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Boomer's math problem

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  • Boomer's math problem

    Here's my weigh-in on the proper formula for calculating your expectation on reverse
    middles:

    (((1-f)*((.5*ovP)+(.5*unP)))-(f*fcost))/i

    "f" is your chance of the total landing flat, expressed as a decimal.
    "fcost" is how much landing flat will cost you.
    "ovP" is your profit if the game goes over.
    "unP" is your profit if the game goes under.
    "i" is your total wager (both sides).

    To plug Boomer’s example’s numbers into this, and assuming an 8% hit rate for even
    numbered totals (12% would be better for odd numbered totals):
    (((1-.08)*((.5*60)+(.5*56)))-(.08*644))/1144 = .0016 ROI

    using a less conservative estimate of the hit rate, 6% (10% for odd numbers) = .01388

    It’s not a pretty picture. 1% ROI might be plausible, but that would mean getting 5 dimes down a side to average $100 bucks a bet, and assumes no slippage and a zero stiff rate.

    Of course, I’m just guessing what the hit rates are and with research you might come up
    with a less conservative number, and you might also sometimes find better line
    differences than in Boomer’s example.


  • #2
    pay2play- Question about your formula? Why when multiplying your profit if the game goes over or under do you use .5? I would think in the above example it would be about.46 as far as probability of game landing either way. In your example if I am reading your fromula correctly 50% of the time the game would go over, 50% would go under and 8% of the time it would land on the number. Well I don't understand how the probabilties can be over 100%.

    Comment


    • #3
      I would think also that the plays are primarily a low risk way to put large chunks of the bankroll in action. With bonuses of upwards of 20% you clearly will show a profit, but I still can't expose myself to that "bad number" falling.

      Comment


      • #4
        BBuster,
        The .5's are multiplied by .92 to get your .46
        .46 twice is .92, + the .08 chance of failure, =1
        Certainly you could write it as (((1-f)/2)*ovP)+(((1-f)/2)*unP), or a number of other ways.
        The important thing is that you have the chance of profit*the profit, less the chance of failure*the cost of failure.

        randyrohm, yes you could put a lot of money in play, and in fact would need to to make it worthwhile. I have no idea what bonuses are out there. I have no desire to sit in front of a screen and look for them. But if I was scalping/middling, if I was in front of that screen anyway, I think I'd keep an eye out for outstanding reverse middles, especially on even-numbered totals. And I do think the chance of failure is probably slightly less than 10% for odd numbered totals and 6% for even numbered ones. I used the conservative estimate because I haven't seen a large sample study. Certainly if I was going to put any real effort into such a play, you'd want first to do a very large study.
        I don't think its a gold mine. I think to make a go of it you'd be working pretty hard. But the Reno's and Boomers ARE working hard, so for them its another weapon in their arsenal.

        Comment


        • #5
          Here are some stats for you guys.

          First set of numbers are as follows:

          Column 1 is total score in game.
          Column 2 is # times this happened in 92 thru 99 seasons.
          Column 3 is pct.

          Total # games in sample is 17510.

          7 1815 10.36
          8 1310 7.48
          9 1710 9.76
          10 1161 6.63
          11 1424 8.13

          Comment


          • #6
            In the second set of numbers

            Column 1 is the actual total score in game
            Column 2 is the # games where this total is scored and the books have their total posted at the value in column 1 or the value in column 1 plus or minus half a run.
            Column 3 is the total number of games where the books have their total posted at the value in column 1 or the value in column 1 plus or minus half a run.
            Column 4 is the pct.

            So for example the first line reads as follows: There were 2730 games where the posted total was 6.5, 7, or 7.5. Of these games 313 came in at 7 runs. This happened 11.46 pct of the time.

            7 313 2730 11.46
            8 553 6854 8.06
            9 728 7407 9.82
            10 314 4918 6.38
            11 249 2456 10.13

            Note that the value for 9 is essentially the same in both tables. For 8 and 10 it is off a bit. For 7 and 11 it is off a lot. In games where the books think a lot of runs will be scored you find that this usually happens, similarly for low scorng games. They actually do know what they're doing most of the time .

            Comment


            • #7
              The reason that the value for 9 looks good in both tables is that 9 is the approximate average value for runs scored in professional baseball. Games where the posted total is greater than 9 tend to have a higher percentage of 9 run games. Games where the posted total is lower have a lower percentage. The percentages even out. The reason that the percentage for 7 is quite different in the 2 tables is because you virtually never see posted totals less than 6.5 and the # of 7 run games drops quickly as you go to higher and higher posted totals.
              Same for totals of 11 - except the logic is reversed.

              One thing that is obvious in both tables is that even totals occur much less frequently than odd totals. This of course is because of baseball's rule that the home team wins in the 9th or subsequent innings as soon as they take the lead. If you win by 1 run you can't have an even total.

              Comment


              • #8
                In the previous post I meant to say that the value for 9 was the same in both tables - not good.

                For people who will inevitably ask where I got the data from - it was purchased from Computer Sports World. I got most of it years ago when it was cheap. Now it's way overpriced. Their web address is www.cswstats.com

                The data was converted to dbf format and the analyis was done using a simple Visual Foxpro program.

                Cheers

                'mute

                Comment


                • #9
                  thanks mute - there's a real sample

                  looks like if you want to play with it, around 10 is the fun number

                  Comment


                  • #10
                    Just curious Aussie - what's the time difference between you and us? It must be awkward synchronizing your posts with those coming from Morth America - or are you independently wealthy from your sports gambling and can post anytime you want?

                    Comment


                    • #11
                      Morth was a typo - not some kind of pun or play on words. It's hard typing anything coherent in the little boxes that the BW folks give you.

                      'mute

                      Comment


                      • #12
                        Actually there's not much room for making money in reverse scalping baseball totals.

                        You scalpers would be better advised to spend your time doing the same thing but with baseball money lines and run lines. With them it's a matter of knowing the frequencies of 1-run games.

                        There are interesting relationships there - for example the frequency depends on the total.

                        But I'm not going to tell you everything.

                        As the X-files people say

                        "the truth is out there"

                        'mute

                        Comment


                        • #13
                          OK Boomer, since this is a slow baseball day I made your money management calculation. I could actually make some simplifying assumptions due to the ROI being so small. Here is the simple version:

                          Take 100 bets as you laid out in your original example, losing $644 8% of the time and winning $58 92% of the time. Now even though the average number of losses in 100 tries is 8, the 97% probable worst case is to lose 13. This means that only one time in about 30 would the result be worse than this.

                          So losing 13 and winning 87 leaves you with a net loss of $3326. Now I'm assuming that this hypothetical "bankroll" is not all your wagering bankroll, but rather only a portion set aside for this particular wagering opportunity. In this case you could wager three times your example with a bankroll of $10,000.

                          A loss would cost you $1932, and a win would get you $174.

                          Comment


                          • #14
                            Not far off from my guess that you could risk 20% of your "bankroll" on such a wager.

                            Comment


                            • #15
                              Thanks Sportshobby. I had started using 3 dimes for the last few bets. My average win rate has been 260 to 360 roughly, depending on the numbers. After about 6 wagers, I lost 3 dimes.

                              I've read everything everyone said and hardly understood any of the math or formulas that were presented, but I understand one thing: I will NOT use the numbers 7 and 9 in my strategy. Just a gut feeling for a mathematically challenged dude.

                              I'm actively working on some other strategies which will hopefully advance me from a 2nd class gambling citizen to a 1st class gambling citizen (gambling and winning).

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