West Concord, MN United States Member #21 December 7, 2001 3942 Posts Online

Posted: April 15, 2014, 8:09 pm - IP Logged

Quote: Originally posted by LottoMetro on April 15, 2014

Chi-squared analysis. It's the same exact test used by lottery auditors, whose sole purpose, by the way, is to determine if the drawings are rigged/biased.

Happyland United States Member #146344 September 1, 2013 1139 Posts Offline

Posted: April 15, 2014, 8:54 pm - IP Logged

Quote: Originally posted by JADELottery on April 15, 2014

Χ^{2} ah, yes.

test, hmm, good.

within the test, what are you using as an expected frequency?

or better yet, how are you deriving the expected frequency?

observations we have, it's that pesky expectedness we like to see.

Since PowerPlay is only 1 digit it's pretty simple: the expected frequency is derived from the proportion of digits that are theoretically supposed to come up. For the case of x2, the odds are 1 in 2. This means the expected frequency is 50% of drawings. For x3, the expected frequency is about 30% of drawings.

For x2

24 drawings = 12 expected

26 drawings = 13 expected

For x3

24 drawings = 7.21 expected

26 drawings = 7.81 expected

etc.

Obviously, 7.21 exact appearances is impossible, but that doesn't matter. It's just the expected theoretical frequency. When you do the chi-squared test it basically just compares the observed frequency versus the 'expected' frequency discussed above. You calculate a p-value for the x^{2} statistic and check it against significance values (i.e. 0.05, 0.10). Anything outside these levels indicates that there is insufficient evidence for a bias; on the other hand, if you see values within these levels, there's a pretty good chance you've got a biased drawing on your hands.

If the chances of winning the jackpot are so slim, why play when the jackpot is so small? Your chances never change, but the potential payoff does. If a crystal ball showed you the future of the rest of your life, and in that future you will never win a jackpot, would you still play?

West Concord, MN United States Member #21 December 7, 2001 3942 Posts Online

Posted: April 15, 2014, 9:00 pm - IP Logged

Quote: Originally posted by LottoMetro on April 15, 2014

Since PowerPlay is only 1 digit it's pretty simple: the expected frequency is derived from the proportion of digits that are theoretically supposed to come up. For the case of x2, the odds are 1 in 2. This means the expected frequency is 50% of drawings. For x3, the expected frequency is about 30% of drawings.

For x2

24 drawings = 12 expected

26 drawings = 13 expected

For x3

24 drawings = 7.21 expected

26 drawings = 7.81 expected

etc.

Obviously, 7.21 exact appearances is impossible, but that doesn't matter. It's just the expected theoretical frequency. When you do the chi-squared test it basically just compares the observed frequency versus the 'expected' frequency discussed above. You calculate a p-value for the x^{2} statistic and check it against significance values (i.e. 0.05, 0.10). Anything outside these levels indicates that there is insufficient evidence for a bias; on the other hand, if you see values within these levels, there's a pretty good chance you've got a biased drawing on your hands.

You're sure you want to go with those values?

We see a problem.

Presented 'AS IS' and for Entertainment Purposes Only. Any gain or loss is your responsibility. Use at your own risk.

Order is a Subset of Chaos Knowledge is Beyond Belief Wisdom is Not Censored Douglas Paul Smallish Jehocifer

Happyland United States Member #146344 September 1, 2013 1139 Posts Offline

Posted: April 15, 2014, 9:04 pm - IP Logged

Quote: Originally posted by JADELottery on April 15, 2014

You're sure you want to go with those values?

We see a problem.

There are 10 total "virtual" PowerPlay balls:

1 for 5X

1 for 4X

3 for 3X

5 for 2X

5X = 1/10 = 10%

4X = 1/10 = 10%

3X = 3/10 = 30%

2X = 5/10 = 50%

Total 100%

Aside from the simple method, using percentages above, you can use the E_{i} = N/n like so: E(2X) = 24 (drawings) * 5 (x2 balls) / 10 (total balls) = 12

Calculating chi for 2X where observed = 16 and expected = 12:

x^{2} = (16-12)^{2}/12 = 1.3333

p-value is essentially 1, Excel says 0.9982

Other chi results:

x^{2} = 0.0667 for 5X

x^{2} = 0.0667 for 4X

x^{2} = 1.4222 for 3X

Naturally, none of which are significant.

