ksdjdj
Joined: Oct 20, 2013
• Posts: 894
December 19th, 2019 at 12:36:35 AM permalink
For the Powerball draw on the 12/18/19, I put 3,562.5 down to win 375.

Opening Bal: 25,005
Finishing Bal: 25,380
P/(L) for all bets resolved to date: (620)
Chance of no winner (based on actual tickets sold): 95.26...%
Odds Taken: -950
Actual EV: 5.29...%
Tickets Sold***: 14,175,783
*** http://lottoreport.com/ticketcomparison.htm

----
Other:

If anyone wants to know how I lock in the odds that I am taking "safely", in other words "without over-betting my bankroll/budget", then read below:

At 5dimes there is a thing called "if bets" and "rolling if bets", they are similar to having a "parlay", but compared to a "parlay" you have more control over how much to bet "on the 2nd leg" of an "if bet" (see example and steps below):

Example:

Note: you are betting "full kelly" in this example just so that there are "less things to do in working out the figures" (I am still planning to do about 1/3 kelly for all my "real-life" bets)

Say you had \$1000 budget and you saw the odds below for the next two "lotto draws" on the "no winner bet" (for both draws the odds are +100 just to make it "simple").

You estimate that you have a 60% chance of the "no winner" bet winning in each draw
You are "nearly certain" the "no winner odds" are going to get worse/ shorten, but you also want to get more money down if your first bet wins (see steps below on how to do it).

1. Bet 20% of your current bank roll of \$1000, on the first lottery (lotto 1) @ +100
2. Have 20% of your (new) current bank roll of \$800, on the second lottery (lotto 2) @ +100.
3. There should be a "heading" that has "RIF" as part of the name on the side of the screen, click on it.
4. Select "lotto 2", then click on "no winner", and then when you go to place the bet there should be an icon that says something like "IF" or "RIF" next to where you place the bet, click on that icon.
5. Then you should see a list of bets that you are allowed to use, and one of them should be your bet on "lotto 1", select that one.
6. For the "if bet" you have to "pretend that the bet in step 1" had won, and use that figure as part of your working out (in this case, your balance would have been \$1200 if you only bet on lotto 1, and the bet won).
7. Take away the figure in step 6. from step 2., in this case you should get \$400 (1200 minus 800).
8. Multiply the figure in step 7. by the "Kelly bet %", In this case you should select \$80 as your "if bet", from lotto 1 into lotto 2 (since \$80 is 20% of your "extra money" if the lotto 1 bet wins).
9. If done correctly your total bet on lotto 2 should be:
(a) \$240, if the lotto 1 bet wins
(b) \$160, if the lotto 1 bet loses

Also, In this example one of the four outcomes listed below would occur:
(i) "both bets lose": you lose \$360 (\$200 from the first bet and \$160 from the 2nd bet)
(ii) "bet 1 wins and bet 2 loses": you lose \$40 (you win \$200 on the first bet, and lose \$240 on the 2nd).
(iii) "bet 1 loses and bet 2 wins": you lose \$40 (you lose \$200 on the first bet, and win \$160 on the 2nd).
(iv) "bet 1 and 2 win": you win \$440 (win \$200 on bet 1 and win \$240 on bet 2).

Hope I explained the steps well enough on how to use an "if bet/ rolling if bet" effectively (for when you want to "lock in the early prices, because you think the odds will be worse in the future").

Note: if you think the odds will "stay the same" or "get better", then you can forget about the "if bet" and bet "the more normal way" (by just having "two single bets" closer to the draws).

----
Other 2:

Here is the analysis for the next MM draw:

I just used 10 million tickets as the estimate for the tables below (If i was betting, I would probably use a range of 8.5 to 10.5 million tickets)

No Winner (odds @ -4000)

Outcome Pays Probability Return
No winner 0.025 0.9674... 0.0241...
Winner -1 0.0325... -0.0325...
Total 1.0000 -0.0083...

Winner (odds +2000)

Outcome Pays Probability Return
Winner 20 0.0325... 0.6501...
No winner -1 0.9674.. -0.9674...
Total 1.0000 -0.3173...

So the estimated HOUSE EDGE^^^ is about 0.8% for the "no winner prop", and 31.7% for the "winner prop", in the next Mega Millions draw.

^^^: I wrote "house edge" in capitals because most of the time I have been using terms like "EV", in previous posts.

Note: I don't like doing tables, but I also think the "info looks nicer in a table" (so don't expect me to do it like this too often).

