Playing the NC Lottery

[R0CKnR0LLA]

um, it's true, although for such low odds, it doesn't make a huge difference.

 

Let's say there were only 10 sets of number combinations for simplicity sake.

 

 

If you draw 4 sets of numbers once, your odds of drawing the winning set of numbers are 4/10, or 40% for that one drawing.

 

 

Let's say, in another scenario, you draw 1 set of numbers, 4 times, all independently events.

 

For this scenario, you have 3 options: not winning, winning once, and winning more than once.

 

To figure out the odds of winning at least once, you need to figure out the odds of not winning at all.  This will be the product of not winning in each instance.

 

ie P(1st attempt) * P(2nd attempt) * P(3rd attempt) * P(4th attempt)

 

(0.9*0.9*0.9*0.9)= 65.6% chance of not drawing the winning number combination in 4 tries.

 

This puts the chance drawing the winning numbers at least once at 34.4% in 4 attempts.

 

Vs a 40% chance of winning if you drew the 4 numbers all at once.

 

mmmmmmmmmmm math

 

[Happy Panther]

um, it's true, although for such low odds, it doesn't make a huge difference.

 

Let's say there were only 10 sets of number combinations for simplicity sake.

 

 

If you draw 4 sets of numbers once, your odds of drawing the winning set of numbers are 4/10, or 40% for that one drawing.

 

 

Let's say, in another scenario, you draw 1 set of numbers, 4 times, all independently events.

 

For this scenario, you have 3 options: not winning, winning once, and winning more than once.

 

To figure out the odds of winning at least once, you need to figure out the odds of not winning at all.  This will be the product of not winning in each instance.

 

ie P(1st attempt) * P(2nd attempt) * P(3rd attempt) * P(4th attempt)

 

(0.9*0.9*0.9*0.9)= 65.6% chance of not drawing the winning number combination in 4 tries.

 

This puts the chance drawing the winning numbers at least once at 34.4% in 4 attempts.

 

Vs a 40% chance of winning if you drew the 4 numbers all at once.

 

What sorcery is this?

 

Your assumption is that the odds on one drawing go up which is may be fine, however your expected winnings remain exactly the same. What you are showing is that the variance is lower in the first case but you aren't increasing your expected payout. You are capping your winnings to one drawing while the second example allows for the possibility of 4 winnings at a much lower probability. 

 

To see the extreme example give yourself 10 chances* in the first example vs 1 chance in 10 drawings. Or a more extreme example where there are 2 possibilities and you can either have 2 picks in one go or 1 pick in 2.

 

In other words you can increase your frequency all you want but it doesn't help as a lottery strategy.

 

*You are also assuming no replacement. I think if you add in replacement the frequency odds will then be identical again. 

[Kurb]

Guys you are both wrong.

 

 

see this formula here:

 

 

39052X342358^2 + x22235230523852-(x3953-55x^9) x MC^2x[25253525253677=2534-] 235255 qg7s09g7s09g7sg09s7gs09g709t709720t97eq-7et-qe08t=tq-t7q-qt7q-qtq=t0et8q=etq8et=qt8q8=t(w23778679)-32syg98g7gsg0sg7sgs0s9dg7s0g97sdg0s97g0s34*55)) x45356456))) (((33r32etwtwrtwt4y43664363736w462365568578696t868dsgs97u9g)))

 

 

=100%

 

 

 

So if you buy a ticket you have a 100% greater chance of winning than not buying a ticket.

 

You're welcome in advance.

[Happy Panther]

Get...Over...It.

 

Get over your specious math?

 

Done.

[Happy Panther]

 

 

39052X342358^2 + x22235230523852-(x3953-55x^9) x MC^2x[25253525253677=2534-] 235255 qg7s09g7s09g7sg09s7gs09g709t709720t97eq-7et-qe08t=tq-t7q-qt7q-qtq=t0et8q=etq8et=qt8q8=t(w23778679)-32syg98g7gsg0sg7sgs0s9dg7s0g97sdg0s97g0s34*55)) x45356456))) (((33r32etwtwrtwt4y43664363736w462365568578696t868dsgs97u9g)))

 

I see an error in there

[Kurb]

Prove it sir.

