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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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ArizonaDream's avatar - Lottery-009.jpg

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Posted: February 26, 2016, 9:00 pm - IP Logged

What are the odds of matching 2 numbers, any position, any order playing a pick 3 game?


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    Posted: February 26, 2016, 9:04 pm - IP Logged

    What are the odds of matching 2 numbers, any position, any order playing a pick 3 game?

    I have 7 combo from which always 2 digit matches so maybe 1:7

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      Lincoln, California
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      Posted: February 26, 2016, 9:14 pm - IP Logged

      With 10 choices for 2 numbers wouldn't Your Odds Be 10 Squared or 1:100?

        ArizonaDream's avatar - Lottery-009.jpg

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        Posted: February 26, 2016, 9:20 pm - IP Logged

        With 10 choices for 2 numbers wouldn't Your Odds Be 10 Squared or 1:100?

        That would be the odds for a pick 2 game ,  but I think adding the 3rd number makes it easier. 

         

        My last prob/stat course was too long ago to remmber how to figure it out. It was my least favorite math course anyway.

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          Burlington, VT
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          Posted: February 26, 2016, 11:28 pm - IP Logged

          What are the odds of matching 2 numbers, any position, any order playing a pick 3 game?

          What state offers this game?

          Nevertheless, let's assume a 50% house edge. The probability of winning this type of bet is 8/1,000. That means for a $1 bet, the state will award you 1,000/8/2=$62.50.

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            Madison, WI
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            Posted: February 27, 2016, 12:16 am - IP Logged

            What state offers this game?

            Nevertheless, let's assume a 50% house edge. The probability of winning this type of bet is 8/1,000. That means for a $1 bet, the state will award you 1,000/8/2=$62.50.

            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.

              ArizonaDream's avatar - Lottery-009.jpg

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              Posted: February 27, 2016, 1:15 am - IP Logged

              Not a real game. I'm just interested in knowing if something I'm watching is happening more often than one would expect based on probability.

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                Burlington, VT
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                Posted: February 27, 2016, 1:18 am - 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 messed up but not in that way. What I did was run a query on my table of 1,000 pick 3 numbers and found 8 that have a 5 and a 7 (my two chosen numbers) and have 3 different digits. The last requirement is not necessary (e.g., 5-5-7 should be a winner) and the revised query produces 10, which is 10/1,000 = 1 in 100 = 1%.

                n1n2n3
                057
                157
                257
                357
                457
                557
                567
                577
                578
                579
                  ArizonaDream's avatar - Lottery-009.jpg

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                  Posted: February 27, 2016, 1:53 am - IP Logged

                  I messed up but not in that way. What I did was run a query on my table of 1,000 pick 3 numbers and found 8 that have a 5 and a 7 (my two chosen numbers) and have 3 different digits. The last requirement is not necessary (e.g., 5-5-7 should be a winner) and the revised query produces 10, which is 10/1,000 = 1 in 100 = 1%.

                  n1n2n3
                  057
                  157
                  257
                  357
                  457
                  557
                  567
                  577
                  578
                  579

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

                    jimjwright's avatar - Yellow 3.png
                    Park City, UT
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                    Posted: February 27, 2016, 5:54 am - IP Logged

                    Any pair made of 2 distinct digits equates to 54 straight combinations in Pick 3.  It would be made up of 8 6-way combinations and 2 3-way combinations.

                    So I would say 1000/54 equates to 1 in 18.5

                    Jimmy

                      ArizonaDream's avatar - Lottery-009.jpg

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                      Posted: February 27, 2016, 6:44 am - IP Logged

                      Any pair made of 2 distinct digits equates to 54 straight combinations in Pick 3.  It would be made up of 8 6-way combinations and 2 3-way combinations.

                      So I would say 1000/54 equates to 1 in 18.5

                      Jimmy

                      The pairs need not be distinct digits. Even if they were distinct, I don't see how you'd have 54.  There are 100 pairs, 00 through 99, 10 of which are doubles.

                        Tialuvslotto's avatar - Jailin
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                        Posted: February 27, 2016, 8:49 am - IP Logged

                        There are 100 pairs in the P3 game: 90 pairs with 2 distinct digits and 10 same-digit pairs.

                        The 90 pairs with 2 different digits can be summarized as 45 "box" or "any-order" pairs. 

                        Each winning combination has 3 pairs (if it is unmatched, 2 if it is a Double).

                        So you have 3 chances to match one of your 45 any-order pairs for unmatched (.72) or 1 chance for doubles (.27) or 2.43 chances overall.

                        45/2.43 = 18.5 like the J-man said.

                        "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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                          Madison, WI
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                          Posted: February 27, 2016, 9:32 am - IP Logged

                          There are 100 pairs in the P3 game: 90 pairs with 2 distinct digits and 10 same-digit pairs.

                          The 90 pairs with 2 different digits can be summarized as 45 "box" or "any-order" pairs. 

                          Each winning combination has 3 pairs (if it is unmatched, 2 if it is a Double).

                          So you have 3 chances to match one of your 45 any-order pairs for unmatched (.72) or 1 chance for doubles (.27) or 2.43 chances overall.

                          45/2.43 = 18.5 like the J-man said.

                          Although I think it would make more sense to look at odds seperately for distinct and pairs, as when you make your prediction you have do decide which you are choosing.

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                            Posted: February 27, 2016, 10:43 am - IP Logged

                            Nm pays $5 for a pair.

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                              Posted: February 27, 2016, 11:52 am - IP Logged

                              What are the odds of matching 2 numbers, any position, any order playing a pick 3 game?

                              14.4% is the best E(s=2), you can have for 2 digits correct for a pick 3 drawing.