If the chances of winning the jackpot are so slim, why play when the jackpot is so small? Your chances never change, but the potential payoff does. If a crystal ball showed you the future of the rest of your life, and in that future you will never win a jackpot, would you still play?

West Concord, MN United States Member #21 December 7, 2001 3942 Posts Online

Posted: April 15, 2014, 10:15 pm - IP Logged

Quote: Originally posted by LottoMetro on April 15, 2014

There are 10 total "virtual" PowerPlay balls:

1 for 5X

1 for 4X

3 for 3X

5 for 2X

5X = 1/10 = 10%

4X = 1/10 = 10%

3X = 3/10 = 30%

2X = 5/10 = 50%

Total 100%

Aside from the simple method, using percentages above, you can use the E_{i} = N/n like so: E(2X) = 24 (drawings) * 5 (x2 balls) / 10 (total balls) = 12

Calculating chi for 2X where observed = 16 and expected = 12:

x^{2} = (16-12)^{2}/12 = 1.3333

p-value is essentially 1, Excel says 0.9982

Other chi results:

x^{2} = 0.0667 for 5X

x^{2} = 0.0667 for 4X

x^{2} = 1.4222 for 3X

Naturally, none of which are significant.

Hmm.

Hmm, hmm.

Hmm, hmm, hmm.

Yes.

Presented 'AS IS' and for Entertainment Purposes Only. Any gain or loss is your responsibility. Use at your own risk.

Order is a Subset of Chaos Knowledge is Beyond Belief Wisdom is Not Censored Douglas Paul Smallish Jehocifer

NEW YORK United States Member #90535 April 29, 2010 12138 Posts Offline

Posted: April 16, 2014, 12:10 pm - IP Logged

Quote: Originally posted by mypiemaster on April 5, 2014

Been all pi$$ed off about the PB not having a multiplier for 2nd prize, and the stupid 2x multiplier which makes absolutely no sense. When I think about it, I get all-pi$$ed-off again, so I quit thinking about it. Now I'm all pi$$ed off again.

Time to take a break from playing the lottery? You can still buy 1 Powerplay ticket monthly or yearly.

WARREN BUFFETT AND BILL GATES Do Not Play The Lottery, AND THEY ARE BILLIONAIRES.

West Concord, MN United States Member #21 December 7, 2001 3942 Posts Online

Posted: April 20, 2014, 11:17 am - IP Logged

Alright, here we go.

Throughout this topic odds have made an appearance and have been used to explain the occurrences of the various multiplier selections.

The Megaplier does have problems, but it seems to benefit the player more than the Power Play multiplier does.

For now we'll focus on the Power Play multiplier.

There are 3 different odds to work with: Stated, Calculated and Estimated.

The Stated and Calculated should always be in exact agreement with each other.

The Estimated odds should be very close to the Stated and Calculated odds.

If we state the odds of something is 1 in 50 and a 100 selections later we estimate the odds to be 1 in 3, something is wrong.

Overall, all 3 of these odds need to be in general agreement of each other.

The Stated odds for the Power Play are fixed by the lottery and they are:

Power Play

Odds

2

1 in 2

3

1 in 3.33

4

1 in 10

5

1 in 10

LottoMetro has given us some virtual bias values for a possible explanation of the high number of x2 selections, with the Power Play in ascending order:

Power Play

Number of Virtual "Power Play" Balls

2

5

3

3

4

1

5

1

From these values we can Calculate the odds of each Power Play.

A = 5, B = 3, C = 1, D = 1

Power Play

Calculated Odds

Stated Odds

2

1 in (B + C + D) / A

1 in 2

3

1 in (A + C + D) / B

1 in 3.33

4

1 in (A + B + D) / C

1 in 10

5

1 in (A + B + C) / D

1 in 10

Substitute the values it becomes:

Power Play

Calculated Odds

Stated Odds

2

1 in (3 + 1 + 1) / 5

1 in 2

3

1 in (5 + 1 + 1) / 3

1 in 3.33

4

1 in (5 + 3 + 1) / 1

1 in 10

5

1 in (5 + 3 + 1) / 1

1 in 10

This becomes:

Power Play

Calculated Odds

Stated Odds

2

1 in 5 / 5

1 in 2

3

1 in 7 / 3

1 in 3.33

4

1 in 9 / 1

1 in 10

5

1 in 9 / 1

1 in 10

Finally, make the fractions decimal values.

