Powerball simulator plays 1,000 draws a second.

[JohnnyRingo]

http://neal.fun/powerball/

First, a bit about odds. It's understood that if I shuffle a regular deck of cards I will see a result that has never before been seen in history. Considering the number of times a 52 card deck has been shuffled even this year in Las Vegas, I questioned that fact at first. But then I started thinking.

What are the odds of thouroughly shuffling a deck and finding the cards in the exact same order as when the were taken from a fresh pack? The answer is, the same odds as any other combination. Imagine shuffling the deck and having them in exact ace to king order except the spades and clubs are switched. So close!

Of course in a lottery we don't have to pick the order of all 52 cards, only the first five or six in any order, but the odds are easier to understand in such context. Also, while I admit I play once in a while, I only buy one ticket. People mistakenly believe they reduce their odds to half by playing twice. No, it's just another chance of 1 in 200,000,000 or so. If I'm going to be so incredibly lucky, I only need to try once.

How many years did it take to win a million in that simulator?

[Roland99]

1. Odds are pretty damn slim!

You have bought 63,830 tickets and have not won the Powerball
Your Numbers
40
66
9
18
41
16
Buy TicketAuto Buy Tickets
Win Breakdown
$4 prize
Won 1,772 times ($7,088)
1 in 36 win rate
$4 prize
Won 599 times ($2,396)
1 in 107 win rate
$7 prize
Won 23 times ($161)
1 in 2,775 win rate
$7 prize
Won 6 times ($42)
1 in 10,638 win rate
$100 prize
Won 1 times ($100)
1 in 63,830 win rate
$100 prize
Won 0 times ($0)
1 in Infinity win rate
$50,000 prize
Won 0 times ($0)
1 in Infinity win rate
$1,000,000 prize
Won 0 times ($0)
1 in Infinity win rate
Grand Prize
Won 0 times
1 in Infinity win rate
Total Winnings: $9,787
Loss Breakdown
At $2 per ticket, you have spent $127,660
Subtracted from your winnings your net gain is $-117,873
Your return on investment is -92%

[Oneironaut]

2. That's known as the Gambler's Fallacy, I think.

I think a lot of people think of the lotto tickets like marbles drawn from a jar, which is incorrect. Buying additional tickets does not multiply your chances. Every ticket you buy has a 1/X chance (with X being some absurdly high, but costant number).

[getagrip_already]

3. not quite true....

There iis an equation for this but your chances do go up.

A simple example is a single six sided die. If you bet on the number 2, your odds are one in six. But if you bet on number 2 "or" number 4, your odds are no longer 1/6. Sure, it's 1/6 for either outcome, but you now have two winning outcomes.

An even simpler way to describe it is you buy all six numbers. You now have a 1/1 roll chance that one of your numbers will win.

So buying more tickets does increase your chances of winning, but each number combo has the same odds.

[Oneironaut]

13. Correct. I was incorrectly thinking of each combination in a separate set.

What I said would be true for a different lotto drawing, not two cards in the same lotto drawing.

If there are 8546875467 number combinations, for example, your odds would go from 1/X to 2/X. That is indeed double.

[getagrip_already]

29. what is interesting.....

is that when someone went back and looked at all the winning numbers since the game (I think it was powerball, but it was definitely one with balls) began, a few pop up far more frequently than others.

I don't have the article handy, but you can google it.

It's still within statistical norms though, just not truly random.

Is that because some of the balls are heavier than others, or the way they start them out, or some other non random factor, or just plain old chance?

Dunno. But it would be smart money to play those numbers more frequently, at least some of the time.

[Vox Populi]

8. No it isn't the gambler's fallacy

If you buy two tickets in a lottery your chances are double what they would be if you bought one ticket. If you fail to understand this, you need remedial arithmetic.

[Iggo]

14. You're confusing chances and odds, and I'm going to leave it at that. (n/t)

[Adrahil]

19. Nope.

If there is a finite number of possible outcomes (and there are), then it would be theoretically possible to bet on every possible combination and guarantee a win.

It follows that betting on more than one possible outcome does improve your odds. Each INDIVIDUAL combination is still at the same odds, but because all bets apply to a single outcome, all the individual odds are additive. Basic probability.

