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rsactuary
rsactuary
Joined: Sep 6, 2014
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June 1st, 2019 at 7:30:09 AM permalink
Quote: rsactuary

They sold about 22.3 million tickets for last draw. This is a very rough approximation by taking number of tickets hitting just the powerball and dividing by the probability of winning said amount.

I have to think given the size and the attention that this drawing is getting that they will sell well in excess of 25.5 million tickets.



I think I had my math wrong because they've changed the number of megaballs. There is now 25.

rough estimate of ticket sales for last night's draw is 842,135 x 25 = 21,053,375 ----> number of winners of the easiest prize (matching the megaball) times the number of megaballs.
AxelWolf
AxelWolf
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June 1st, 2019 at 8:41:38 AM permalink
Quote: ksdjdj

You can bet to win $2,000 for less than even money odds (eg $12,000 @ -600) and you can bet $2000 on odds greater than even money (eg $2000 @ +450).

Note: if the odds change, you can have another bet at the new odds.

Note 2: About a year ago, the limits were $500, but I am happy that they increased it to $2000.

I think it is just available at 5d.

----------------
For yesterday's Mega Millions I put 3000 down to win 500 (roughly 14.5% of bank roll).

Starting Bal: 20,950

Finishing Bal: 21,450

Chance of no winner (based on actual tickets sold): 90.544...%

Odds Taken: -600

EV: +5.635...%

Tickets Sold***: 30,053,111^^^ (inc "just the jackpot")

*** http://www.lottoreport.com/ticketcomparison.htm

^^^: rsactuary posted earlier that my estimate was probably low, and they were right.

---------------------------
Update:

"How much can you reasonably get down on this bet before thy stop you or change the line dramatically?"

In another post on this thread, the odds changed from $3.65 (when i first notice the odds on offer for that game) to $2.95 (got my first $2000 on at this price) then they went to $2.55 (had another $1400 at this price) and closed at $2.45.

Thanks for the great details.
♪♪Now you swear and kick and beg us That you're not a gamblin' man Then you find you're back in Vegas With a handle in your hand♪♪ Your black cards can make you money So you hide them when you're able In the land of casinos and money You must put them on the table♪♪ You go back Jack do it again roulette wheels turinin' 'round and 'round♪♪ You go back Jack do it again♪♪
ksdjdj
ksdjdj
Joined: Oct 20, 2013
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June 1st, 2019 at 5:23:52 PM permalink
Quote: rsactuary

I think I had my math wrong because they've changed the number of megaballs. There is now 25.

rough estimate of ticket sales for last night's draw is 842,135 x 25 = 21,053,375 ----> number of winners of the easiest prize (matching the megaball) times the number of megaballs.


I think you were right with what you said earlier but wrong this time.
I think There is a ~1/37 chance of getting the easiest prize and a 1/25 of getting any prize with the mega ball

So 842,145 x 37 = ~ 31 million

I think the lotto report website had it as ~ 30 million tickets; so that is pretty close to 31.

Written by phone and I am at work, so didnít have time to check the figures above
AxelWolf
AxelWolf
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Thanks for this post from:
ksdjdj
June 1st, 2019 at 8:07:48 PM permalink
Quote: ksdjdj



Written by phone and I am at work, so didnít have time to check the figures above

Please stop using four-letter words around here.
♪♪Now you swear and kick and beg us That you're not a gamblin' man Then you find you're back in Vegas With a handle in your hand♪♪ Your black cards can make you money So you hide them when you're able In the land of casinos and money You must put them on the table♪♪ You go back Jack do it again roulette wheels turinin' 'round and 'round♪♪ You go back Jack do it again♪♪
Romes
Romes
Joined: Jul 22, 2014
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June 1st, 2019 at 11:41:43 PM permalink
Quote: ksdjdj

...As I have said in other posts, communication isn't a strength of mine, so feel free to correct me if I misread your post.

