Any rocket scientists on here???

Dumb question right? So I’m curious if someone way smarter than me would like to tackle some “simple” math equations?
My local watering hole runs some football boards for the local Jaycees. I’ve spent money on them and won occasionally, not nearly what I’ve spent, but it’s fun. So I’m on 3 boards currently for the SuperBowl.
Board A: one square for $10. Total of 25 squares. One winner collects 200 for matching #s. I purchased 2 squares and the way I figure each square has a 4% chance to win, so do I should have an 8% chance of winning, or 2 separate 4% chances?
Board B: One square for $5. Total of 100 squares. One winner collects $375. I purchased 4 squares. Each square would have a 1% chance of winning, so I should have a 4% chance right?
Board C: One square for $10. Total of 100 squares. A winner at the end of each quarter will collect $250 for matching #s. All squares are good for all quarters. Or if game goes to overtime, 1st, 2nd, 3rd, and final scores will pay. I purchased 2 squares. So I should have a 2% chance to win any quarter.

So what would my odds of winning all 4 payouts on Board C be? 2%? .05%?
What would my odds of winning all 6 payouts on all 3 boards be? For $1575.
What are my odds that I lost $60?

It was quite the conversation at work today. Much better than usual job site crap. But I don’t think anyone really knew a real mathematical answer.

I have a hard enough time figuring out if I can keep a pike or not, something like ten under something and one or two over something else?

all i can tell you is you want crooked numbers for superbowls…..at least for the final. lots of 2 point attempts

Whoa, Whoa, Whoa I know the answer to this.

$17

How are the #’s assigned? For you to win all 6, wouldn’t you need the same #’s across all boards? That complicates the probability, unless you can pick them to be the same.

Then, not all #’s have the same probability. The scoring system coupled with team and player dynamics will give certain squares (#’s) better odds.

If the outcome was purely random, your chance of winning all 4 payouts on board c would be .02 to the 4th power.

The easy answer is slim to none to win all 6 or even all 4 on board C. For board C that would be 2% times 2% times 2% times 2% which is .00016%… so for winning all 6 it would be that same .00016% times 8% board A times 4% of that board B which is .0000008192% Im honestly not sure on the method to figure out losing all $60. Granted im not sure that the math is right on winning everything either either way its for good fun

C .04*.04*.04*.04 = no chance (1 in like half a million) (assuming that the winning entry can redraw on subsequent draws)
1-(.96^4)=15% chance you win something.

.92*.96*.85=75% chance you lost 60 bucks.

Looks like whoever is running board A and B has a 100% chance of winning without even having a number?

Maybe start your own A and B board, save your purchase of a # and still be winner!

riverruns wrote:

Looks like whoever is running board A and B has a 100% chance of winning without even having a number?

A and B are run by local Jaycees, so they are making money for their community.
Board C is run by a local guy who’s been doing it for years. Doesn’t make a penny. Just has fun.

______________ wrote:

How are the #’s assigned? For you to win all 6, wouldn’t you need the same #’s across all boards? That complicates the probability, unless you can pick them to be the same.

#s for A and B are hidden along side of columns of squares by scratch off “paint” like a lotto ticket. #s are scratched once board is full. #s for C are picked by cards at a “draft” style party the day before at the bar by whoever wants to be there.
Everything is random and you have no say or control in your #(s), except for how many squares you buy.

A and B are run by local Jaycees, so they are making money for their community.
Board C is run by a local guy who’s been doing it for years. Doesn’t make a penny. Just has fun

Good for the Jaycees too do this then.

I got to the 3rd sentence in board a and my brain hurt.

Deuces wrote:

I got to the 3rd sentence in board a and my brain hurt.

After four attempts to read it I gave up

What’s the old adage…..fools and their money are soon parted.

Take the money for the boards and pamper the wife with something. The payout will be worth more.

Board A – you have 2 separate .04% chances to win. (Or 1/25 x 2) Even if you bought all 25 squares.

Board B – odds are 1/100

Board C – odds are 1/100. If you buy 2, you have 2 1/100 chance to win.

However, since these are based on the score of football games, your actual odds of winning decrease significantly if you draw a crappy number like 2-2 or 8-8. Since these number at not likely to ever be at the end of a quarter, half or final score. You odds increase when you draw a number like 0-0, 7-0, 3-0 since these numbers are more likely to hit.

-J.

years back i participated in a super bowl board that was called a crazy board. it was crazt all right.

you picked a pull type ticket, you got a number. say 2-9.
some of the ways you could win where, number on the referrees’s shirt, first kicker to kick a field goal, numbder of the player to score the first TD start of each half, on the kickoff what yardline the ball was downed at. quarter scores hardly mattered, end of games first downs……..

yea it was crazy.

Pailofperch wrote:

#s for A and B are hidden along side of columns of squares by scratch off “paint” like a lotto ticket. #s are scratched once board is full. #s for C are picked by cards at a “draft” style party the day before at the bar by whoever wants to be there.
Everything is random and you have no say or control in your #(s), except for how many squares you buy.