[WeekiWater]

26. Pretty sure you increase your odds as well.

What doesn’t increase is the odds on each individual ticket. You are conflating the odds of individual tickets with the odds of an individual in possession of two tickets. Two very different statistical models.

[Iggo]

53. You double your chances with a second ticket.

You nowhere near double your odds with a second ticket. You barely move the needle.

[Oneironaut]

58. Here's proof you double your chances.

You have X total possible number combinations. You choose one number combination. That gives you 1/X chance to win. If you buy two tickets with a unique number combination (as everybody would), that gives you 2/X percent chance.

Remember, this is the same drawing. See my cards example below for another analogy. The lottery is one drawing, and has a finite number of combinations possible.

However, multiplying your chances doesn’t mean they’re in any way good. If you bought 1000 powerball tickets, even, and your chance of one ticket winning is 1000 out of 88 quadrillion, your odds are basically zero (0.000000001136364%). To even make it to 1%, you would need to buy some absurdly high amount of tickets (880 billion tickets). I didn’t look this up, but we probably have more of a chance of an asteroid hitting earth and killing everyone than someone with 1000 powerball tickets winning the jackpot.

[Iggo]

62. I will accept that double of statistical nothing is nothing.

[Oneironaut]

63. Absolutely. By buying 10,000 tickets, for example, you're making terrible odds better, but still

absolutely terrible.

[WeekiWater]

83. So my statement is accurate. N/T

[underpants]

76. The Gambler's Fallacy is that previous outcomes in random chance affect future outcomes

https://www.logicallyfallacious.com/tools/lp/Bo/LogicalFallacies/98/Gambler-s-Fallacy

[Iggo]

4. Second ticket? That's another chance at the same odds.

I've explained this to my coworkers, and they nod their heads, but I don't think they get it.

[Vox Populi]

6. what?

If you buy a second ticket, you double your odds. What don't you understand about this?

Imagine there are 100 different combinations. If you buy 100 tickets each with a unique combination you will win with certainty.

If you buy 1 ticket you a have a 1/100 chance of winning.
If you buy another ticket with a different combination you now have 2/100 chances of winning.

How is this not doubling your chances? Each ticket has the same 1/100 chance but you have 2 tickets.

[Ms. Toad]

9. You're both right.

When you buy "another chance at the same odds," you aren't tossing away your first chance - you are adding the chance of winning on your second ticket to the "same odds" of winning on the first one - i.e. doubling it.

Instead of having 1/gazillion (your first ticket), you still have 1/gazillion (your first ticket) and you are adding 1/gazillion (your second ticket). So you now have 2/gazillion (which is double the 1/gazillion chances of your single ticket)

[Adrahil]

20. exactly correct....

because all individual bets apply to a single outcome.

This does require that each individual bet is on a DIFFERENT unique outcome. Buying multiple bets on a single combination does you no good.

[Ms. Toad]

21. Equally true.

I was merely trying to clarify for the two posters who seemed to be talking past each other that they were both saying the same thing.

[JohnnyRingo]

10. You absolutely do not "double your odds".

If the odds are 1 in 300m, buying another ticket doesn't mean your odds of winning will be 1 in 150m. It means you have two chances to win at 1 in 300m. The advantage is infinitesimal enough as to be disregarded.

If you bet you'd draw an ace of spades from a deck of cards the odds are 1 in 52. If you want to try twice, you put the card back, shuffle, and try again at 1 in 52. One could keep doing that over and over for hours at the same odds.

[Oneironaut]

11. Actually, wouldn't a better analogy be that you have 6 card decks, and you choose a card.

Someone picks a card from each card deck, and if they choose your combination, you win. If you choose two cards from each card deck in the same drawing, don’t your chances double (1/52 in each set to 2/52).

A lottery reading isn’t multiple drawings (my mistake above). It wouldn’t be the same as pulling a card from a deck and putting it back in.

[JohnnyRingo]

16. No

Your odds aren't 2 in 52. It's one in 52 twice. There's a huge mathematical difference.

[muriel_volestrangler]

18. If, for one draw, you buy 2 tickets, and they are different, then you do have double the odds

of winning that draw than if you had one ticket. If it's a deck of cards, it is indeed 2 in 52.

"One in 52 twice" would describe if cards are drawn twice.

