1. #1
    Iwinyourmoney
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    Gambling MATH people, question for ya (GANCH)

    So I was just thinking about this in the shower. For this deal, throw the spreads out the window.

    1) A NFL weekend after byeweeks, there are 16 games a week. If I wanted to do a 16-team parlay, and have to pick every game (no O/U), ect.). Just one pick per game. How many different cards would I have to fill out to get EVERY SINGLE possible combination, which would give a guarentee win. Obviously I am not going to do it, its way to much to even try. Just wondering.

    2). If I wanted to play the Mega Millions Lottery, and wanted every combination to guarentee a win, how many would there be? (If your not fimailar, the lottery has balls 1-75. There is also a "Mega Ball" number 1-75. You need to match 5/5 of the regular lottery balls, the you have to match you mega ball, the mega ball drawing is in a seperate container then the regular lottery, so # repeat is possible).


    ................Good luck

  2. #2
    etothep
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    haha, you testing them?

  3. #3
    pokernut9999
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    seems like the mega millions is like 175,711,536 combos
    Last edited by pokernut9999; 12-06-07 at 02:08 PM.

  4. #4
    pico
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    Quote Originally Posted by Iwinyourmoney View Post
    So I was just thinking about this in the shower. For this deal, throw the spreads out the window.

    1) A NFL weekend after byeweeks, there are 16 games a week. If I wanted to do a 16-team parlay, and have to pick every game (no O/U), ect.). Just one pick per game. How many different cards would I have to fill out to get EVERY SINGLE possible combination, which would give a guarentee win. Obviously I am not going to do it, its way to much to even try. Just wondering.

    2). If I wanted to play the Mega Millions Lottery, and wanted every combination to guarentee a win, how many would there be? (If your not fimailar, the lottery has balls 1-75. There is also a "Mega Ball" number 1-75. You need to match 5/5 of the regular lottery balls, the you have to match you mega ball, the mega ball drawing is in a seperate container then the regular lottery, so # repeat is possible).


    ................Good luck
    you need 8 for 3 teamer. 16 or 4 teamer, 32 for 5 teamer, 64 for 6 teamer...just multiply 2 for every extra team. you have a claculator

  5. #5
    pico
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    Quote Originally Posted by pokernut9999 View Post
    seems like the mega millions is like 17+ million combos
    for mega million 56 pick 5 and 1 in 42 for mega ball, the odds is about 1/178million

  6. #6
    SBR Lou
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    Just you'll get the same result without the suspense.

  7. #7
    Louisvillekid1
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    Ha, Hope your not thinking this is a way to beat the system. . .

  8. #8
    pico
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    Quote Originally Posted by Louisvillekid1 View Post
    Ha, Hope your not thinking this is a way to beat the system. . .
    i do 8 bets for 3 teams parlays for my free plays. that is one way to beat the system.

  9. #9
    HedgeHog
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    I believe the answer to question one is 65,536 combinations. Each game offers you 2 choices and there are 16 games. So the answer is 2 to the 16th power or 65,536.

  10. #10
    pico
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    Quote Originally Posted by HedgeHog View Post
    I believe the answer to question one is 65,536 combinations. Each game offers you 2 choices and there are 16 games. So the answer is 2 to the 16th power or 65,536.
    i can gurantee you that the parlay odds for 16 games is less than +6,553,600

  11. #11
    pokernut9999
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    ?????/
    Last edited by pokernut9999; 12-06-07 at 02:34 PM.

  12. #12
    HedgeHog
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    To answer question #2, I need to correct your given info. There are actually 56 regular balls, of which 5 are chosen, and 46 mega balls (sounds dirty) of which one is chosen.

    The formula is [56!/ (5! x 51!) ] x 46 = 175,711,536 possibilities.

    P.S. 5! = 5x4x3x2x1

    I confirmed the answer at the MM site. Do the math for yourself...it works.
    Last edited by HedgeHog; 12-06-07 at 03:01 PM.

  13. #13
    pokernut9999
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    Quote Originally Posted by HedgeHog View Post
    To answer question #2, I need to correct your given info. There are actually 56 regular balls, of which 5 are chosen, and 46 mega balls (sounds dirty) of which one is chosen.

    The formula is [56!/ (5! x 51!) ] x 46 = 175,711,536 possibilities.

    P.S. 5! = 5x4x3x2x1

    I confirmed the answer at the MM site. Do the math for yourself...it works.
    wow , same answer i posted an hour ago.

  14. #14
    HedgeHog
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    Quote Originally Posted by pokernut9999 View Post
    wow , same answer i posted an hour ago.
    I know. I just wanted to show the math behind your answer. Then I went to the Mega Millions site to confirm your devine genius

  15. #15
    pokernut9999
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    Quote Originally Posted by HedgeHog View Post
    I know. I just wanted to show the math behind your answer. Then I went to the Mega Millions to confirm your devine genius

  16. #16
    Ganchrow
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    Quote Originally Posted by Iwinyourmoney View Post
    1) A NFL weekend after byeweeks, there are 16 games a week. If I wanted to do a 16-team parlay, and have to pick every game (no O/U), ect.). Just one pick per game. How many different cards would I have to fill out to get EVERY SINGLE possible combination, which would give a guarentee win. Obviously I am not going to do it, its way to much to even try. Just wondering.
    Just as Hedgehog and Picoman explained, you would need 216 = 65,536 different cards to cover all possible outcomes.