All good, I just meant odds would be:

P(MM Hit) = (1/259,000,000)*NumEntries = (1/259,000,000)*~30,000,000 = ~11.58%, which means the no is about 89.42%, but of course I'm "rounding" the odds and ticket sales... I was just looking to confirm this was also what you were doing to get the proper "No" odds.
Playing it correctly means you've already won.
ksdjdj
ksdjdj
Joined: Oct 20, 2013
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June 2nd, 2019 at 4:33:29 AM permalink
Quote: Romes

All good, I just meant odds would be:

P(MM Hit) = (1/259,000,000)*NumEntries = (1/259,000,000)*~30,000,000 = ~11.58%, which means the no is about 89.42%, but of course I'm "rounding" the odds and ticket sales... I was just looking to confirm this was also what you were doing to get the proper "No" odds.


No, i think you are wrong, but I am not good at explaining why, I think it has something to do with binomial distribution ?

see links below:

https://en.wikipedia.org/wiki/Binomial_distribution

https://www.mathsisfun.com/data/binomial-distribution.html

note: on the wikipedia page, scroll down to the "biased coin example" (about 20% of the way down the page).
tringlomane
tringlomane
Joined: Aug 25, 2012
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ksdjdjRomes
June 2nd, 2019 at 3:41:42 PM permalink
Quote: Romes

All good, I just meant odds would be:

P(MM Hit) = (1/259,000,000)*NumEntries = (1/259,000,000)*~30,000,000 = ~11.58%, which means the no is about 89.42%, but of course I'm "rounding" the odds and ticket sales... I was just looking to confirm this was also what you were doing to get the proper "No" odds.



This can give a rough estimate if ticket sales are low, but will be significantly off if ticket sales are higher.

The more precise way is:

Chance of "rollover" = (1 - "probability of winning")^number of tickets sold"

Using the numbers in your example this is:

(258,999,999/259,000,000)^30,000,000 = 89.06%.
Ace2
Ace2
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June 2nd, 2019 at 10:12:34 PM permalink
Quote: tringlomane

This can give a rough estimate if ticket sales are low, but will be significantly off if ticket sales are higher.

The more precise way is:

Chance of "rollover" = (1 - "probability of winning")^number of tickets sold"

Using the numbers in your example this is:

(258,999,999/259,000,000)^30,000,000 = 89.06%.

Assuming these figures are correct, the chance of a rollover will be higher than 89.06 % since the 30 million picks are not all random (some people make picks based on birthdays, lucky numbers etc). ..There will be many more duplicate picks than if they were all random.
ksdjdj
ksdjdj
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June 3rd, 2019 at 2:53:58 PM permalink
Yeah I read somewhere (can't remember where) that about 70% of the tickets are automatic/machine picks and 30% are "self-picked".

------------------
For the next mega millions draw there is a probable +EV on the "no jackpot winner" prop.

Book: www.5dimes.eu
Prop: "2 No Jackpot Winner 06/04/19"
Odds: - 550
Estimated Tickets: 32 million to 35 million
Estimated Chance: ~89.08% to ~89.96%
Estimated EV: ~ +5.27% to ~ +6.32% (middle ~ +5.8%)
"Staking Method": 1/3 Kelly
Bet Amount: ~ 10.6% of bank roll
Will I have a bet***: yes

***: Even though it doesn't quite make it, using my new "rule" and the "average Estimated EV", I had a bet because there is still "plenty of margin of error^^^" and also the Tuesday draw is usually not as big, in terms of tickets sales (based on similar jackpots when compared to the other drawing day)

^^^: The "break-even" point @ odds of -550, is more than 50 million tickets.
Ace2
Ace2
Joined: Oct 2, 2017
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June 3rd, 2019 at 7:27:17 PM permalink
On the first MM draw after a jackpot win, there are about 11 million (t) tickets sold on average. There are 302,575,350 (c) combinations.

((c - 1) / c)^t =~ 96.4% chance of a rollover. So fair odds would be about -2700.

I must be missing something or -550 is a very +ev opportunity

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