Sow your work!

[Happy Panther]

Prove it sir.

Sow your work!

 

What if I just say Q.E.D. and walk away?

[CarolinaKid704]

Get over your specious math?

 

Done.

 

fug. gg.

[Jase]

What sorcery is this?

 

Your assumption is that the odds on one drawing go up which is may be fine, however your expected winnings remain exactly the same. What you are showing is that the variance is lower in the first case but you aren't increasing your expected payout. You are capping your winnings to one drawing while the second example allows for the possibility of 4 winnings at a much lower probability. 

 

To see the extreme example give yourself 10 chances* in the first example vs 1 chance in 10 drawings. Or a more extreme example where there are 2 possibilities and you can either have 2 picks in one go or 1 pick in 2.

 

In other words you can increase your frequency all you want but it doesn't help as a lottery strategy.

 

*You are also assuming no replacement. I think if you add in replacement the frequency odds will then be identical again. 

 

All I care about is the odds.

Why would I include replacement, assuming one is competent enough to not select duplicate numbers?  That's the whole point of playing one drawing.  Only one set of winning numbers, and a selection of numbers that will only be right or wrong only once.

 

drawing multiple numbers one time = no replacement

drawing one number multiple times = multiple replacements

 

I may be a little rusty on my probability/statistics, but this just seems like common sense to me.

[Jangler]

bought a maga millions ticket for tonights drawing. Winning numbers too.

[bleys]

my dad buys lottery tickets here and there...   and then he says, "I'm selling the business and becoming a millionaire."

 

 

 

 

works every time...

[Happy Panther]

All I care about is the odds.

Why would I include replacement, assuming one is competent enough to not select duplicate numbers?  That's the whole point of playing one drawing.  Only one set of winning numbers, and a selection of numbers that will only be right or wrong only once.

 

drawing multiple numbers one time = no replacement

drawing one number multiple times = multiple replacements

 

I may be a little rusty on my probability/statistics, but this just seems like common sense to me.

 

You may indeed be rusty ;)

 

You seem to be suggesting that by grouping your plays into one drawing you are increasing your advantage. This is simply false and basic probability.

 

You cannot gain any monetary advantage from a profit and loss perspective by varying the timing of your plays. Not in any possible way. You could play 1000 times in the next lottery or play 1 time in the next 1000 lotteries and your expected profit and loss does not change by a penny either way. Whether you use replacement or not.

 

What your are doing is increasing your "odds" for the next play by increasing your bet. Getting better odds by playing more tickets is not an advantage. Instead of buying 50 tickets for the next lottery, buy 50,000,000. That might give you a 50% of hitting. Is that a better economic decision? Or buy every ticket. Doesn't matter.

 

People do the same thing in roulette which is very similar to the lottery. Cover 80% of the board  to increase your odds. Or play red and black. Doesn't matter. You can vary your bet however you want in roulette and it makes no difference.

 

What you can do to increase your expected value is save up your tickets for when the pot is large. Then your expected return does increase. I have played lotteries where I was expected to make a profit. IF your odds are 300M to 1 to hit and the pot is $400M then technically you are in an arbitrage situation.

 

If you had unlimited funds and expected to live for a few billion years.

[Anybodyhome]

Did you know you can play the Virginia lotteries online? It includes the Euro Millions, Mega Millions, PowerBall, New York, Irish and the UK National lotteries.

 

http://www.virginialotteryonline.net/

 

And you can subscribe to MegaMillions and Win for Life weekly drawings here:

 

https://www.valottery.com/subscribe.aspx

 

Does NC allow that?

 

 

[Doyle]

Ever heard of the stupid tax?