    Were you to receive true parlay odds, each parlay would pay out at US odds of about +3,113,234.

    By betting $1 on each of the 65,536 parlays (and assuming no pushes) you would lock in a loss of $65,535 - $31,132.34 = $34,402.66 (or $34,402.66 / $65,535 ≈ 52.49% of the initial investment).



    Quote Originally Posted by Iwinyourmoney View Post
    2). If I wanted to play the Mega Millions Lottery, and wanted every combination to guarentee a win, how many would there be? (If your not fimailar, the lottery has balls 1-75. There is also a "Mega Ball" number 1-75. You need to match 5/5 of the regular lottery balls, the you have to match you mega ball, the mega ball drawing is in a seperate container then the regular lottery, so # repeat is possible).
    Based on the game as you've descibed it (which is a bit different from the actual Mega MillionsWK), the probability of hitting the first 5 balls would be 1/combin(75,5) = 1/17,259,390 (where combin() refers to the Excel combinatorial function).

    The probability of hitting the first 5 balls and the Mega Ball would then be 1/17,259,390/75 = 1 / 1,294,454,250 ≈ 0.00000007725%.

    So to answer your actual question, if you wanted to guarantee a win you'd need to buy 1,294,454,250 tickets.

  17. #17
    HedgeHog
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    Ganch:

    Premise on problem #2 was incorrect.

    There are actually 56 reg balls, 5 chosen.

    And 46 mega balls, 1 chosen.

  18. #18
    Ganchrow
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    Quote Originally Posted by HedgeHog View Post
    Ganch:

    Premise on problem #2 was incorrect.

    There are actually 56 reg balls, 5 chosen.

    And 46 mega balls, 1 chosen.
    Yeah, I was just answering his question based on the game he described.

    The answer given by you and pokernut is of course correct for the real Mega Millions game.

  19. #19
    Iwinyourmoney
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    Sorry, had my megamillions messed up. I was thinking of bingo (lol). And no, I am not planning on trying to beat the system on the NFL picks. Was just wondering

  20. #20
    pico
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    Quote Originally Posted by Iwinyourmoney View Post
    Sorry, had my megamillions messed up. I was thinking of bingo (lol). And no, I am not planning on trying to beat the system on the NFL picks. Was just wondering
    just thought about a joke, "how do you make 100 old ladies say **** at the same time?"

  21. #21
    BuddyBear
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    Everyone on here has thought of these questions at least once at one point or another in their life.....

  22. #22
    VegasDave
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    Yes like the following (martingale alert)...

    If you start with $8000 and are gunning to make $10 each try, every time you lose wagering enough to win back all of your losses...

    (Numbers are rounded)

    1. Bet 11 to win 10 (-11)
    2. Bet 23.1 to win 21 (-34.1)
    3. Bet 48.5 to win 44.1 (-82.6)
    4. Bet 101.9 to win 92.6 (-184.5)
    5. Bet 214 to win 194.5 (-398.5)
    6. Bet 449.4 to win 408.5 (-847.9)
    7. Bet 943.7 to win 857.9 (-1791.6)
    8. Bet 1981.8 to win 1801.6 (-3773.4)
    9. Bet 4161.8 to win 3783.4 (-7935.2)

    So if you DON'T lose 9 bets in a row, you win $10. If you do, you lose $8000. But assuming that each bet you make has a 48% chance of being a winner, that means it has a 52% chance of being a loser... and the odds of hitting 9 losers in a row would be; .52^9 = .0028, or roughly 3 in 1000.

    So how many of you would consider doing this? With a 99.997% success rate you should be fine!
    Last edited by VegasDave; 12-07-07 at 04:14 PM.

  23. #23
    HedgeHog
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    U31:

    If I can only hit 48% on straights, I wouldn't bet at all--let alone using the martingale system. I'm more interested in the "reverse martingale", where I start with a $10 bankroll and shoot for an $8000 score!

  24. #24
    VegasDave
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    Well I was just using a conservative number as an estimate.

    At 50% winners, .00195 chance of losing 9 straight
    (or about 1 in 500).

    At 52% winners, .00135 chance of losing 9 straight
    (less then 1.5 in 1000)

    At 55% winners, .000757
    (less then 1 in 1250)

    The problem of course is practicality... If you've got the $8000 to put up, what good is $10 a pop to you?

  25. #25
    HedgeHog
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    Quote Originally Posted by usckingsfan31 View Post
    Well I was just using a conservative number as an estimate.

    At 50% winners, .00195 chance of losing 9 straight
    (or about 1 in 500).

    At 52% winners, .00135 chance of losing 9 straight
    (less then 1.5 in 1000)

    At 55% winners, .000757
    (less then 1 in 1250)

    The problem of course is practicality... If you've got the $8000 to put up, what good is $10 a pop to you?
    Exactly. I got it and was just having fun with your example.

  26. #26
    Ganchrow
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    Quote Originally Posted by usckingsfan31 View Post
    Well I was just using a conservative number as an estimate.

    At 50% winners, .00195 chance of losing 9 straight
    (or about 1 in 500).

    At 52% winners, .00135 chance of losing 9 straight
    (less then 1.5 in 1000)

    At 55% winners, .000757
    (less then 1 in 1250)

    The problem of course is practicality... If you've got the $8000 to put up, what good is $10 a pop to you?
    Way more than you ever wanted to read about Martingale.

  27. #27
    VegasDave
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    Thanks Ganch.

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