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• United States
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March 28, 2019
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Say two people each buy one quick pick for Powerball. (No powerplay or double play add-on, just a normal \$2 ticket) What is the probability they win the same amount?

The reason for the question is I recently bought qp's for two people and one of the wanted to bet the other they would win the same amount. At first I thought it was a bad bet, then I realized the probability of both of them winning \$0 is pretty high. I could run some simulations to estimate the probability, but I wanted to know what it is exactly -- some huge numerator over an even bigger denominator.

• United States
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March 28, 2019
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This morning it dawned on me that the answer depends on whether or not you live in California, where the prize amounts are not fixed from drawing to drawing. In most of the country 4+0 and 3+1 both pay \$100, 3+0 and 2+1 both pay \$7, and 1+1 and 0+1 both pay \$4.

I figured out the exact probability of two random quick picks earning the same prize in a normal state is

39,385,084,830,437,945/42,690,810,964,495,122 ≈ 92.3%

(That includes the likelihood they both win \$0)

Unfortunately, you can't really do an exact calculation for California because sometimes the lower prize amounts happen to be equal and sometimes they are off from each other by a dollar. ¯\_(ツ)_/¯

• Belgium
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April 17, 2021
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Quote: Originally posted by cottoneyedjoe on Apr 28, 2022

This morning it dawned on me that the answer depends on whether or not you live in California, where the prize amounts are not fixed from drawing to drawing. In most of the country 4+0 and 3+1 both pay \$100, 3+0 and 2+1 both pay \$7, and 1+1 and 0+1 both pay \$4.

I figured out the exact probability of two random quick picks earning the same prize in a normal state is

39,385,084,830,437,945/42,690,810,964,495,122 ≈ 92.3%

(That includes the likelihood they both win \$0)

Unfortunately, you can't really do an exact calculation for California because sometimes the lower prize amounts happen to be equal and sometimes they are off from each other by a dollar. ¯\_(ツ)_/¯

" In most of the country 4+0 and 3+1 both pay \$100,"

4+0 Euromillions paid 48.40 euros and costs 2.5 euros.

€48.40/2.5
=€19.36

Game: 5/50*2/12

• United States
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March 28, 2019
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Quote: Originally posted by Speler on Apr 30, 2022

" In most of the country 4+0 and 3+1 both pay \$100,"

4+0 Euromillions paid 48.40 euros and costs 2.5 euros.

€48.40/2.5
=€19.36

Game: 5/50*2/12

Hello Speler,

How are you doing? I agree your Euromillions 4+0 payout is pretty paltry in relation to Powerball's. However, the odds of Euromillions 4+0 are 1 in 13,000-ish, whereas the odds for Powerball 4+0 are 1 in 36,000. Considering your game's 4+0 is easier to win, the payouts are comparable. I see they also do pari-mutuel prizes like California.

Thank you for not posting about Africans and for not bringing Garyo as your plus-one. I know you will feign not getting that joke due to language barrier.

• Ruler Of Flat Earth - King Under The Dome
United States
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November 27, 2014
969 Posts
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Quote: Originally posted by cottoneyedjoe on Apr 28, 2022

Say two people each buy one quick pick for Powerball. (No powerplay or double play add-on, just a normal \$2 ticket) What is the probability they win the same amount?

The reason for the question is I recently bought qp's for two people and one of the wanted to bet the other they would win the same amount. At first I thought it was a bad bet, then I realized the probability of both of them winning \$0 is pretty high. I could run some simulations to estimate the probability, but I wanted to know what it is exactly -- some huge numerator over an even bigger denominator.

Why would anybody still play the PB?

If you can get 4 decent pick 3 odds bets vs 1 astronomical PB odds bet...than thats how you define smart.

Thats a Gail Howard maxim, i think...

The whole idea of a 5 dollar ticket is throw away money.

The mathematics behind the draw and the skills applied are mostly scoffed at...

Well our money is mostly thrown towards MUGA these days.  MAGA is racist.  MUGA is divine.

SLAVA MUGA!!!

• United States
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March 28, 2019
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Quote: Originally posted by JeetKuneDoLotto on May 11, 2022

Why would anybody still play the PB?

If you can get 4 decent pick 3 odds bets vs 1 astronomical PB odds bet...than thats how you define smart.

Thats a Gail Howard maxim, i think...

The whole idea of a 5 dollar ticket is throw away money.

The mathematics behind the draw and the skills applied are mostly scoffed at...

Well our money is mostly thrown towards MUGA these days.  MAGA is racist.  MUGA is divine.

SLAVA MUGA!!!

For some players it's just an old habit they can't break. What is MUGA, btw? Mice Under Garyo's Apartment?