[Oneironaut]

22. Isn't that only true for different drawings, though?

If you chose two cards from a card deck, and a random card was selected from a 52 card deck, the odds of the selected card being yours would be 2/52. The proof is, if you chose 52 cards from a 52 card deck in the same drawing, you would have a 100% chance of picking the right card. You have two cards out of 52 you chose in the same drawing.

If it were a different drawing, your odds for each would be 1/52. This was the mistake I made above.

[Ms. Toad]

24. But you're changing the setup.

As long as you are choosing distinct numbers, the lottery problem is the non-replacement (dependent) scenario, in which 52 draws guarantee that you will pick every single card in the deck. The odds, when you pull two cards, is 2/52 = .03846 (or, in the pick 6, choosing from the numbers 1-49: 2/13,983,816 - and you probably don't want me to walk through how we know there are 13,983,816 cards in that particular deck )

You seem to be discussing the replacement scenario - once a card is drawn, you replace it in the deck so it is possible for it to be drawn again. You could play the lottery that way (by letting yourself choose the same number again), but it would be a dumb way to do it

The odds in a replacement scenario are, as you suggest, different: 1-(51/52)^2 = .03809 (roughly) {100% less the probability on each draw of NOT drawing the desired card raised to the power of the number of draws you make; if you are aiming for a particular card, the probability of not drawing it is 51/52. In probabiliyt, 100% = 1, so the formula for two draws is 1-(51/52)^2; for different numbers of draws with replacement, use a different exponent}

So, as you can see, the odds are different - but not as hugely different as you suggest.

(The difference between the two increases with the more cards drawn, because you have to reach 100% for the without replacement scenario, and you never hit 100% probability with replacement).

At 10 cards here is the probability for each: 10/52 = .1923; with replacement: .1765;
At 26 draws without replacement 26/52=.5; with replacement .3964)

[Adrahil]

17. There are a finite number of possible outcomes...

... so it stands to reason that if you bet on every possible combination, you would win. 1:1 odds.

Likewise, if you bet on half the possible combinations, you would 1 in 2 odds of winning.

Following this on down, betting on unique combinations does reduce you odds.

This is in contast to the “law of averages” argument that says if you bet on one number of a six sided die roll, that rolling the die six times means there is a 1:1 probability that your number will come up. That is false.

[qazplm135]

28. yes they go from 1 in 292 million

to 2 in 292 million.

[Adrahil]

33. Yes.... which is double....

2 in 292 million is the same as 1 in 146 million.

[JohnnyRingo]

37. Think about that.

If the odds can be halved from 1 in 292m to 1 in 146m by buying another ticket, then one could get it down to 1 in 100 just by buying a couple dozen tickets: 1 in 73m, 1 in 35m, 1 in 17m. See where that's going?

Buying a second ticket is another chance at the same 1 in 292m odds. Every ticket you buy is a 1 in 292 chance.

[fishwax]

39. Buying a second ticket doubles the odds. Buying a third doesn't double them again.

A couple dozen tickets would give you 24 combinations out of a possible 292 million (assuming that's the right number ... I don't really know the ins and outs of the game, so I'm just running with the numbers I've seen thrown around here ) which works out to about 1 in 12.17 million.

To double your odds from 2 in 292 million to 4 in 292 million (1 in 73 million) you have to buy TWO more tickets. Which sounds like a bargain, I suppose, until you consider that to then double them AGAIN you'll have to buy four tickets, giving you 8 unique combinations out of 292 million (or 1 in 36.5 million). These are still horrible odds, but of course you can see why you can't quickly get it down to 1 in 100 very easily: in order to continue doubling your chances you have to continue doubling your investment. To get it down to 1 in 100 you have to buy one combination out of every 100, which would cost you 2.92 million.

[Adrahil]

51. +1 NT

[Adrahil]

42. NO.

I just took a graduate level probability course for my MSSE. Let's check some facts:

Do you agree that there are a finite number of possible outcomes?

Assuming you do, do you then agree that if one were to bet on every possible outcome, one would guarantee a win (1:1 odds)?

I'll assume we agree so far.

So, it follows, that each individual possible outcome you bet on increases your odds of winning. IN fact, you odds are additive, not multiplicative. So if you buy 3 tickets, you odds are 3 in 292m. If you buy a "couple dozen tickets," your odds are 24:292m or about 1 in 12,166,666.