Also note: So that I could put up the table faster, I just "copied and pasted" the table, and a lot of the "wording" from the link below (pretty much just changed the relevant figures, where necessary):

https://wizardofvegas.com/forum/gambling/sports/31799-no-jackpot-winner-powerball-prop-bet/#post682169
Last edited by: ksdjdj on Dec 19, 2019
ksdjdj
Joined: Oct 20, 2013
• Posts: 894
December 19th, 2019 at 11:54:19 AM permalink
Betting Summary:

Game: Powerball
Book: www.5dimes.eu
Prop: "No Jackpot Winner 12/21/19"
Odds: -1000
Estimated Chance: 94.36...% (based on an estimate of 16.94*** million tickets sold)
Estimated EV: +3.80...%
Bet Amount (%): ~ 12.7% of bank roll (about 1/3 Kelly)
(New###) Bet Amount (\$): \$3,220
When will I have a bet: now

***: I used the link>>>> https://wizardofodds.com/games/lottery/ticket-sales-calculator/ >>> then typed in 171 for "Jackpot size (in millions):" >>> lastly I clicked on "Powerball" to get the ticket estimate for this draw.

###: I have decided to add this line so that you know what figure I am using in \$'s before the draw. Remember, the figures used in the "results summary page" are currently "simulated".

----
Other (Not necessary for the bets we are having, but I still think it is interesting):

From when the "jackpot is reset", you would expect about 15 or 16 draws on average^^^, if a bookie put up a market for "how many rolls/draws will it take for the next Jackpot to hit in the Powerball?".

^^^: In other words about half the time it will be hit above that many draws, and half the time the jackpot will be hit below that many draws.

Important: This does NOT change the "chances in each draw", as all draws are independent of each other. In other words, knowing this doesn't change what we "plan to bet" for each draw (for the bets we are having, all we need to know are: the odds offered, the estimated tickets, the estimated EV, and the amount we plan to bet)

I just used the site below to work it out "very rough"
https://www.txlottery.org/export/sites/lottery/Documents/jackpotestimates/pb20191218.pdf

I may post a similar one for the Mega Million later, and I would expect it to be a larger number of "average rolls between jackpots hit", mainly because the chance for hitting the Jackpot in the MM is slightly lower, also tickets sales for the MM are usually less than the PB (at least for jackpots under about \$200 million).

Once a new "jackpot roll" has started, the "rough average" number of draws for the MM jackpot to be hit is: 18.
Last edited by: ksdjdj on Dec 19, 2019
Wizard
Joined: Oct 14, 2009
• Posts: 21934
December 20th, 2019 at 11:33:02 AM permalink
Quick post today.

I found no winner for the Mega Millions to be a negative bet at 5Dimes.

On the PowerBall I bet \$500 to win \$50, with a perceived player advantage of 3.81%.
It's not whether you win or lose; it's whether or not you had a good bet.
ksdjdj
Joined: Oct 20, 2013
• Posts: 894
December 21st, 2019 at 11:16:23 PM permalink
For the Powerball draw on the 12/21/19, I put 3,220 down to win 322.

Opening Bal: 25,380
Finishing Bal: 25,702
P/(L) for all bets resolved to date: (298)
Chance of no winner (based on actual tickets sold): 94.51...%
Odds Taken: -1000
Actual EV: 3.96...%
Tickets Sold***: 16,491,521
*** http://lottoreport.com/ticketcomparison.htm

----
Other:

The next draw is the Christmas draw, so I will only be putting up estimated tickets on the day/night of the draw^^^, as that is when I think I will get the "best estimate".

^^^: I plan to either bet late or not at all for this draw.

Also, my plan to "bet later" is mainly for "subjective reasons###" , but I will still put up a ticket estimate on the day/night of the draw (whether I have a bet or not)

### : As I think I said in earlier posts, things like:
1. "Christmas/New Years' Bump": not enough data to prove, especially since i only think it would be significant for draws above \$200 or \$300 million (again subjective, as there is not enough data to prove conclusively).
2. "Saturday vs Wednesday": Wednesday draws usually have less tickets sold than Saturday draws
(technically subjective, even I think I could eventually prove or disprove this theory, since there is a lot more data available to test it, when compared to the "...Bump" in point "1.")