Do they even guarantee a winner for each event? The calculated odds assumes there’s a winner on the board. It will come down the to details of each game. As pointed out earlier in the thread not all numbers are created equal in football (namely 0,3,4,7 will likely have a higher probability of occurring than 2,5,8,9).

Pailofperch wrote:

What would my odds of winning all 6 payouts on all 3 boards be? For $1575.

First, you would have to get the exact same numbers on each board, then hit the numbers. Figure out how many different number combinations there are for the 3 boards… that would be a lot. So idk.. might have better odds playing the powerball.

Considering each # has it’s own probability of hitting, you will also need a brain surgeon and a MLB analytics department to calculate it. Or as my Dad says, you have a 50% chance, either you win or you don’t.

Reading a few of these.. probability of a successful launch are 0%

My answers in caps: Also I’m an Advanced Placement AP Stats teacher!

Board A: one square for $10. Total of 25 squares. One winner collects 200 for matching #s. I purchased 2 squares and the way I figure each square has a 4% chance to win, so do I should have an 8% chance of winning, or 2 separate 4% chances? STRAIGHT 8% CHANCE OF WINNING
Board B: One square for $5. Total of 100 squares. One winner collects $375. I purchased 4 squares. Each square would have a 1% chance of winning, so I should have a 4% chance right? YES
Board C: One square for $10. Total of 100 squares. A winner at the end of each quarter will collect $250 for matching #s. All squares are good for all quarters. Or if game goes to overtime, 1st, 2nd, 3rd, and final scores will pay. I purchased 2 squares. So I should have a 2% chance to win any quarter. CORRECT.

So what would my odds of winning all 4 payouts on Board C be? 2%? .05%?
(.02)*(.02)*(.02)*(.02) VERY SLIM; DONT GET YOUR HOPES UP HAHA
What would my odds of winning all 6 payouts on all 3 boards be? For $1575.
What are my odds that I lost $60? AGAIN, YOU CAN MULTIPLY THESE PROBABILITIES AS THEY ARE INDEPENDENT. (.08)*(.04)*(.02)*(.02)*(.02)*(.02)

It was quite the conversation at work today. Much better than usual job site crap. But I don’t think anyone really knew a real mathematical answer.

IN TERMS OF THINKING ABOUT WHAT YOU PLAY, YOU SHOULD THINK OF IT FROM AN EXPECTED VALUE STANDPOINT.
BOARD A: THEY ARE COLLECTING 10*25 = $250, BUT ONLY PAYING OUT $200. SO THE PAYOUT IS 200/250 = 80%. I KNOW ITS FOR CHARITY, BUT REALIZE IN THE LONG RUN YOU ARE GETTING BACK 80 CENTS ON EVERY DOLLAR

BOARD B: THEY ARE COLLECTING 5*100 = 500, BUT PAYING OUT $375, SO THE EXPECTED VALUE LONG RUN PAYOUT IS 375/500 = 75% PAYOUT; SLIGHTLY WORSE.

BOARD C. NOT SURE IF THEY JUST LIKE DOING THIS FOR RUN, OR DIDN’T REALLY THINK ABOUT IT, BUT THE PAYOUT HERE IS 100% THE CHARITY PUTTING THIS ON WILL NOT MAKE ANYTHING ON BOARD C AS THEY ARE COLLECTING $10*100 = $1000, AND PAYING OUT $1000 (250 EACH OF THE 4 QUARTERS). NOW, THEY MIGHT NOT SELL ALL THE TICKETS AND NOBODY WINS A CERTAIN QUARTER AND THE MAKE MONEY, OR THEY MIGHT NOT SELL ALL THE TICKETS AND SOMEONE WINS EACH QUARTER AND THEY LOSE MONEY, BUT IN THE LONG RUN THIS GAME WILL NOT MAKE THE CHARITY ANY MONEY WHICH I FIND ODD.

LONG STORY SHORT; PUT ALL YOUR MONEY IN BOARD C FOR THE MOST BANG FOR YOUR BUCK!

My advice: Stick to raffle drawings, or ones that are limited tickets sold. At least if you’re not the winner out of 100, your money went to a good cause(charity)

phoyem wrote:

NOT SURE IF THEY JUST LIKE DOING THIS FOR RUN, OR DIDN’T REALLY THINK ABOUT IT, BUT THE PAYOUT HERE IS 100% THE CHARITY PUTTING THIS ON WILL NOT MAKE ANYTHING ON BOARD C

‘

Excuse me Mr. Phoyem but isn’t the first instruction for a word problem to read the whole problem? Board C is not for charity and is put on just for fun. F- and make sure your mother signs your report card and bring it back on Monday!

I’m an aquanaut, not an astronaut, so I agree with the above.

BigWerm – I read the original problem! Just skipped over all the posts after it; gotta be worth a C+ at least