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Old 12-06-2007, 01:59 PM   #1 (permalink)
Iwinyourmoney
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Default 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
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Old 12-06-2007, 02:04 PM   #2 (permalink)
etothep
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haha, you testing them?
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Old 12-06-2007, 02:05 PM   #3 (permalink)
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seems like the mega millions is like 175,711,536 combos

Last edited by pokernut9999 : 12-06-2007 at 02:08 PM.
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Old 12-06-2007, 02:06 PM   #4 (permalink)
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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
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Old 12-06-2007, 02:06 PM   #5 (permalink)
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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
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Old 12-06-2007, 02:07 PM   #6 (permalink)
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Just you'll get the same result without the suspense.
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Old 12-06-2007, 02:10 PM   #7 (permalink)
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Ha, Hope your not thinking this is a way to beat the system. . .
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Old 12-06-2007, 02:13 PM   #8 (permalink)
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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.
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Old 12-06-2007, 02:17 PM   #9 (permalink)
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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.
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Old 12-06-2007, 02:19 PM   #10 (permalink)
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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
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Old 12-06-2007, 02:30 PM   #11 (permalink)
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?????/

Last edited by pokernut9999 : 12-06-2007 at 02:34 PM.
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Old 12-06-2007, 02:57 PM   #12 (permalink)
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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 : 12-06-2007 at 03:01 PM.
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Old 12-06-2007, 03:02 PM   #13 (permalink)
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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.
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Old 12-06-2007, 03:05 PM   #14 (permalink)
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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
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Old 12-06-2007, 03:06 PM   #15 (permalink)
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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
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Old 12-06-2007, 03:09 PM   #16 (permalink)
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