Power Play

Calculated Odds

Stated Odds

2

1 in 1

1 in 2

3

1 in 2.33

1 in 3.33

4

1 in 9

1 in 10

5

1 in 9

1 in 10

Based on LottoMetro's biased virtual ball counts, there's a problem with the calculated odds matching up with the stated odds.

Both of these Must be exactly equal for each of the Power Play odds.

So, which is correct in values.

Simple, test them using the formula we gave earlier: w·x·y·z - w·x - w·y - w·z - x·y - x·z - y·z - 2·w - 2·x - 2·y - 2·z - 3 .

This formula will always be 0 if the odds have been stated or calculated correct.

We have already done Power Play's stated odds using the formula and it does not become 0.

Let's run LottoMetro's virtual odds.

If w = 1, x = 7 / 3, y = 9, z = 9, then the formula works out to the following:

Looks like LottoMetro's assertion the virtual biased ball counts are correct, however, the odds do not match the stated odds.

Therein lay the problem, the stated and calculated odds do not match Exactly.

It's our assertion, you will never find any combination of virtual ball counts that can produce the exact same odds as the ones stated by the lottery.

Since we have shown the test for odds correctness works when odds are calculated and stated correctly, it becomes fairly obvious, the stated Power Play odds appear to be bogus.

This is also demonstrated by the Estimated odds.

As of 2014-04-20, the frequency of the Power Play for this run of the game's version is:

Power Play

Frequency

2

16

3

5

4

2

5

3

The estimated odds based on this frequency and compared to the stated odds is:

Power Play

Frequency

Estimated Odds

Stated Odds

2

16

1 in 0.63

1 in 2.00

3

5

1 in 4.20

1 in 3.33

4

2

1 in 12.00

1 in 10.00

5

3

1 in 7.67

1 in 10.00

The odds are fairly close except Power Play 2 appears stick out as being bogus.

When we look and compare LottoMetro's virtual odds to the sampled estimated odds we can see they are a much better fit:

Power Play

Frequency

Estimated Odds

LottoMetro's Odds

2

16

1 in 0.63

1 in 1

3

5

1 in 4.20

1 in 2.33

4

2

1 in 12.00

1 in 9

5

3

1 in 7.67

1 in 9

The x2 value is much closer to 1 in 1 than 1 in 2.

So, this boils down to a simple assertion.

The stated odds are most likely Not Correct.

With more drawings, we can see how the Estimated odds fair against both the Stated and LottoMetro's odds.

We'll monitor this for a bit and may post a reply or two as there are more drawings to estimate odds.

For now, you can debate this among yourselves.

We've said all we have to up to this point.

Presented 'AS IS' and for Entertainment Purposes Only. Any gain or loss is your responsibility. Use at your own risk.

Order is a Subset of Chaos Knowledge is Beyond Belief Wisdom is Not Censored Douglas Paul Smallish Jehocifer

Happyland United States Member #146344 September 1, 2013 1139 Posts Offline

Posted: April 20, 2014, 12:22 pm - IP Logged

I've never seen a lottery that had a number frequency matching its odds of appearance exactly. Randomness keeps this from happening. As such anything is game for the short run.....streaks, skips, whatever.

There's only been 26 draws of Power Play and you already believe it's rigged/bogus/biased....despite me providing statistical evidence to the contrary. Honestly I don't think there have been enough draws to reach this conclusion. Last night Power Play number 5 came up.

On an unrelated note, yesterday I heard via recording of a panelist in November that they are going to change Powerball again. Apparently an average of 14 jackpots per year is too many, so they will probably raise the bonus ball pool from 35 to 45 but keep the price $2. Don't know when this will take place though, but it came from the horse's mouth.

If the chances of winning the jackpot are so slim, why play when the jackpot is so small? Your chances never change, but the potential payoff does. If a crystal ball showed you the future of the rest of your life, and in that future you will never win a jackpot, would you still play?

Los Angeles, California United States Member #103813 January 5, 2011 1530 Posts Offline

Posted: April 20, 2014, 12:53 pm - IP Logged

Quote: Originally posted by JADELottery on April 20, 2014

Alright, here we go.

Throughout this topic odds have made an appearance and have been used to explain the occurrences of the various multiplier selections.

The Megaplier does have problems, but it seems to benefit the player more than the Power Play multiplier does.

