Powerball hits $1 Billion
Quote: Ace2You’re missing something fundamental in your calculation.
link to original post
I might be...the original calculation includes the 1/e "possibility" that you have one of the zero winning tickets. Er...
Then again, if you change it to (the probability of you having a winning ticket) / (the expected number of other winning tickets), then, by my earlier formula, that should be 1, but then the value of the ticket is 1/2.
The problem may be, "the jackpot divided by the expected number of winning tickets" and "the expected value of your share of the jackpot" are two different things, when you take into account that you only get 1/N of the jackpot if there are N winners.
When I calculate this:
P(winning ticket) / (P(nobody else has one) + P(one other ticket) / 2 + P(two other tickets) / 3 + ... + P(N-1 other tickets) / (N-1))
then I get 1 - 1/e.
Any mathoids out there know if there is a simplified exact solution to either of these (the second one is just the first one with every term multiplied by N):
Q^(N-1)
+ 1/2 C(N-1,1) P Q^(N-2)
+ 1/3 C(N-1,2) P^2 Q^(N-3)
+ 1/4 C(N-1,3) P^3 Q^(N-4)
+ ...
+ 1/N P^(N-1)
C(N,1) Q^(N-1)
+ C(N,2) P Q^(N-2)
+ C(N,3) P^2 Q^(N-3)
+ C(N,4) P^3 Q^(N-4)
+ ...
+ C(N,N) P^(N-1)
(1) A PB or Mega Millions play costs $2, $3 if you want the Powerplay or the Megaplier. In Idaho and Montana, Powerball is bundled with Power Play for a minimum purchase price of $3 per play. I keep seeing assumptions based on $1 play, and there is no play for either game that's $1.
(2) How does multiple plays for the same game/draw impact these calculations? For example, last PB I play 3 draws. That was $6. I won $2.
Using the 300 million scenario, look at it this way. You have one ticket and there are 299,999,999 others. That 299.9M is the same as 300 million for all practical/calculation purposes
The drawing is made. Before even considering your one ticket, there is:
e^-1/0! chance of zero other winners giving you a 100% share of the jackpot IF you win. e^-1/0! * 100% is 0.368
e^-1/1! chance of one other winner giving you a 50% share of the jackpot IF you win. e^-1/1!* 50% is 0.184
e^-1/2! chance of two other winners giving you a 33.3% share of the jackpot IF you win. e^-1/2! * 33% is 0.061
Etcetera. Sum those through infinity and you”ll get an expected jackpot share of 63 % and a jackpot only ticket value of 1/e = 63 cents. Edit: Should say 1 - 1/e = 63 cents
I believe you’re a little off on this part of the calculation
Quote: GenoDRPhI don't have a dog in this discussion, other than casually playing PB and MM when prizes get super high. But 2 comments, if I may:
(1) A PB or Mega Millions play costs $2, $3 if you want the Powerplay or the Megaplier. In Idaho and Montana, Powerball is bundled with Power Play for a minimum purchase price of $3 per play. I keep seeing assumptions based on $1 play, and there is no play for either game that's $1.
(2) How does multiple plays for the same game/draw impact these calculations? For example, last PB I play 3 draws. That was $6. I won $2.
link to original post
Are you saying that in Idaho and Montana the minimum purchase price for PB and MM is $3? I have never heard that and it would surprise me. I am pretty sure I bought a ticket in Montana and didn't know that.
Quote: Ace2
Etcetera. Sum those through infinity and you”ll get an expected jackpot share of 63 % and a jackpot only ticket value of 1/e = 63 cents
I believe you’re a little off on this part of the calculation
link to original post
I followed you to here. I think you mean the ticket value is:
Jackpot / (chance of winning * e * cost of ticket)
ETA:
This ignores non jackpot prizes which per Google are worth 0.25 for mega millions and 0.32 for power ball (each a $2 ticket).
Quote: DRichQuote: GenoDRPhI don't have a dog in this discussion, other than casually playing PB and MM when prizes get super high. But 2 comments, if I may:
(1) A PB or Mega Millions play costs $2, $3 if you want the Powerplay or the Megaplier. In Idaho and Montana, Powerball is bundled with Power Play for a minimum purchase price of $3 per play. I keep seeing assumptions based on $1 play, and there is no play for either game that's $1.
