Gambling MATH people, question for ya (GANCH)

**etothep** · 12-06-07, 03:04 PM

haha, you testing them?

**pokernut9999** · 12-06-07, 03:05 PM

seems like the mega millions is like 175,711,536 combos

**pico** · 12-06-07, 03:06 PM

Originally posted by Iwinyourmoney

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

**pico** · 12-06-07, 03:06 PM

Originally posted by pokernut9999

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

**SBR Lou** · 12-06-07, 03:07 PM

Just

you'll get the same result without the suspense.

**Louisvillekid1** · 12-06-07, 03:10 PM

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

**pico** · 12-06-07, 03:13 PM

Originally posted by Louisvillekid1

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.

**HedgeHog** · 12-06-07, 03:17 PM

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.

**pico** · 12-06-07, 03:19 PM

Originally posted by HedgeHog

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

**pokernut9999** · 12-06-07, 03:30 PM

?????/

**HedgeHog** · 12-06-07, 03:57 PM

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.

**pokernut9999** · 12-06-07, 04:02 PM

Originally posted by HedgeHog

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.

**HedgeHog** · 12-06-07, 04:05 PM

Originally posted by pokernut9999

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

**pokernut9999** · 12-06-07, 04:06 PM

Originally posted by HedgeHog

I know. I just wanted to show the math behind your answer. Then I went to the Mega Millions to confirm your devine genius

**Ganchrow** · 12-06-07, 04:09 PM

Originally posted by Iwinyourmoney

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).

Originally posted by Iwinyourmoney

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 Millions^WK), 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.

**HedgeHog** · 12-06-07, 04:14 PM

Ganch:

Premise on problem #2 was incorrect.

There are actually 56 reg balls, 5 chosen.

And 46 mega balls, 1 chosen.

**Ganchrow** · 12-06-07, 04:16 PM

Originally posted by HedgeHog

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.

**Iwinyourmoney** · 12-07-07, 01:55 AM

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

**pico** · 12-07-07, 02:29 AM

Originally posted by Iwinyourmoney

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?"

**BuddyBear** · 12-07-07, 03:09 AM

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

**VegasDave** · 12-07-07, 03:44 AM

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!

**HedgeHog** · 12-07-07, 07:38 AM

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!

**VegasDave** · 12-07-07, 05:12 PM

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?

**HedgeHog** · 12-07-07, 05:22 PM

Originally posted by usckingsfan31

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.

**Ganchrow** · 12-07-07, 10:55 PM

Originally posted by usckingsfan31

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.

**VegasDave** · 12-08-07, 06:07 AM

Thanks Ganch.