      February 25, 2016
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      Posted: February 27, 2016, 12:39 pm - IP Logged

      When I posted 10 results, I was getting ahead of myself. I was thinking of how you would actually play this given that a bet like that is not one of the games. But you can build a bet just like that by buying box tickets. So how many box tickets would you need to buy in order to cover all possible winning numbers that include the two you are betting on (5 and 7 in my example)? It is 10. In other words, 10 tickets purchased as a box will win if any one of the 54 are drawn (5.4%, or 1 in 18.5). If you bet 50 cents each (so $5 in total), if you win, you'll win either $42 or $83 depending on whether the 3 winning digits are distinct ($42 win) or are there just two (e.g., 5-7-5 is $83 win). 48 of the winning combos are $42 wins and 6 are $83 wins (i.e., 4.2% chance of winning $42 and 0.6% chance of winning $83).

      Interesting! Thanks!

        ArizonaDream's avatar - Lottery-009.jpg

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        Posted: February 27, 2016, 5:59 pm - IP Logged

        I'm on board with those who said 54 in 1000 now.  I had to map it out in a way I could see the matching pairs easily and then cross out the dups.  6 sets or 10, and then subtract the 6 duplicates.  1000/54 or 1 in 18.52 or 5.4%

        057 075
        157 175
        257 275
        357 375
        457 475
        557 575
        657 675
        757 775
        857 875
        957 975

        507 705
        517 715
        527 725
        537 735
        547 745
        557* 755
        567 765
        577 775*
        587 785
        597 795

        570 750
        571 751
        572 752
        573 753
        574 754
        575* 755*
        576 756
        577* 757*
        578 758
        579 759

          ArizonaDream's avatar - Lottery-009.jpg

          United States
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          Posted: February 27, 2016, 6:04 pm - IP Logged

          When I posted 10 results, I was getting ahead of myself. I was thinking of how you would actually play this given that a bet like that is not one of the games. But you can build a bet just like that by buying box tickets. So how many box tickets would you need to buy in order to cover all possible winning numbers that include the two you are betting on (5 and 7 in my example)? It is 10. In other words, 10 tickets purchased as a box will win if any one of the 54 are drawn (5.4%, or 1 in 18.5). If you bet 50 cents each (so $5 in total), if you win, you'll win either $42 or $83 depending on whether the 3 winning digits are distinct ($42 win) or are there just two (e.g., 5-7-5 is $83 win). 48 of the winning combos are $42 wins and 6 are $83 wins (i.e., 4.2% chance of winning $42 and 0.6% chance of winning $83).

          Interesting! Thanks!

          Illinois also has a front pair and back pair option, it would only take 4 lines to play a pair front and back and either order.  But you loose when the pair splits. I was thinking I had something worthwhile, but I'm not beating the odds on picking a pair by as much as I thought. Back to the drawing board.    Wait a minute, I need to look at this again, may be something there after all. 

          Thanks all!

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            Kentucky
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            Posted: February 27, 2016, 9:15 pm - IP Logged

            I'm on board with those who said 54 in 1000 now.  I had to map it out in a way I could see the matching pairs easily and then cross out the dups.  6 sets or 10, and then subtract the 6 duplicates.  1000/54 or 1 in 18.52 or 5.4%

            057 075
            157 175
            257 275
            357 375
            457 475
            557 575
            657 675
            757 775
            857 875
            957 975

            507 705
            517 715
            527 725
            537 735
            547 745
            557* 755
            567 765
            577 775*
            587 785
            597 795

            570 750
            571 751
            572 752
            573 753
            574 754
            575* 755*
            576 756
            577* 757*
            578 758
            579 759

            Though my method of calculation the number of combos the 5-7 pair creates is different than yours, the results (54) are the same. I multiplied the other 8 digits times the 6 possible ways each creates and added the two 3 ways using the same digits creates. But what about the odds of another pair: 1-1?

            I came up with nine 3 ways plus one triple for 28. The odds are obviously different on which "2 out of 3 digits" are drawn.

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              Lincoln, California
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              Posted: February 27, 2016, 10:39 pm - IP Logged

              I don't think you're seeing this one right, Tucker. The odds should be much better. I believe it looks like this:

              (10X10)/(3!).

               

              1 in 16.7

               

              Pretty sure anyway.