Below are the two best links to analyze the next Powerball draw (eg, to estimate tickets sold etc)

https://wizardofodds.com/games/lottery/ticket-sales-calculator/

Note: the advertised jackpot is currently 183 million for the draw on the 12/25/19.

https://www.txlottery.org/export/sites/lottery/Documents/jackpotestimates/pb20191221.pdf

And on the day of the next PB, this one below would probably be the "single best one" to use.

https://www.txlottery.org/export/sites/lottery/Documents/jackpotestimates/pb20191225.pdf

Note: For the "txlottery" pages. to get the estimated tickets remember to divide the figure(s) in the "Sales" column by 2 (since it costs \$2 per ticket).

Note 2: The link that says "...pb20191225.pdf" at the end, won't work until the 12/25/19.
Last edited by: ksdjdj on Dec 21, 2019
weezrDASvegas
Joined: Feb 2, 2018
• Posts: 69
December 22nd, 2019 at 4:28:55 AM permalink
It is a matter of precision calculatin the odds of winning any lotto including powerball, mega mils. We need to know number of tickets sold - theose are estimates only. We can do some approx. good calculations by converting estimated jackpots to estimated number of tickets sold.

There is a resource with many cases regarding such calculations with plenty formulae. When the jackpot exceeds the odds the player has an advantage – no more house advantage! The cases are rare and the odds still extreme.

\if this editor doesn’t allow active links, google instead Odds to Win Powerball, Mega Millions Jackpot, Tickets Sold – you’ll find the link with the same name.
https://saliu.com/powerball-jackpot-odds.html
Odds to Win Powerball, Mega Millions Jackpot, Tickets Sold
ksdjdj
Joined: Oct 20, 2013
• Posts: 894
December 22nd, 2019 at 5:42:49 AM permalink
Quote: weezrDASvegas

It is a matter of precision calculatin the odds of winning any lotto including powerball, mega mils. We need to know number of tickets sold - theose are estimates only. We can do some approx. good calculations by converting estimated jackpots to estimated number of tickets sold.

There is a resource with many cases regarding such calculations with plenty formulae. When the jackpot exceeds the odds the player has an advantage – no more house advantage! The cases are rare and the odds still extreme.

\if this editor doesn’t allow active links, google instead Odds to Win Powerball, Mega Millions Jackpot, Tickets Sold – you’ll find the link with the same name.
https://saliu.com/powerball-jackpot-odds.html
Odds to Win Powerball, Mega Millions Jackpot, Tickets Sold

Hi weezrDASvegas,
I think there is some interesting stuff in the link you provided.
But I don't agree with a lot of the claims on the site because I think the "wording is wrong/misleading" (at least in relation to this thread in general).
I am not saying this is done intentionally and don't really want to get into a long/heated argument about it, because I am quite often not very good at getting my point across (I am sure other people on this site would probably confirm that my "lotto math" posted in this thread is solid though).

Below is an example of one section of the site that i think got it "wrong" (it is about half-way down the page, in the link that you provided >>> https://saliu.com/powerball-jackpot-odds.html )

"... The Powerball game has higher odds beginning October 2015: 5 regular numbers from 1 to 69 and 1 Power Ball from 1 to 26. The odds to hit the jackpot (5 regulars AND the Power Ball) are 1 in 292201338 (the worst in the lottery business).
The January 13, 2016 drawing has an estimated jackpot of 1.5 billion dollars — staggering, the largest jackpot ever in the lottery business.
The winnings pool allocated to the grand prize (jackpot) is set to 68% of ticket sales. That means that some 1500000000 / .68 = 2.2 billion tickets sold.
The calculations that the Powerball jackpot will be won on January 16, 2016 —
p = 1 / 292,201,338
N = 2,200,000,000

Doing the calculations in SuperFormula, we get the most accurate result:

DC = 99.95%. ... "

I agree that IF (big if) the tickets sold in a single draw were 2.2 billion tickets, then yes "there would be about a 99.95% chance of at least 1 winner" (but that is where the agreement ends for the above example).
The reason for this disagreement is, according to the link here >>> http://lottoreport.com/ticketcomparison16.htm <<<
635,103,137*** tickets were sold in the draw on January 13, 2016.
***: For "100% randomly chosen tickets" that is about an 11.4% chance of not being hit, so in other words, that is an ~88.6% chance of at least 1 winner.