• 100
Texas
United States
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October 23, 2007
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For what it's worth, I play MM and PB because I also play Texas Lotto. The odds hitting the jackpot on Texas Lotto are 1 in 25,827,165. The odds of hitting the \$1,000,000 2nd place prize on MM or PB are approximately half of Texas Lotto's odds. Granted the Lotto's jackpot is much higher, but in my case (age), \$1 million will do me fine until I croak.

To each his own.

CAN'T WIN IF YOU'RE NOT IN

A DOLLAR AND A DREAM (OR \$2)

• United States
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March 28, 2019
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I play PB/MM when the jackpots are ginormous, but I Superlotto Plus has better odds and costs less, and sometimes the jackpot gets up to the high double-digit millions. I suspect Jeet plays PB/MM too regardless of the weird act he puts on.

• Sugar Land
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August 28, 2019
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The answer is dependent on the subtle nuance of how the question is asked. There are two possibilities: (a) the unconditional (a priori) probability that two randomly chosen tickets will have the same, winning outcome, and (b) the conditional probability that two randomly chosen tickets will have the same winning outcome given that both tickets are winners. "Winning" and "winners" here mean one of the outcomes that pays out a prize, not limited to the Jackpot.

Firstly, we need to recognize that there are two pools of balls. One has numbers 1 to 69, and the second has numbers 1 to 26. Five numbers are chosen at random from the first (main) pool, and one is chosen from the second (Powerball) pool. The total number of combinations is 292201338. To calculate the cases (a) and (b) described above we must first calculate the number of combinations that result in prize payouts.

Out of 292201338, there are a total of 11750538 that result in some payout ("winners"). Using the notation Main/Pball for (number of main balls matched) / (Powerball matched or not) here is the number of combinations breakdown for winning outcomes. "0" for Powerball means not matched, and "1" means matched.

0/1 = 7624512

1/1 = 3176880

2/1 = 416640

3/0 = 504000

3/1 = 20160

4/0 = 8000

4/1 = 320

5/0 = 25

5/1 = 1

Total = 11750538

To calculate the "unconditional" probability that two randomly and independently chosen tickets will have the same winning outcome, we need to square each possibility above and divide by the total number of combinations squared. Thus,

P(0/1; 2 winners) = 58133183238144 / 85381621928990200

P(1/1; 2 winners) = 10092566534400 / 85381621928990200

P(2/1; 2 winners) = 173588889600 / 85381621928990200

P(3/0; 2 winners) = 254016000000 / 85381621928990200

P(3/1; 2 winners) = 406425600 / 85381621928990200

P(4/0; 2 winners) = 64000000 / 85381621928990200

P(4/1; 2 winners) = 102400 / 85381621928990200

P(5/0; 2 winners) = 625 / 85381621928990200

P(5/1; 2 winners) = 1 / 85381621928990200

P(Total) = 68653825190770 / 85381621928990200 = 0.000804081998440691

To calculate the conditional probability of the same winning outcome, given both tickets are winners, we take same numerators in the fractions above and instead divide by the square of 11750538, which is 138075143289444.

P(0/1 | Both are winners) = 0.421025695524941

P(1/1 | Both are winners) = 0.0730947388064133

P(2/1 | Both are winners) = 0.00125720593485903

P(3/0 | Both are winners) = 0.00183969390831999

P(3/1 | Both are winners) = 2.94351025331199E-06

P(4/0 | Both are winners) = 4.63515723940537E-07

P(4/1 | Both are winners) = 7.41625158304859E-10

P(5/0 | Both are winners) = 4.52652074160681E-12

P(5/1 | Both are winners) = 7.24243318657089E-15

P(Total | Both are winners) = 0.497220741946669

• Sugar Land
United States
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August 28, 2019
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I might add that the probability that both tickets are winners of any kind is P = (11750538 / 292201338) ^ 2 = 0.00161715296769927

• Ruler Of Flat Earth - King Under The Dome
United States
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November 27, 2014
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Quote: Originally posted by Orange71 on May 23, 2022

I might add that the probability that both tickets are winners of any kind is P = (11750538 / 292201338) ^ 2 = 0.00161715296769927

Numbers with lots of decimals, and carryovers,

is great.

However the visualizaiton of these numbers MUCH less than great.

Where is serge...he likes to spooge his senseless numbers everywhere.

• Sugar Land
United States
Member #200,559
August 28, 2019
77 Posts
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Well, if you can tell me how to paste an image object into this website's crappy edit window, I'll be happy to post visuals for you. Apparently other posters must have figured this out. By the way, I have a nasty visual of "spooge".

• Reno, NV
United States
Member #173,296
February 25, 2016
337 Posts
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Quote: Originally posted by Orange71 on May 23, 2022

The answer is dependent on the subtle nuance of how the question is asked. There are two possibilities: (a) the unconditional (a priori) probability that two randomly chosen tickets will have the same, winning outcome, and (b) the conditional probability that two randomly chosen tickets will have the same winning outcome given that both tickets are winners. "Winning" and "winners" here mean one of the outcomes that pays out a prize, not limited to the Jackpot.

