Settling a lottery bet

IDLEchild

IDLEchild

Well-Known Member
So my sibling and I had a discussion. As you know the Mega Millions jackpot is 333 million now...I didn't win (surprise surprise)...

...I said, since you can only pick a # between 1-56, after enough drawings a certain winning number will repeat itself again. Now granted the number of drawings would have to be an astronomical amount but eventually it'll happen.

Sibling disagrees...

who is right?

Also has any mathematician ever taken all the past winning numbers and noticed a pattern emerging ala movie Pi?
 
The total number of combinations is y=C(n,56), n being the number of numbers you must pick. y is also a good reference about the number of drawings needed in order to get a repeated result, provided that the sets of winning numbers are uniformly distributed.

If the are uniformly distributed, then you can't predict what the next result will be, however you can be almost confident that it won't be any of the y last winning sets.
 
Jackpot probability/odds (Payout varies)
The number of ways the first 5 numbers on your lottery ticket can match the 5 White numbers is COMBIN(5,5) = 1. The number of ways your final number can match the Mega number is: COMBIN(1,1) = 1. The product of these is the number of ways you can win the Jackpot: COMBIN(5,5) x COMBIN(1,1) = 1. The probability of success is thus: 1/175,711,536 = 0.00000000569114597006. If you express this as “One chance in ???”, you just divide “1” by the 0.00000000569114597006, which yields “One chance in 175,711,536”.
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isn't that enough?
 
I tried it a couple of times, just for the heck of it...picking the same winning numbers,
but not anymore.
cash3, and cash 4, sure, I'll go for it.
 
Yes, after a certain number of draws the EXACT same set of winning numbers from the past will HAVE to repeat itself. Sure, that number is HUGE but there are a finite number of combinations that can be made with 56 numbers.

That said, the same set of numbers COULD repeat itself multiple times in a short time period as well, but that is unlikely.
 
I marvel at people who can understand math. Much respect. Thanks for helping me win the bet.

Ok question 2 regarding something similar.

A Billiards table: You set them up and a machine, perfectly calibrated to make a consistent break with the same amount of force and angle breaks. Let the balls stop and record the pattern. After enough breaks (again a mind boggling astronomical number) eventually you'd end up a same exact pattern as before...

right or wrong? I say right. The machine never moves and let's say never breaks down or wears down so it's angle and force remains consistent. Let's just say the balls never wear down either, neither does the consistency of the felt or the integrity of the table structure. Not plausible in real life but let's imagine they never do. After enough breaks all the balls will fall exactly in a previously recorded pattern.
 
Since there is a specified amount of space, it'd have to be true. Do the same thing on a living room floor. The number would be astronomical.
 
Well, I think the assumption of how the balls are set up needs to come into play in that second question. Are you taking into account the human error with the placement of the balls? Small changes in the exact placement of the balls before the break?
 
To put an analogy, you can add 2 exactly to 2 exactly and you'll always get 4 exactly.

If you put all the balls in the exact same position, the white ball in the exact same position and you make the opening shot in the exact same way, then you should get the exact same pattern over and over again. That of course is assuming no external elements or the exact same external elements. All of the aforementioned scenarios are of course, impossible.

In the real world, there is no such thing as exactitude, you can only bet on precision. You might claim to have obtained the same pattern, whether you are using your own eyes, cameras or anything else, your measurements are limited in precision and will never be exact, thus making it even impossible to determine if two patterns are equal.
 
Luis G said:
To put an analogy, you can add 2 exactly to 2 exactly and you'll always get 4 exactly.

If you put all the balls in the exact same position, the white ball in the exact same position and you make the opening shot in the exact same way, then you should get the exact same pattern over and over again. That of course is assuming no external elements or the exact same external elements. All of the aforementioned scenarios are of course, impossible.

In the real world, there is no such thing as exactitude, you can only bet on precision. You might claim to have obtained the same pattern, whether you are using your own eyes, cameras or anything else, your measurements are limited in precision and will never be exact, thus making it even impossible to determine if two patterns are equal.
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Your second paragraph is exactly what I was getting at. If you're assuming exactness in every aspect then it's not a matter of eventually getting the same pattern, you'd get the same pattern every time. That said, if you're taking into account human error in placing the balls as well as other various influencing factors then who really knows...
 
The way the balls break depends on the microscopic interactions between the solid molecular structures of adjacent balls. It's an electromagnetic force that makes pool balls move - nuclear repulsion. Lots of people make the mistake of defining physical forces (such as two objects colliding) as separate from electromagnetic forces. Then, they realized that when two objects collide, it's actually an electromagnetic repulsion force between the atomic nuclei of the different objects. Pool balls are mostly empty space - a bunch of nuclei spread out far, far away from each other, with electron probability clouds between them. These electrons obey schroedinger's uncertainty principle, which means that there is an unavoidable error of 6.63 * 10^-34 Js over 4 pi in the measurement of the electron's momentum and its position. Therefore, it would be impossible to accurately position the electrons in the same locations as they were during the initial experiment, due to a fundamental uncertainty in that information. The result of this is that it is completely impossible for any two separate pool games to proceed in exactly the same way. It might be close, but there will always be a tiny but finite uncertainty in the position of the pool balls.
 
Altron said:
The way the balls break depends on the microscopic interactions between the solid molecular structures of adjacent balls. It's an electromagnetic force that makes pool balls move - nuclear repulsion. Lots of people make the mistake of defining physical forces (such as two objects colliding) as separate from electromagnetic forces. Then, they realized that when two objects collide, it's actually an electromagnetic repulsion force between the atomic nuclei of the different objects. Pool balls are mostly empty space - a bunch of nuclei spread out far, far away from each other, with electron probability clouds between them. These electrons obey schroedinger's uncertainty principle, which means that there is an unavoidable error of 6.63 * 10^-34 Js over 4 pi in the measurement of the electron's momentum and its position. Therefore, it would be impossible to accurately position the electrons in the same locations as they were during the initial experiment, due to a fundamental uncertainty in that information. The result of this is that it is completely impossible for any two separate pool games to proceed in exactly the same way. It might be close, but there will always be a tiny but finite uncertainty in the position of the pool balls.
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But that would fall under "the balls wearing down." By that, he meant the balls changing slightly, which is what happens if the electrons move.

But I would be curious to see if you set up the pool table, machine, etc. up in Canada, then again at the equator, and again in Argentina, if you'd get the same results then, all else being equal.
 
Coriolis effect would have some say.

Also, the balls would go farther at the equator, due to centripetal acceleration making everything weigh less (and therefore have less friction) at the equator versus at the poles.

Minor changes in atomic structure are not "wearing down". Unless the experiment is conducted at absolute zero (in which case you might get some superfluid or crazy shit like that), the molecular structure of the balls is always in a state of change. Not enough to see with the naked eye, but the molecules are still flying around all over the place, even in a room temperature solid. And the electrons will move no matter what - their orbital energy comes from the potential well created by the nucleus, not by temperature. At room temp, your nuclei and your electrons are moving. At absolute zero (-273.15 degrees celcius), the nuclei aren't moving at all, but the electrons should still move, since temperature is not a parameter within the schroedinger wave function. These fluctuations are not "wearing down" in the sense that they are changes due to use - they're fundamental inconsistencies in matter.
 
Here, let's consult the eight ball....

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Nixy said:
Well, I think the assumption of how the balls are set up needs to come into play in that second question. Are you taking into account the human error with the placement of the balls? Small changes in the exact placement of the balls before the break?
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Let's see this is ran as a simulation on a super computer. But LuisG makes an interesting point. If all the factors are exactly the same as before the same pattern should occur.
 
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