(2) How does multiple plays for the same game/draw impact these calculations? For example, last PB I play 3 draws. That was $6. I won $2.
link to original post
Are you saying that in Idaho and Montana the minimum purchase price for PB and MM is $3? I have never heard that and it would surprise me. I am pretty sure I bought a ticket in Montana and didn't know that.
link to original post
The PB website states this verbatim: "Powerball® costs $2 per play. In Idaho and Montana, Powerball is bundled with Power Play® for a minimum purchase price of $3 per play."
The ID state lottery states this verbatim "How to Play
Each play costs $3. Pick five numbers between 1 and 69. Pick one Powerball number between 1 and 26. If you'd like random numbers, ask for a quick pick. Multi-draw tickets can be produced for up to 15 consecutive draws. Drawings are Monday, Wednesday and Saturday. The PowerPlay option is automatically included in every Powerball purchase. PowerPlay for Powerball is a way to multiply your prize of $50,000 or less by 2, 3, 4, 5, or even 10 times. IMPORTANT: The 10 times multiplier is ONLY in play when the advertised annuity jackpot is $150 million or I less. The Match 5 prize with PowerPlay is always $2,000,000. If you match all the numbers you win the jackpot!"
The MT state lottery states this verbatim: "Cost per Play: $3.00
Drawing Days: Mondays, Wednesdays and Saturdays at 8:59 p.m. Mountain Time. You have until 8:00 p.m. on draw days to purchase your Powerball tickets.
Multiple Draws: You may play the same set of numbers for up to 24 consecutive drawings.
Power Play is included with every Powerball purchase! Power Play doubles the Match 5 prize from $1 million to $2 million and increases other prizes according to the chart below. The Power Play number will be drawn separately, but at the same time as the winning numbers for every draw.
Overall odds of winning a prize in Powerball: 1:24.9
Overall Odds: 1:24.9"
Near as I can tell, MM is still 2 bucks everywhere.
The example we’re using here is 300 million tickets sold for a $300 million jackpot and 300 million possible combinations. So those numbers cancel and we’re left with 1 - 1/e = 63 cents. This recent discussion has always been about a “jackpot only” ticket, before tax. The non-jackpot ev is constant and easy to calculateQuote: unJonQuote: Ace2
Etcetera. Sum those through infinity and you”ll get an expected jackpot share of 63 % and a jackpot only ticket value of 1/e = 63 cents
I believe you’re a little off on this part of the calculation
link to original post
I followed you to here. I think you mean the ticket value is:
Jackpot / (chance of winning * e * cost of ticket)
ETA:
This ignores non jackpot prizes which per Google are worth 0.25 for mega millions and 0.32 for power ball (each a $2 ticket).
link to original post
I posted the formula yesterday:
The pretax cash value of a “jackpot-only” ticket is:
j/k * (1 - e^(-k/c))
Where j is jackpot value, k is tickets sold and c is number of ticket combinations.
the cost of a megamillions ticket can be $1.50 if you do the just the jackpot option. You get 2 quick pick for $3 but the only thing you can win is the jackpot. 5 numbers correct but incorrect megaball pay zilch.Quote: GenoDRPhQuote: DRichQuote: GenoDRPhI don't have a dog in this discussion, other than casually playing PB and MM when prizes get super high. But 2 comments, if I may:
(1) A PB or Mega Millions play costs $2, $3 if you want the Powerplay or the Megaplier. In Idaho and Montana, Powerball is bundled with Power Play for a minimum purchase price of $3 per play. I keep seeing assumptions based on $1 play, and there is no play for either game that's $1.
(2) How does multiple plays for the same game/draw impact these calculations? For example, last PB I play 3 draws. That was $6. I won $2.
link to original post
Are you saying that in Idaho and Montana the minimum purchase price for PB and MM is $3? I have never heard that and it would surprise me. I am pretty sure I bought a ticket in Montana and didn't know that.
link to original post
The PB website states this verbatim: "Powerball® costs $2 per play. In Idaho and Montana, Powerball is bundled with Power Play® for a minimum purchase price of $3 per play."
The ID state lottery states this verbatim "How to Play
Each play costs $3. Pick five numbers between 1 and 69. Pick one Powerball number between 1 and 26. If you'd like random numbers, ask for a quick pick. Multi-draw tickets can be produced for up to 15 consecutive draws. Drawings are Monday, Wednesday and Saturday. The PowerPlay option is automatically included in every Powerball purchase. PowerPlay for Powerball is a way to multiply your prize of $50,000 or less by 2, 3, 4, 5, or even 10 times. IMPORTANT: The 10 times multiplier is ONLY in play when the advertised annuity jackpot is $150 million or I less. The Match 5 prize with PowerPlay is always $2,000,000. If you match all the numbers you win the jackpot!"