Firstly, we need to recognize that there are two pools of balls. One has numbers 1 to 69, and the second has numbers 1 to 26. Five numbers are chosen at random from the first (main) pool, and one is chosen from the second (Powerball) pool. The total number of combinations is 292201338. To calculate the cases (a) and (b) described above we must first calculate the number of combinations that result in prize payouts.

Out of 292201338, there are a total of 11750538 that result in some payout ("winners"). Using the notation Main/Pball for (number of main balls matched) / (Powerball matched or not) here is the number of combinations breakdown for winning outcomes. "0" for Powerball means not matched, and "1" means matched.

0/1 = 7624512

1/1 = 3176880

2/1 = 416640

3/0 = 504000

3/1 = 20160

4/0 = 8000

4/1 = 320

5/0 = 25

5/1 = 1

Total = 11750538

To calculate the "unconditional" probability that two randomly and independently chosen tickets will have the same winning outcome, we need to square each possibility above and divide by the total number of combinations squared. Thus,

P(0/1; 2 winners) = 58133183238144 / 85381621928990200

P(1/1; 2 winners) = 10092566534400 / 85381621928990200

P(2/1; 2 winners) = 173588889600 / 85381621928990200

P(3/0; 2 winners) = 254016000000 / 85381621928990200

P(3/1; 2 winners) = 406425600 / 85381621928990200

P(4/0; 2 winners) = 64000000 / 85381621928990200

P(4/1; 2 winners) = 102400 / 85381621928990200

P(5/0; 2 winners) = 625 / 85381621928990200

P(5/1; 2 winners) = 1 / 85381621928990200

P(Total) = 68653825190770 / 85381621928990200 = 0.000804081998440691

To calculate the conditional probability of the same winning outcome, given both tickets are winners, we take same numerators in the fractions above and instead divide by the square of 11750538, which is 138075143289444.

P(0/1 | Both are winners) = 0.421025695524941

P(1/1 | Both are winners) = 0.0730947388064133

P(2/1 | Both are winners) = 0.00125720593485903

P(3/0 | Both are winners) = 0.00183969390831999

P(3/1 | Both are winners) = 2.94351025331199E-06

P(4/0 | Both are winners) = 4.63515723940537E-07

P(4/1 | Both are winners) = 7.41625158304859E-10

P(5/0 | Both are winners) = 4.52652074160681E-12

P(5/1 | Both are winners) = 7.24243318657089E-15

P(Total | Both are winners) = 0.497220741946669

This is good but there is an error in your calculation. The question was what is the probability two tickets win the same amount (not, match the same number of balls).

0/1 and 1/1 win the same amount -- \$4

2/1 and 3/0 win the same amount -- \$7

3/1 and 4/0 win the same amount -- \$100

The calculation should be done with those three groups of two equal prizes added together. The probability of two Powerball tickets winning the same amount (given that both are winners) is very high, 85%. This is because the vast majority of winning tickets (92%) are \$4, so it's not unusual for two winning tickets to win \$4.

The probability that two tickets win the same amount (without the condition that both are winners of some amount) is still low, 0.1376% (or 1 in 727).

• Sugar Land
United States
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August 28, 2019
77 Posts
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Tucker, you are absolutely right. Very keen eyes!

This is the corrected tabulation of the conditional probabilities:

P(0/1 OR 1/1 | Both Are Winners) = (10801392/11750538)^2 = 0.844975180602138

P(2/1 OR 3/0 | Both Are Winners) = (920640/11750538)^2 = 0.00613852710493474

P(3/1 OR 4/0 | Both Are Winners) = (28160/11750538)^2 = 5.74314522591283E-06

P(4/1 | Both are Winners) = (320/11750538)^2 = 7.41625158304859E-10

P(5/0 | Both are Winners) = (25/11750538)^2 = 4.52652074160681E-12

P(5/1 | Both are Winners) = (1/11750538)^2 = 7.24243318657089E-15

Summed:

P(Both Win Same Amount | Both Are Winners) = 0.851119451598458

And now for the Unconditional Probabilities (without the prior condition that both are winners):

P(0/1 OR 1/1) = (10801392/292201338)^2 = 0.00136645412094298

P(2/1 OR 3/0) = (920640/292201338)^2 = 9.92693732504765E-06

P(3/1 OR 4/0) = (28160/292201338)^2 = 9.28754434601285E-09

P(4/1) = (320/292201338)^2 = 1.19932132567315E-12

P(5/0) = (25/292201338)^2 = 7.32007645064176E-15

P(5/1) = (1/292201338)^2 = 1.17121223210268E-17

Summed:

P(Both Win Same Amount) = 0.00137639034701902

• Reno, NV
United States
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February 25, 2016
337 Posts
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