Each INDIVIDUALpossible outcome has the same (very low) chance of occurring, but because all you bets apply to a SINGLE outcome, your odds are increased. Otherwise, betting on every possible outcome would not guarantee a win, and it does.

[Drahthaardogs]

45. Yep, you got it except that SECOND ticket really does Double

Your odds. From 1 in a million to 2 in a million. Then "DOUBLING" your odds would require exponentially growing your ticket purchases.

[Adrahil]

47. That's is a unique situation.

IN the same way that 1+1=2 and 1X2=2, buying a second ticket DOES double you odds. However, buying a third ticket does not then double THOSE odds.

SO the "doubling" only applies from when going from one to two tickets. As I point out, the increase in odds is actually additive, not muliutplicative.

[Drahthaardogs]

48. Exactly, but that is where people are getting confused

Only that SECOND ticket DOUBLES your odds.

I'm a toxicologist, so I use stats a lot. There is a movement to shift tobacco users to moust snuff and OFF of cigarettes. The American Lung Association says "Moist snuff more than doubles your chances of oral cancer, so it's not a safe alternative to smoking"

The truth is, you have about a 1 in 2,000,000 chance of developing oral cancer each year with no risk factors. Moist snuff user's risk MORE than DOUBLE to 3 in 2,000,000. Still, not a huge risk. Compared to smoking which increases it to 20 in 2,000,000 or those who drink and smoke it raises to over 80 in 2,000,000.

Statitics really don't lie, you just have to know what you are looking at.

[Adrahil]

49. Well said. NT

[Oneironaut]

59. The odds of winning the jackpot aren't 1 in 292m, though.

I read that they’re 1 in 88 trillion. Even if you keep halving the chances, no person on earth could ever even realistically have a .0005% chance of winning.

[Ms. Toad]

66. Depends on the game

1 in 292 m is correct for pick 6, using the numbers 1-49.

[Kaleva]

31. Why do you put the card back?

When buying multiple tickets, one doesn't use the same number on all tickets. In your analogy, a person with two chances would bet on the ace of spades and another card.

[JohnnyRingo]

38. Because the stock of numbers stays the same for every play.

Just because someone plays five particular numbers, it doesn't mean they're removed from the draw. Every play has the same odds as the first. Of course the odds are slightly better because you're trying again, but so mathematically slight as to be ignored.

Otherwise, there'd be a lot more billionaires.

[Adrahil]

43. Not sure what you are getting at here...

As I understand, each ball is unique, and every number that appears cannot be repeated in a single play. That is, the ball is not returned to the pool for the next ball draw.

[Kaleva]

77. But the play is done only once.

Here is what you said earlier:

"If you bet you'd draw an ace of spades from a deck of cards the odds are 1 in 52. If you want to try twice, you put the card back, shuffle, and try again at 1 in 52. One could keep doing that over and over for hours at the same odds."

In the above game, what are my odds of winning if I bet on 2 cards? It is not 1 in 52. It is 1 in 26. If I bet on 4 cards, the odds are now 1 in 12. If I bet on all 52 cards, the odds are now 1 in 1. In the previous examples, If the winning card isn't picked the first time, the card isn't reinserted in to the deck and the second hand is played with the odds remaining at 1 in 52 until 52 hands are played.

This is what the person said that you had responded to:

"If you buy a second ticket, you double your odds. What don't you understand about this?

Imagine there are 100 different combinations. If you buy 100 tickets each with a unique combination you will win with certainty.

If you buy 1 ticket you a have a 1/100 chance of winning.
If you buy another ticket with a different combination you now have 2/100 chances of winning. "

[Takket]

40. That's not correct

The lottery isn’t a “put the card back” drawing. Once you check your first ticket and see it is a loser that combination is gone forever and now you check your second ticket. Since the first set of numbers isn’t “put back” you know that ticket cannot lose because of the first combination being redrawn and applied to your second ticket.

So your odds of either ticket hitting are basically 1 in 300 million EACH so the total is 2 in 300 million or double the odds (actually your second ticket’s odds are really 1 in 299,999,999 since my first ticket eliminated one wrong combination, but as you mentioned this is infinitesimally small).