The MT state lottery states this verbatim: "Cost per Play: $3.00
Drawing Days: Mondays, Wednesdays and Saturdays at 8:59 p.m. Mountain Time. You have until 8:00 p.m. on draw days to purchase your Powerball tickets.
Multiple Draws: You may play the same set of numbers for up to 24 consecutive drawings.
Power Play is included with every Powerball purchase! Power Play doubles the Match 5 prize from $1 million to $2 million and increases other prizes according to the chart below. The Power Play number will be drawn separately, but at the same time as the winning numbers for every draw.
Overall odds of winning a prize in Powerball: 1:24.9
Overall Odds: 1:24.9"
Near as I can tell, MM is still 2 bucks everywhere.
link to original post
Quote: Ace2The example we’re using here is 300 million tickets sold for a $300 million jackpot and 300 million possible combinations. So those numbers cancel and we’re left with 1 - 1/e = 63 cents. This recent discussion has always been about a “jackpot only” ticket, before tax. The non-jackpot ev is constant and easy to calculateQuote: unJonQuote: Ace2
Etcetera. Sum those through infinity and you”ll get an expected jackpot share of 63 % and a jackpot only ticket value of 1/e = 63 cents
I believe you’re a little off on this part of the calculation
link to original post
I followed you to here. I think you mean the ticket value is:
Jackpot / (chance of winning * e * cost of ticket)
ETA:
This ignores non jackpot prizes which per Google are worth 0.25 for mega millions and 0.32 for power ball (each a $2 ticket).
link to original post
I posted the formula yesterday:
The pretax cash value of a “jackpot-only” ticket is:
j/k * (1 - e^(-k/c))
Where j is jackpot value, k is tickets sold and c is number of ticket combinations.
link to original post
Your formula needs cost of ticket.
He tried to talk me out of buying chewing tobacco.
Touché.
Quote: Mission146I tried to talk someone out of buying $20 in tickets today, but I was unsuccessful.
He tried to talk me out of buying chewing tobacco.
Touché.
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Love it. A real chuckle. Trying to stay away from the universal LOL.
tuttigym
Quote: avianrandythe cost of a megamillions ticket can be $1.50 if you do the just the jackpot option. You get 2 quick pick for $3 but the only thing you can win is the jackpot. 5 numbers correct but incorrect megaball pay zilch.Quote: GenoDRPhQuote: DRichQuote: GenoDRPhI don't have a dog in this discussion, other than casually playing PB and MM when prizes get super high. But 2 comments, if I may:
(1) A PB or Mega Millions play costs $2, $3 if you want the Powerplay or the Megaplier. In Idaho and Montana, Powerball is bundled with Power Play for a minimum purchase price of $3 per play. I keep seeing assumptions based on $1 play, and there is no play for either game that's $1.
(2) How does multiple plays for the same game/draw impact these calculations? For example, last PB I play 3 draws. That was $6. I won $2.
link to original post
Are you saying that in Idaho and Montana the minimum purchase price for PB and MM is $3? I have never heard that and it would surprise me. I am pretty sure I bought a ticket in Montana and didn't know that.
link to original post
The PB website states this verbatim: "Powerball® costs $2 per play. In Idaho and Montana, Powerball is bundled with Power Play® for a minimum purchase price of $3 per play."
The ID state lottery states this verbatim "How to Play
Each play costs $3. Pick five numbers between 1 and 69. Pick one Powerball number between 1 and 26. If you'd like random numbers, ask for a quick pick. Multi-draw tickets can be produced for up to 15 consecutive draws. Drawings are Monday, Wednesday and Saturday. The PowerPlay option is automatically included in every Powerball purchase. PowerPlay for Powerball is a way to multiply your prize of $50,000 or less by 2, 3, 4, 5, or even 10 times. IMPORTANT: The 10 times multiplier is ONLY in play when the advertised annuity jackpot is $150 million or I less. The Match 5 prize with PowerPlay is always $2,000,000. If you match all the numbers you win the jackpot!"