I hope I have made my point clearly enough, but if I haven't I am sure someone "better at writing" will eventually tell you why I am probably more "correct" (at least in the context of what has been written about previously in this thread).
Wizard
Joined: Oct 14, 2009
• Posts: 21934
Thanks for this post from:
December 22nd, 2019 at 6:49:48 PM permalink
Let's look at the drawings for 12/24 and 12/25

Date: Dec 24
Lottery: Mega Millions
Jackpot: 45M
Estimated sales: 9.74M
Probability of winner: 3.17%
Probability no winner: 96.83%
Fair line no winner: -3055
5 Dimes line: -3500
Wiz bet: \$0

Date: Dec 25
Lottery: PowerBall
Jackpot: 183M
Estimated sales: 17.49M
Probability of winner: 5.81%
Probability no winner: 94.19%
Fair line no winner: -1621
5 Dimes line: -1000
Wiz bet: \$400 to win \$40.

All due respect to ksd for the Christmas bump theory, but I'm in a gambling mood and am betting it anyway! May nobody win (hahahahha!!!). I am such a Christmas scrooge.
It's not whether you win or lose; it's whether or not you had a good bet.
ksdjdj
Joined: Oct 20, 2013
• Posts: 894
December 22nd, 2019 at 8:42:23 PM permalink
Quote: Wizard

Let's look at the drawings for 12/24 and 12/25

Date: Dec 24
Lottery: Mega Millions
Jackpot: 45M
Estimated sales: 9.74M
Probability of winner: 3.17%
Probability no winner: 96.83%
Fair line no winner: -3055
5 Dimes line: -3500
Wiz bet: \$0

Date: Dec 25
Lottery: PowerBall
Jackpot: 183M
Estimated sales: 17.49M
Probability of winner: 5.81%
Probability no winner: 94.19%
Fair line no winner: -1621
5 Dimes line: -1000
Wiz bet: \$400 to win \$40.

All due respect to ksd for the Christmas bump theory, but I'm in a gambling mood and am betting it anyway! May nobody win (hahahahha!!!). I am such a Christmas scrooge.

Good luck, there is probably a nearly 100% chance that I will bet on it too, just going to wait closer to the draw.
There is even a good chance that it may be a "Christmas dip" (at least this time) because according to the Powerball websites' own "early estimate" on the Dec 21, they are expecting to sell 11.48 million*** tickets ??? (if it is not a typo or other error?)

***: \$22,964,768 (estimated PB sales in \$) from the link here >>>> https://www.txlottery.org/export/sites/lottery/Documents/jackpotestimates/pb20191221.pdf

Merry Christmas, and "may nobody win" to you too. : )
Last edited by: ksdjdj on Dec 22, 2019
Wizard
Joined: Oct 14, 2009
• Posts: 21934
December 23rd, 2019 at 7:56:38 AM permalink
Quote: ksdjdj

Merry Christmas, and "may nobody win" to you too. : )

You too!

I was listening to the radio on the drive home yesterday and it spoke of a world-record being broken in Spain for the largest lottery jackpot -- El Gordo.

‘El Gordo’ Lottery in Spain Spreads Riches Worth \$2.6 Billion.

What I don't get is the article says the winning "number" was 26590. Wikipedia says the draw is 5 numbers from 1 to 54 and one from 0 to 9. The odds of winning the jackpot are 1 in 31,625,100.
It's not whether you win or lose; it's whether or not you had a good bet.
weezrDASvegas
Joined: Feb 2, 2018
• Posts: 69
December 23rd, 2019 at 9:02:05 AM permalink
You right – don’t put hard on yourself. aint easy to explain these things. Only the exact number of tickets could calculate the most precise odds of winning the jackpot. and that’s not enough cos you see that 95% odds do not have a winner sometimes.

So everything is a guestimate like the site you refer to. You need know the percentage of sales allocated to prizes, percentage of prize money allocated to jackpot – then estimate total tickets sold. Then there is one more problem if one ticket consists of two combos… know what I mean?

If you could play all possible combos then you’re guaranteed to win. but who could possibly fill millions of tickets by hand??? Maybe one day they shall allow playing online state lotteries wher you could download a file with all combos and deposit the right amount of monies. But hey one more problem – there more winning tickets not only yours!!! You paid 300 mil to cover all powerball combos but there are 4 winners 250 mil per ticket say.

The aussies who won Virginia loto in 1980s were lucky… they had the only winning ticket of some 100 mil and they paid some 14 mil tickets. They thought they covered all combos but they played quick pix or randomly chosen combos that covered some 63% of possibilities cos of repeats.

Anyway this is the best oportunity to play the big lotos when the jackpot is higher than the odds. You got a players advantage. But you need play millions such situations if we live that long.

So methinks I beat you in getting the point across… wink…wink!