In your deck example I have a 1/52 chance on draw #1. Then I throw that loser card away. My next chance is 1/51. Total odds are 103/2652 which is actually just a TINY bit better then 2/52.

[Adrahil]

44. Yup.

Although rather than thinking of the ball draws as individual "chances" (which is a legit way to view it), I prefer to think of the entire draw as a single outcome from among many possible outcomes, and each ticket represents a bet on one of those possible outcomes.

[JohnnyRingo]

50. You're doubling your chances, not your odds.

You have a chance at 1 / 300m odds. You're right that the second play is 1 /299m, but that doesn't help much. Indeed it would help in a 52 card deck.

People are confusing chances and odds, but I was arguing with someone here who thinks the second try is at 1 /150m odds. That'd be a great reason for them to buy as many tickets as possible.

[Hav]

56. All what matters for a player is the chance to win

In a game with 10 possible outcomes, what is the chance to win when betting on one outcome? What is the chance to win when you can bet on two outcomes for the same single trial?
Don't you agree that a game with one bet and 1/10 odds basically becomes a game with 1/5 odds when you can make two bets?

[JohnnyRingo]

75. No

A game with 1/10 odds becomes another game with 1/10 odds when a second bet is made. That's two chances at the same odds.

Your chance of winning improves, but the odds that either bet will win do not.

[Hav]

82. I think I'm starting to see where one part of the confusion comes from

Not being a native speaker, I used the term odds wrong. The odds change from 1/9 to 2/8 if we see odds as the ratio of favorable outcomes to unfavorable outcomes. But you got it wrong by this definition as well by arguing that the odds don't change at all because you fail to see that we don't have two separate draws where the chosen number gets returned after the first draw as you seemed to indicate in post 10. You have two bets on the same draw. Your scenario of having two bets with the same odds would be buying lottery tickets for two separate events and not the same one.

A player doesn't care which one of his bets will win, he only cares that any one of his bets wins. This is why the chances and odds to win the game change by definition. This is what the whole point about buying multiple tickets was about. Despite multiplying your chances to win, you will still have almost the same 300million-x outcomes against you.

[JohnnyRingo]

86. I agree.

If the odds improved substantially with each new bet, (ie by half) the game wouldn't roll over each week. Some people buy hundreds of tickets thinking it makes them they very likely to win. The result of such deluded tactics provides the billion and a half jackpot we see today.

I was wrong in saying the odds don't improve, if only slightly.

[Oneironaut]

61. Even if you bought 1000 tickets at those odds,

your chances would be 0.000333333333333%. Those are abysmal odds.

You aren’t halving the denominator each time. You’re adding to the numerator.

[Ms. Toad]

65. I think he's confused because

Adding one to the numerator DOES halve the denominator when you buy the second ticket. That is the one and only time that buying one more ticket does that - but he's latched onto the incorrect pattern to get from 1/300 m to 1/150 m, since there are two ways to get there - adding one to the numerator and reducing, or dividing the denominator by two. The former is correct, the latter is not.

[Ms. Toad]

64. It does.

First draw = 1/300 m
Second draw = 2/300 m = 1/150 m
Third draw = 3/300 m = 1/100 m (note -not 1/75, as you seem to be suggesting - no one who knows anything about math was arguing that EVERY draw doubles your chances. Each draw ADDS one more chance at the sane odds. It just happens that 1+1 is the same as 1 x 2, so adding one is the same as multiplying by 2; after the second draw, adding one more doesn't double your chances again. It adds one more chance, and 2 + 1 is not equal to 4 (2 x 2)

[Iggo]

12. .

[roamer65]

5. IMO... if one single person is going to consistently play a lotto type game...

Lucky for Life is the best one.

It has better payouts at the lower end than most state lottos, MegaMillions and Powerball.

I win much more often on it.

Mega and Power are an “all or nothing” scheme, really meant for lottery pool participation

[Kaleva]

32. The odds at winning the Michigan Fantasy 5 is 1: 575,757

"5 of 5 1: 575,757 Jackpot
4 of 5 1: 3,387 $100
3 of 5 1: 103 $10
2 of 5 1: 10 $1
Overall Odds of Winning: 1 in 9"


https://www.michiganlottery.com/games/fantasy-5

[JI7]

35. how much does it cost ?

im in california and never played any of these things before.

but was thinking it would be better to play something where the payout might be less but chances of winning much higher .