The MT state lottery states this verbatim: "Cost per Play: $3.00
Drawing Days: Mondays, Wednesdays and Saturdays at 8:59 p.m. Mountain Time. You have until 8:00 p.m. on draw days to purchase your Powerball tickets.
Multiple Draws: You may play the same set of numbers for up to 24 consecutive drawings.
Power Play is included with every Powerball purchase! Power Play doubles the Match 5 prize from $1 million to $2 million and increases other prizes according to the chart below. The Power Play number will be drawn separately, but at the same time as the winning numbers for every draw.
Overall odds of winning a prize in Powerball: 1:24.9
Overall Odds: 1:24.9"
Near as I can tell, MM is still 2 bucks everywhere.
link to original post
link to original post
Your mileage may vary. I know for a fact that MA does not offer the $1.50 jackpot only play.
Quote: Mission146I tried to talk someone out of buying $20 in tickets today, but I was unsuccessful.
He tried to talk me out of buying chewing tobacco.
Touché.
link to original post
I quit chewing 15 years ago, still miss it.
Bought 3 tickets today what can I say it;s the only sucker bet I intentionally make and only
after it reaches 1B.
Quote: unJonYour formula needs cost of ticket.
link to original post
No - the "value" is what the ticket is worth once you own it. I think "return" is the amount of profit (i.e. value minus cost).
Obviously, if the value is less than the cost, then it is not a good play.
Quote: rainman
I quit chewing 15 years ago, still miss it.
Bought 3 tickets today what can I say it;s the only sucker bet I intentionally make and only
after it reaches 1B.
I hate to say it, but I bought a lottery ticket too.
Quote: DRichQuote: rainman
I quit chewing 15 years ago, still miss it.
Bought 3 tickets today what can I say it;s the only sucker bet I intentionally make and only
after it reaches 1B.
I hate to say it, but I bought a lottery ticket too.
link to original post
The sacrilege! The blasphemy!
Some lottery commissions offer these.
For the cost of a stamp, you may get to play again.
If you can get non-winning tickets for a good price...
Quote: Mission146Quote: DRichQuote: rainman
I quit chewing 15 years ago, still miss it.
Bought 3 tickets today what can I say it;s the only sucker bet I intentionally make and only
after it reaches 1B.
I hate to say it, but I bought a lottery ticket too.
link to original post
The sacrilege! The blasphemy!
link to original post
I know, I am embarrassed. At least I don't play negative EV slots.
Rounded numbers using j/k * (1 - e^(-k/c)) to get $1.78
So it could potentially and theoretically be slightly +EV for a foreigner with no tax on winnings. That said, I bet the US federal government would find some justification to withhold tax on a foreigner. We’re talking over $200 million in tax.
Quote: Ace2Assuming a $614 million jackpot, 100 million tickets sold (85 million last drawing) and 292 million ticket combinations, the pre-tax expected value of a ticket is $1.78. Take 40% tax off and it’s $1.07 or 53% of the $2 ticket cost.
Rounded numbers using j/k * (1 - e^(-k/c)) to get $1.78
link to original post
If you are going to do a real example you should add the $0.32 cents for non jackpot winnings.
I just did, I was still editingQuote: unJon
If you are going to do a real example you should add the $0.32 cents for non jackpot winnings.
link to original post
e^(-115/292) =~ 67%
That chance is 50% at Ln(2) * 292 million = 202 million tickets sold
Quote: DRichQuote: rainman
I quit chewing 15 years ago, still miss it.
Bought 3 tickets today what can I say it;s the only sucker bet I intentionally make and only
after it reaches 1B.
I hate to say it, but I bought a lottery ticket too.
link to original post
No winner it's over 1.5 I'm counting on you to re-up so I won't be the only sucker here.
Quote: Ace2I posted the formula yesterday:
The pretax cash value of a “jackpot-only” ticket is:
j/k * (1 - e^(-k/c))
Where j is jackpot value, k is tickets sold and c is number of ticket combinations.
link to original post
For those of you AHEM those of us that's better interested in where that formula comes from:
Let p = the probability of winning = 1 / c
Assuming you have a winning ticket, the amount you win = j divided by (1 + the number of other winning tickets)
In a Poisson distribution, the probability of exactly n winners under the given conditions (k tickets, winning probability c) is:
P(n winners) = (kp)^n / (e^(kp) n!)
In this case, your share of the money would be j / (n + 1) * (kp)^n / (e^(kp) n!)
