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Are the following odds correct?

Topic closed. 12 replies. Last post 5 years ago by RJOh.

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Spain
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August 4, 2011
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Posted: September 28, 2011, 5:45 am - IP Logged

Hello,

If playing a 5if5 100% in 16 numbers wheel, CM  in the detailed report says that there is a 10.4% of all numbers landing in one line. My question is, are really the odds 10 in 100 or 1 in 10 that all 6 numbers will land in one line? If this is the case, this would seem a good wheel for a group play.

 

Also, is there a chart that states the odds of getting all 6 numbers in  a 14,16 etc number wheel?

 

Thanks

Spy

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    Chief Bottle Washer
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    Posted: September 28, 2011, 9:36 am - IP Logged

    <Moved to Mathematics forum>

    Please post in the appropriate forum ... thank you.

      RJOh's avatar - chipmunk
      mid-Ohio
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      Posted: September 28, 2011, 2:07 pm - IP Logged

      Hello,

      If playing a 5if5 100% in 16 numbers wheel, CM  in the detailed report says that there is a 10.4% of all numbers landing in one line. My question is, are really the odds 10 in 100 or 1 in 10 that all 6 numbers will land in one line? If this is the case, this would seem a good wheel for a group play.

       

      Also, is there a chart that states the odds of getting all 6 numbers in  a 14,16 etc number wheel?

       

      Thanks

      Spy

      If you were playing a 5/16 game then everything you said would be true but chances are you're playing a 5/39 game so you have to figure all 39 numbers when you calculate odds.  If you have all 5 winning numbers in your 16 number pool then it makes no difference.

      A 5/39 game have 575,757 possible combinations and with a 5/16 wheel you're covering 4,368 of those combinations and hoping one or more of them will have 3 or more of the winning numbers.  Also you odds of have all 5 winning numbers in one line are the number of combinations in your wheel divided by 575,757.

       * you don't need to buy more tickets, just buy a winning ticket * 
         
                   Evil Looking       

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        Spain
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        Posted: September 28, 2011, 3:02 pm - IP Logged

        Hello, thank you for your reply.Please let me know if the following train of thought is wrong. Lets say I am playing a 6/49 game, I play a 5if5in6 16 numbers 840 lines. At this point, lets say I hit all 6 numbers in my wheel of 16 and I already have a 5# hit in one line. What will be the odds of that 6th number landing in that winning line? I don't think it would  be 840/ 13,983,816, as I already have a 5# hit. This means I need the 6th number to land in one of the 840 lines, so that would be 1in 840? 

         

        Even I  do the 840/13983816 calculation I get a 6% - which is 4% lower than what CM states.

        Thank again

        Spy

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          Kentucky
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          Posted: September 28, 2011, 9:24 pm - IP Logged

          Hello, thank you for your reply.Please let me know if the following train of thought is wrong. Lets say I am playing a 6/49 game, I play a 5if5in6 16 numbers 840 lines. At this point, lets say I hit all 6 numbers in my wheel of 16 and I already have a 5# hit in one line. What will be the odds of that 6th number landing in that winning line? I don't think it would  be 840/ 13,983,816, as I already have a 5# hit. This means I need the 6th number to land in one of the 840 lines, so that would be 1in 840? 

           

          Even I  do the 840/13983816 calculation I get a 6% - which is 4% lower than what CM states.

          Thank again

          Spy

          The odds against matching 5 numbers in 6/49 games is 54.201 to 1. Since you have 840 chances, your odds of your wheel matching 5 numbers is 64.5 to 1.

          "Even I  do the 840/13983816 calculation I get a 6% - which is 4% lower than what CM states."

          It's 0.00006 or 0.006%, which happens to be the percentage of all they possible combos you bought.

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            Spain
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            Posted: September 29, 2011, 8:43 am - IP Logged

            Thank you for your reply. "Since you have 840 chances, your odds of your wheel matching 5 numbers is 64.5 to 1". How can this be? I am playing a 5 if 5 wheel which guarantees a 5 winning line 100% if I hit 5 numbers in my chosen of 16. So I am essentially playing simply against the odds of getting the 5 numbers in my chosen of 16. Isn't this the case?

             

            Also, is there a chart anywhere that lists the odds of getting all 6 numbers in a wheel of 16,18,20,24 numbers in a 6/49 game?

             

            Thanks

            Spy

              RJOh's avatar - chipmunk
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              Posted: September 30, 2011, 1:03 am - IP Logged

              Hello, thank you for your reply.Please let me know if the following train of thought is wrong. Lets say I am playing a 6/49 game, I play a 5if5in6 16 numbers 840 lines. At this point, lets say I hit all 6 numbers in my wheel of 16 and I already have a 5# hit in one line. What will be the odds of that 6th number landing in that winning line? I don't think it would  be 840/ 13,983,816, as I already have a 5# hit. This means I need the 6th number to land in one of the 840 lines, so that would be 1in 840? 

               

              Even I  do the 840/13983816 calculation I get a 6% - which is 4% lower than what CM states.

              Thank again

              Spy

              There are 8008 possible combinations of 6 in a pool of 16 numbers and you have 840 of them, do the math.

              840÷8008=.104895105 or ~10%

               * you don't need to buy more tickets, just buy a winning ticket * 
                 
                           Evil Looking       

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                Spain
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                Posted: September 30, 2011, 7:42 am - IP Logged

                Hi, thanks for replying. Now I get it.

                 

                What about a formula for simply getting all six numbers in a chosen wheel? What are the odds lets say of trapping all 6 numbers in a wheel of 20? (not in a single winning line but simply trapping them in the wheel?)

                Thank you for your time

                Spy

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                  Spain
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                  Posted: September 30, 2011, 8:00 am - IP Logged

                  I think I found a website that does the calculation automatically. Still, if you know the formula for getting the odds  for trapping all 6 numbers in an x amount of numbers in a wheel please let me know.

                   

                  Spy

                    RJOh's avatar - chipmunk
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                    Posted: September 30, 2011, 10:04 am - IP Logged

                    I think I found a website that does the calculation automatically. Still, if you know the formula for getting the odds  for trapping all 6 numbers in an x amount of numbers in a wheel please let me know.

                     

                    Spy

                    You simply figure the number of combinations in the pool size you're using for your wheel and divide that number by the number of combinations in a pool of 49 numbers.

                    6/18 = 18564
                    6/20 = 38760
                    6/24 = 134596
                    6/30 = 593775
                    6/40 = 3838380 and etc.

                    6/49 = 13983816

                     * you don't need to buy more tickets, just buy a winning ticket * 
                       
                                 Evil Looking       

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                      Spain
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                      Posted: September 30, 2011, 5:04 pm - IP Logged

                      Thanks RJoh

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                        illinois
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                        Posted: October 21, 2011, 12:01 am - IP Logged

                        Its nice but i dont get it

                          RJOh's avatar - chipmunk
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                          Posted: October 21, 2011, 12:20 am - IP Logged

                          Its nice but i dont get it

                          You only need to get it if you want to know it, other wise it doesn't matter.

                           * you don't need to buy more tickets, just buy a winning ticket * 
                             
                                       Evil Looking