[Kaleva]

78. 1 dollar

You can register to play and purchase on line at this link:

https://www.michiganlottery.com/promotions/offers?MMI=38224#/welcome-offer

It appears one cannot register on a Sunday.

[GemDigger]

7. About that....

… Killjoy...

[nolabear]

15. IMO you're paying for the thrill. And that's 100%.

The dream is fun. Joy. And I guess for some the analysis is the fun. I buy one when it gets really high and hey, for a few hours I can do some mighty fine things. 😄

[Iggo]

23. "The pleasure is to play. Makes no difference what you say."

[Ms. Toad]

25. Nope. Never. The analysis is never any fun -

Especially not more fun than the real property papers begging to be graded. . . . if not for Sputnik, I'da been a math professor.

[qazplm135]

27. the only way to significantly improve your odds

is to play an appreciable number of separate combinations.

And the cost of playing that many combinations is quite a lot.

Of course, I suppose if you had 292 million dollars and the time/ability to play every possible combination, you could guarantee doubling your money after taxes this next time around.

[Ms. Toad]

30. You'd have to have to wait for the pot to hit 584 million,

and then it's only doubling it before taxes.

And then there's the problem of how to play that many combinations. If you bought tickets continuously from the prior drawing to the next at 30 seconds a ticket,it would take you 101,389 days to buy all of the combinations.

Even at a rate of one/second (assuming some sort of app that would generate the numbers and buy them), you'd need 3,380 days to buy all the combinations.

And that's only to guarantee you'll break even, not to double your money . . .

[Adrahil]

34. Right this is mainly a theoretical game.

But if gambling is your game (it's not mine.... I'm too familiar with the math) Then buying at least a few tickets would seems to make sense...

[Ms. Toad]

36. Hardly anyone has done the math -

so I was just pointing out that even if one had $292 million dollars to spend, buying all of the combinations would be pretty much impossible.

Not to mention that if anyone else chose the winning number, you'd have to split the pot.

[grantcart]

84. Isn't it $2 per ticket?

[Ms. Toad]

87. No idea. I don't gamble.

I was addressing the principle that to double your money the payout needs to be twice the cost of buying every combination.

[grantcart]

88. Correct but you undervalued the actual cost by 100% so it is even more

To cover every option

[qazplm135]

41. It's at 1.6 billion now thus my comment

And if you are rich enough and crazy enough to do it I suspect you have the money to hire enough people to buy you tickets to get it done in time.

[Ms. Toad]

60. I don't gamble, so I've got no clue what the prize is at.

But it would be a monumental undertaking -with 10s of thousands of people, probably spread out geographically . . . Then you'd have to worry about the winner going rogue.

[qazplm135]

67. You realize you can buy more than one ticket

At a time right?

It would be difficult but not as impossible as you think, and you just collect the tickets prior to the drawing.

[fishwax]

69. You and your team would have to buy about 850 tickets per second, from the moment after

the Saturday night drawing until the deadline one hour before the Wednesday drawing. You and your team would have to buy 850 tickets per second, and your team would have to be coordinated enough to make sure that none of them purchased duplicate numbers, so purchasing the tickets would take longer than simply walking up and ordering, say 20 tickets.

It would take a LOT of people to buy tickets at a rate of 850 per second and maintain that rate for 95 hours.

On edit: As I understand it you can buy tickets for up to four weeks in advance, but in doing that you run the risk that somebody else wins one of the draws before yours. Even spread over four weeks you would have to buy about 120 tickets per second.

[qazplm135]

70. Given that you can buy more than one at a time

i.e. more than one per second... again difficult but not the Herculean task you make it out to be.

[Ms. Toad]

72. at one second per number

it takes 3380 days - or 9.25 years (working 24/7).

To buy them in a single week, you would have to hire enough people to buy 23 numbers each and every second of the week, 24 hours a day.

Taking into account hours the stores are closed, and the reality that no store is going to let you monopolize their entire business for a week it's pretty hurculean.

Edit: My calculation is for the pick six (numbers 1- 49), which generates 13,983,816 combinations. For a lottery with odds of 1:292 million, the task is even harder.