              I think you are right.  The Odds I posted were for each of the 3 possibilities in a Pick 3 game.  Dividing that by 3 looks right; because, there are 3 chances of hitting.  Not sure what the point is (here in California you get squat for 2 out of 3 in Pick 3 or 4.

                Thoth's avatar - binary
                Findlay, Ohio
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                Posted: February 28, 2016, 6:31 am - IP Logged

                Odds are 1 in 18.518 or 54 favorable against 946 unfavorable (reduces down to 1 : 17.518 as a ratio ) - however you want to look at it.

                There are 45 "boxed no-match pairs" and 10 "double digit" pairs possible...

                Suppose you play 12X boxed where X is unknown,  there are 8 other digits besides 1 and 2 that could be drawn to give you a no-match combo. In this case we will say X = digit 3. All no matches can be drawn 6 different  ways: 123, 132, 213, 231, 312, 321. Since X can be replaced by 8 other digits besides the 1 or 2, there are 8 digits x 6 ways = 48 combinations.

                Of course X could be replaced with a 1 or a 2 to create a double digit combination. Each double can be drawn 3 different ways so 12X could be 122, 212, 221 OR 112, 121. 211. The 6 doubles combine with the 48 no-match to total the 54 possible straights that would have to be drawn to match your pair of 12X in any order or position.

                ~Probability=Odds in Motion~

                  jimjwright's avatar - Yellow 3.png
                  Park City, UT
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                  Posted: February 28, 2016, 1:29 pm - IP Logged

                  Odds are 1 in 18.518 or 54 favorable against 946 unfavorable (reduces down to 1 : 17.518 as a ratio ) - however you want to look at it.

                  There are 45 "boxed no-match pairs" and 10 "double digit" pairs possible...

                  Suppose you play 12X boxed where X is unknown,  there are 8 other digits besides 1 and 2 that could be drawn to give you a no-match combo. In this case we will say X = digit 3. All no matches can be drawn 6 different  ways: 123, 132, 213, 231, 312, 321. Since X can be replaced by 8 other digits besides the 1 or 2, there are 8 digits x 6 ways = 48 combinations.

                  Of course X could be replaced with a 1 or a 2 to create a double digit combination. Each double can be drawn 3 different ways so 12X could be 122, 212, 221 OR 112, 121. 211. The 6 doubles combine with the 48 no-match to total the 54 possible straights that would have to be drawn to match your pair of 12X in any order or position.

                  It's great to see Thoth post again. Smile

                  Jimmy

                    CARBOB's avatar - FL LOTTERY_LOGO.png
                    ORLANDO, FLORIDA
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                    Posted: March 4, 2016, 4:20 am - IP Logged

                    Odds are 1 in 18.518 or 54 favorable against 946 unfavorable (reduces down to 1 : 17.518 as a ratio ) - however you want to look at it.

                    There are 45 "boxed no-match pairs" and 10 "double digit" pairs possible...

                    Suppose you play 12X boxed where X is unknown,  there are 8 other digits besides 1 and 2 that could be drawn to give you a no-match combo. In this case we will say X = digit 3. All no matches can be drawn 6 different  ways: 123, 132, 213, 231, 312, 321. Since X can be replaced by 8 other digits besides the 1 or 2, there are 8 digits x 6 ways = 48 combinations.

                    Of course X could be replaced with a 1 or a 2 to create a double digit combination. Each double can be drawn 3 different ways so 12X could be 122, 212, 221 OR 112, 121. 211. The 6 doubles combine with the 48 no-match to total the 54 possible straights that would have to be drawn to match your pair of 12X in any order or position.

                    Welcome back! Where have you been? Hope you stick around.

                      Tialuvslotto's avatar - Jailin
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                      Posted: March 4, 2016, 7:38 am - IP Logged

                      Welcome back! Where have you been? Hope you stick around.

                      I Agree!

                      I'm a card-carrying member of the Thoth fan club!

                      "There is no such thing as luck; only adequate or inadequate preparation to cope with a statistical universe."

                      ~Robert A. Heinlein

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                        Kentucky
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                        Posted: March 5, 2016, 3:37 pm - IP Logged

                        I Agree!

                        I'm a card-carrying member of the Thoth fan club!

                        Me too!