= j (kp)^n / (e^(kp) n! (n + 1))
= j (kp)^n / (e^(kp) (n + 1)!)
= j ((kp)^(n + 1) / kp) / (e^(kp) (n + 1)!)
= j / (kp) * ((kp)^(n + 1) / (e^(kp) (n + 1)!)
= j / (kp) * P(n + 1) other winners
The expected return if you have a winning ticket = j * (1 * P(0 other winners) + 1/2 P(1 other winner) + 1/3 P (2 other winners) + ...)
= j / (kp) * (P(1 other winner) + P(2 other winners) + P(3 other winners) + ...)
Since P(0 other winners) + P(1 other winner) + P(2 other winners) + ... = 1, this equals j / (kp) * (1 - P(0 other winners)
= j / (kp) * (1 - 1 / e^(kp))
Substitute p = 1 / c and 1 / e^x = e^(-x):
= j / (k/c) * (1 - e^(-(k/c)))
...but you still have to multiply by p (= 1 / c), which is the probability of you having a winning ticket in the first place:
= j / k * (1 - e^(-(k/c)))
The question is:Quote: ThatDonGuyQuote: Ace2I posted the formula yesterday:
The pretax cash value of a “jackpot-only” ticket is:
j/k * (1 - e^(-k/c))
Where j is jackpot value, k is tickets sold and c is number of ticket combinations.
link to original post
For those of you AHEM those of us that's better interested in where that formula comes from:
Let p = the probability of winning = 1 / c
Assuming you have a winning ticket, the amount you win = j divided by (1 + the number of other winning tickets)
In a Poisson distribution, the probability of exactly n winners under the given conditions (k tickets, winning probability c) is:
P(n winners) = (kp)^n / (e^(kp) n!)
In this case, your share of the money would be j / (n + 1) * (kp)^n / (e^(kp) n!)
= j (kp)^n / (e^(kp) n! (n + 1))
= j (kp)^n / (e^(kp) (n + 1)!)
= j ((kp)^(n + 1) / kp) / (e^(kp) (n + 1)!)
= j / (kp) * ((kp)^(n + 1) / (e^(kp) (n + 1)!)
= j / (kp) * P(n + 1) other winners
The expected return if you have a winning ticket = j * (1 * P(0 other winners) + 1/2 P(1 other winner) + 1/3 P (2 other winners) + ...)
= j / (kp) * (P(1 other winner) + P(2 other winners) + P(3 other winners) + ...)
Since P(0 other winners) + P(1 other winner) + P(2 other winners) + ... = 1, this equals j / (kp) * (1 - P(0 other winners)
= j / (kp) * (1 - 1 / e^(kp))
Substitute p = 1 / c and 1 / e^x = e^(-x):
= j / (k/c) * (1 - e^(-(k/c)))
...but you still have to multiply by p (= 1 / c), which is the probability of you having a winning ticket in the first place:
= j / k * (1 - e^(-(k/c)))
link to original post
What part of the formula/calculation did I not explain on Oct 5-6?
I even helped you understand how the multiple-winner discount factor is derived. Perhaps I don’t go into as much detail as you because I assume the audience here has an above average math level. I do not pull such a formula out of thin air and I’m glad to elaborate further if anyone asks
Quote: Ace2The question is:Quote: ThatDonGuyQuote: Ace2I posted the formula yesterday:
The pretax cash value of a “jackpot-only” ticket is:
j/k * (1 - e^(-k/c))
Where j is jackpot value, k is tickets sold and c is number of ticket combinations.
link to original post
For those of you AHEM those of us that's better interested in where that formula comes from:
Let p = the probability of winning = 1 / c
Assuming you have a winning ticket, the amount you win = j divided by (1 + the number of other winning tickets)
In a Poisson distribution, the probability of exactly n winners under the given conditions (k tickets, winning probability c) is:
P(n winners) = (kp)^n / (e^(kp) n!)
In this case, your share of the money would be j / (n + 1) * (kp)^n / (e^(kp) n!)
= j (kp)^n / (e^(kp) n! (n + 1))
= j (kp)^n / (e^(kp) (n + 1)!)
= j ((kp)^(n + 1) / kp) / (e^(kp) (n + 1)!)
= j / (kp) * ((kp)^(n + 1) / (e^(kp) (n + 1)!)
= j / (kp) * P(n + 1) other winners
The expected return if you have a winning ticket = j * (1 * P(0 other winners) + 1/2 P(1 other winner) + 1/3 P (2 other winners) + ...)
