According to the disclaimer, odds of hitting the jackpot drop from 1 in 195mm to 175mm.
The old field pick is 5 in a field of 59 and 1 in a field of 39.
The new field is pick 5 in a field of 59 and 1 in a field of 35.
Now, assuming you would normally buy $2 worth of powerball tickets, your odds off winning under the old method would be 2/195mm= 1 in 97.5mm.
Isn't this far worse now (based purely on hitting the jackpot and not caring about other miscellaneous lower denominated prizes)?
Now here's the biggest question, and the most interesting point of debate. If a person is only willing to buy a $2 worth of powerball tickets under the old method, when the jackpot is in excess of $150mm, how much does the jackpot have to be under the new method to get the same impact.
The way I was thinking about this was that under the old method, nothing really changes about the odds. The x factor is the "willingness to purchase". If you agree with the stated odds of 1 in 195mm and a starting jackpot at 150mm, is it fair to say that the relationship is pro-portional to the adjusted odds? Meaning that true odds for a $2 bet equates to 150mm/97.5mm= $1.53mm per odds. If so, then under the new rules, you would be willing to make a bet only when the jackpot is in excess of $267mm (1.53*175mm)?
What do you all think?
Quote: Asswhoopermcdaddy
Now, assuming you would normally buy $2 worth of powerball tickets, your odds off winning under the old method would be 2/195mm= 1 in 97.5mm.
That's not how that works. Think about what'd it'd mean if you bought more tickets. If I buy 100 tickets, are my odds of winning really 1 in 195,000? The odds of winning on 1 of 2 tickets are approximately 1 in 195,000,000.
Quote: IbeatyouracesI thought that all the winning prize amounts were doubled. The JP does start at $20 million as opposed to $10 million.
The jackpot resets to $40MM now. And the 5+0 prize went up to $1MM (from $200,000). I think the 2+1 prize also adjusted, from $5 to $7 (or maybe that was 3+0?) Anyway, those two outcomes are both $7 now. And the 0+1 prize increased from $3 to $4.
Quote: rdw4potusThat's not how that works. Think about what'd it'd mean if you bought more tickets. If I buy 100 tickets, are my odds of winning really 1 in 195,000? The odds of winning on 1 of 2 tickets are approximately 1 in 195,000,000.
Why isn't it? Though it's 1 in 1,950,000 - you'd need to by 1000 tickets for 195,000 to 1. And to the OP, it's not the NJ Lottery that made the decision, it's the entire Powerball multi-state agency.
100 Tickets that used to cost $100 now cost $200. A Wheel-players lament.
Quote: IbeatyouracesI thought that all the winning prize amounts were doubled. The JP does start at $20 million as opposed to $10 million.
JP are doubled, but many players will not buy a powerball ticket unless the jackpot is much higher. As per this example, a person does not buy a ticket unless the jackpot is 150mm under the old strategy. What would the new benchmark be under the new odds?
I assume mathematically you would weight the odds. Do you agree?
Quote: rdw4potusThat's not how that works. Think about what'd it'd mean if you bought more tickets. If I buy 100 tickets, are my odds of winning really 1 in 195,000? The odds of winning on 1 of 2 tickets are approximately 1 in 195,000,000.
How can the odds of winning not change if you have more tickets? If you play every possible combination, you're guaranteed to win, though at a substantial loss.
The number of tickets purchased impact the odds.
Purchased Combinations*# of ways they can appear in no dependent order / Total number of combinations =~ probability.
In your example, let's do something simpler. I have a six-sided die. You can buy a ticket with the numbers 1 through 6 written on it. The probability of any # being rolled is 1/6. You go and buy 2 distinct tickets. Your probability of winning is now 2/6 = 1/3.
So I calculated the return on all the prizes EXCEPT the jackpot two ways because with so many numbers it's easy to make an inputting mistake.
First, dividing each of the wins by 2 to convert it to a $1 ticket:
500/5154+5/649+50/19100+50/12000+3.5/360+3.5/706+2/110.81+2/55.41=.18
Second, all returns by just the chance to win:
1000000/5153633+10000/648976+100/19088+100/12245+7/360+7/706.4+4/110.8+4/55.42=.36
The expected value of everything except the jackpot is 36 cents for every two dollar ticket, or .18 cents to the dollar. Therefore you need to recover the rest of the 82% ($1.64) from the jackpot. The Wizard analyzed an old version and got around the same value, so this about what the return should be.
1 in 175,223,510 wins the jackpot
$2=.36+J*(1/175223510)
J=287 million
This is before taxes or the annuity. In my opinion, both should be considered.
Currently the jackpot is 250 million with an immediate payment of 156. This comes out to an adjusted jackpot value of .624*(Jackpot won).
.624*(Jackpot won)=287*10^6
(Jackpot won)=460*10^6
We'll call this "Jpa" for "Jackpot post-annuity". So Jpa=460 million
The taxes part is much more difficult to calculate, mainly because I don't know of any loopholes and I've heard you can setup a trust or something. Let's assume they just take a flat 25% and your lawyers can get you out of anything else. Now "Jpa" will be divided by .75 to find your take-home.
.75*x=460 * 10^6
x = 613 million
So by my calculations, when you see the powerball reach 600 million it starts to become an advantage play... but more than likely you'll be having fun with that .18 EV on everything but the jackpot.
If you assume "everything but the jackpot" is the same at .18 EV...
1=.18+x*(1/195,249,054)
Used to start at 160 million.
Ok so about 1.8 times higher with the new rules.