Someone calculated the number per second for the 1:292 million odds above - which would be even more challenging.

[fishwax]

74. You can buy more than one at a time, but in this case you would have to buy *specific* numbers

You can go to a clerk, ask for, say, 100 tickets, and they'll generate the numbers for you and print them out. I have no idea how long that would take, but conceivably with a fast clerk, a fast printer, and a large number you could get the average under one ticket per second.

But you wouldn't be able to do that if you were trying to make sure to cover all the possible combinations. You would have to specify EVERY SINGLE NUMBER on EVERY SINGLE TICKET in order to make sure that you and your team didn't double up on numbers.

So, yeah, I'd call it Herculean.

[Ms. Toad]

71. Did you see my time calcualtion a few posts up?

Somehow each individual number has to be input into the system. I ran one calculation at 1 second per number, which I think is optimistic, and another at 30 seconds per number (assuming a pure manual entry). At one second per number, the time required is 3,380 days (buying tickets 24 hours a day).

Edit: My calculation is for the pick six (numbers 1- 49), which generates 13,983,816 combinations. For a lottery with odds of 1:292 million, the task is even harder.

[SidDithers]

46. I love this thread because of all the math geeks arguing probabilities...

This world needs more math geeks

Sid

[MontanaMama]

52. THAT

I agree with. 👍🏽☺️

[Hav]

54. They are needed indeed

for education. The "math geeks" in this thread aren't arguing among themselves, they are arguing with those who don't understand simple math. I'm not a math geek and it's still astonishing to see these mental barriers some have when it comes to things like probability.

[Adrahil]

79. Yup....

This is the most basic of all probabity. The probability of all possible oucomes must sum to 100%.

[muriel_volestrangler]

68. Nowhere near as epic as this one:

https://www.democraticunderground.com/discuss/duboard.php?az=view_all&address=228x67044

The explanation: https://www.newscientist.com/article/dn18950-magic-numbers-a-meeting-of-mathemagical-tricksters/?full=true

[Adrahil]

81. WOW

Some folks don't understand conditional probability.

[lapfog_1]

55. The odds of picking the exact sequence of cards after shuffling the deck twice

is 1/52 x 1/51 x 1/50 ... x 1/2

It's an amazingly small number.

Has it ever happened in history... who knows, it might be unlikely... but many unlikely things do happen.

In computer science, I have a rule... if something is very unlikely to happen ( the odds are millions to 1 against... or even billions to 1 against ) , it is almost certain to happen ( so long as the "event" isn't based on constant externalities )

In other words, because computers execute the same code millions or billions of times a second... it is certain that the "bad path" or unplanned for event will, in fact, happen.

Which is why the code you write MUST allow for all such events when possible.

The scary thing is that our current data protection schemes (SECDED memory, CRC disk blocks, parity on data transfer paths) is totally insufficient for today's amount of data and number of computers processing it. Look at it this way... gamma rays are known to "flip bits" in very dense memory devices like modern RAM... we check and protect this with Hamming codes calculated on every 32 or 64 bits (or 128 bits)... adding extra bits that are used along with the XOR (exclusive OR) function to check the validity of the data... and yielding truly impressive odds of detection and even correction ( Single Error Correction, Double Error Detection - SECDED )

But with TBs of memory in your modern server... what are the odds that over years of operation that a set of gamma rays hit the exact same string of cells at the exact same time ( before memory refresh cycles ) and in such a way that the resulting bit flips are neither detected nor corrected. At this point... it is an almost certainty. And probably happens with some frequency. A similar situation exists for the PBs and Exabytes we are storing in both iron oxides ( rotating rust ) and bubble memories ( flash ).

[IADEMO2004]

57. Place your bets on Democrats. Better odds. More Winners

[brooklynite]

73. Your math is wrong...

Buying a second ticket DOES reduce your odds by half: from 1 in 300,000,000 to 2 in 300,000,000, or 1 in 150,000,000 (assuming the second ticket is a different set of numbers). The point is that further ticket purchase reduce the odds by increasingly smaller amounts.

[grantcart]

85. +1

[krispos42]

80. To quote Han Solo

For two bucks you get 3 days of happy fantasies. That is a very high value for your entertainment dollar, much more so than a movie ($7-15 for 90-150 minutes).