= j / (kp) * (P(1 other winner) + P(2 other winners) + P(3 other winners) + ...)
Since P(0 other winners) + P(1 other winner) + P(2 other winners) + ... = 1, this equals j / (kp) * (1 - P(0 other winners)
= j / (kp) * (1 - 1 / e^(kp))
Substitute p = 1 / c and 1 / e^x = e^(-x):
= j / (k/c) * (1 - e^(-(k/c)))
...but you still have to multiply by p (= 1 / c), which is the probability of you having a winning ticket in the first place:
= j / k * (1 - e^(-(k/c)))
link to original post
What part of the formula/calculation did I not explain on Oct 5-6?
link to original post
The part where you show that the sum actually adds up to what you say it does.
Gonna buy myself a new comb!
Quote: rxwineI won $4!
Gonna buy myself a new comb!
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If i was fortunate enough to win $4, I would probably opt for a chili dog.
Hopefully I win it tonight haha...Quote: TigerWuPowerball is getting close to a billion again. Probably get there this week or next if no one wins in the next couple drawings.
link to original post
$100 winner, four numbers. Missed 13 and the powerball number. Still on a weird 'hitting at least a few numbers' streak, but can't manage to guess the other couple of numbers. Main strategy is the theory that the physical balls aren't changed out but every so often, and aren't calibrated perfectly so that certain balls will float into the tube more often than other numbers due to those imperfections. Picked a random assortment of numbers that have hit within the past 5 weeks or so.
Quote: ChallengedMilly(I'm aware such analysis/ideas are mathematically wrong...)
$100 winner, four numbers. Missed 13 and the powerball number. Still on a weird 'hitting at least a few numbers' streak, but can't manage to guess the other couple of numbers. Main strategy is the theory that the physical balls aren't changed out but every so often, and aren't calibrated perfectly so that certain balls will float into the tube more often than other numbers due to those imperfections. Picked a random assortment of numbers that have hit within the past 5 weeks or so.
link to original post
I don't know about Powerball, but California's lottery has multiple sets of balls, in part because of the possibility that not every ball in a set is identical.
However, there's no need to be mathematically correct in this instance; there is the (in)famous case of the "devil's lottery," where the Daily Number draw in Pennsylvania on 4/24/1980 came up 666, because somebody had weighted all of the balls not numbered 4 or 6, and bet heavily on those 8 possible results (444, 446, 464, and so on).
Quote:SALEM, Ore. (AP) — One of the winners of a $1.3 billion Powerball jackpot this month is an immigrant from Laos who has had cancer for eight years and had his latest chemotherapy treatment last week.
Cheng “Charlie” Saephan, 46, of Portland, told a news conference held by the Oregon Lottery on Monday that he and his 37-year-old wife, Duanpen, are taking half the money, and the rest is going to a friend, Laiza Chao, 55, of the Portland suburb of Milwaukie. Chao had chipped in $100 to buy a batch of tickets with them. They are taking a lump sum payment, $422 million after taxes.
“I will be able to provide for my family and my health,” he said, adding that he'd “find a good doctor for myself.”
Saephan, who has two young children, said that as a cancer patient, he wondered, “How am I going to have time to spend all of this money? How long will I live?”
After they bought the shared tickets, Chao sent a photo of the tickets to Saephan and said, “We’re billionaires.” It was a joke before the actual drawing, he said, but the next day they won.
Chao, 55, was on her way to work when Saephan called her with the news: “You don't have to go anymore,” he said.
Saephan said he was born in Laos and moved to Thailand in 1987, before immigrating to the U.S. in 1994. He wore a sash at the news conference identifying himself as Iu Mien, a southeast Asian ethnic group with roots in southern China. Many Iu Mein were subsistence farmers and assisted American forces during the Vietnam war; after the conflict, thousands of Iu Mien families fled to Thailand to avoid retribution and eventually settled in the U.S.
h
Tens of thousands of Iu Mien people live along the West Coast, with a sizeable and active community in Portland.
Saephan graduated from high school in 1996 and has lived in Portland for 30 years. He worked as a machinist for an aerospace company.
In the weeks leading up to the drawing, he wrote out numbers for the game on a piece of paper and slept with it under his pillow, he said. He prayed that he would win, saying, “I need some help — I don't want to die yet unless I have done something for my family first.”