For now we'll focus on the Power Play multiplier.

There are 3 different odds to work with: Stated, Calculated and Estimated.

The Stated and Calculated should always be in exact agreement with each other.

The Estimated odds should be very close to the Stated and Calculated odds.

If we state the odds of something is 1 in 50 and a 100 selections later we estimate the odds to be 1 in 3, something is wrong.

Overall, all 3 of these odds need to be in general agreement of each other.

The Stated odds for the Power Play are fixed by the lottery and they are:

Power Play

Odds

2

1 in 2

3

1 in 3.33

4

1 in 10

5

1 in 10

LottoMetro has given us some virtual bias values for a possible explanation of the high number of x2 selections, with the Power Play in ascending order:

Power Play

Number of Virtual "Power Play" Balls

2

5

3

3

4

1

5

1

From these values we can Calculate the odds of each Power Play.

A = 5, B = 3, C = 1, D = 1

Power Play

Calculated Odds

Stated Odds

2

1 in (B + C + D) / A

1 in 2

3

1 in (A + C + D) / B

1 in 3.33

4

1 in (A + B + D) / C

1 in 10

5

1 in (A + B + C) / D

1 in 10

Substitute the values it becomes:

Power Play

Calculated Odds

Stated Odds

2

1 in (3 + 1 + 1) / 5

1 in 2

3

1 in (5 + 1 + 1) / 3

1 in 3.33

4

1 in (5 + 3 + 1) / 1

1 in 10

5

1 in (5 + 3 + 1) / 1

1 in 10

This becomes:

Power Play

Calculated Odds

Stated Odds

2

1 in 5 / 5

1 in 2

3

1 in 7 / 3

1 in 3.33

4

1 in 9 / 1

1 in 10

5

1 in 9 / 1

1 in 10

Finally, make the fractions decimal values.

Power Play

Calculated Odds

Stated Odds

2

1 in 1

1 in 2

3

1 in 2.33

1 in 3.33

4

1 in 9

1 in 10

5

1 in 9

1 in 10

Based on LottoMetro's biased virtual ball counts, there's a problem with the calculated odds matching up with the stated odds.

Both of these Must be exactly equal for each of the Power Play odds.

So, which is correct in values.

Simple, test them using the formula we gave earlier: w·x·y·z - w·x - w·y - w·z - x·y - x·z - y·z - 2·w - 2·x - 2·y - 2·z - 3 .

This formula will always be 0 if the odds have been stated or calculated correct.

We have already done Power Play's stated odds using the formula and it does not become 0.

Let's run LottoMetro's virtual odds.

If w = 1, x = 7 / 3, y = 9, z = 9, then the formula works out to the following:

Looks like LottoMetro's assertion the virtual biased ball counts are correct, however, the odds do not match the stated odds.

Therein lay the problem, the stated and calculated odds do not match Exactly.

It's our assertion, you will never find any combination of virtual ball counts that can produce the exact same odds as the ones stated by the lottery.

Since we have shown the test for odds correctness works when odds are calculated and stated correctly, it becomes fairly obvious, the stated Power Play odds appear to be bogus.

This is also demonstrated by the Estimated odds.

As of 2014-04-20, the frequency of the Power Play for this run of the game's version is:

Power Play

Frequency

2

16

3

5

4

2

5

3

The estimated odds based on this frequency and compared to the stated odds is:

Power Play

Frequency

Estimated Odds

Stated Odds

2

16

1 in 0.63

1 in 2.00

3

5

1 in 4.20

1 in 3.33

4

2

1 in 12.00

1 in 10.00

5

3

1 in 7.67

1 in 10.00

The odds are fairly close except Power Play 2 appears stick out as being bogus.

When we look and compare LottoMetro's virtual odds to the sampled estimated odds we can see they are a much better fit:

Power Play

Frequency

Estimated Odds

LottoMetro's Odds

2

16

1 in 0.63

1 in 1

3

5

1 in 4.20

1 in 2.33

4

2

1 in 12.00

1 in 9

5

3

1 in 7.67

1 in 9

The x2 value is much closer to 1 in 1 than 1 in 2.

So, this boils down to a simple assertion.

The stated odds are most likely Not Correct.

With more drawings, we can see how the Estimated odds fair against both the Stated and LottoMetro's odds.

We'll monitor this for a bit and may post a reply or two as there are more drawings to estimate odds.

For now, you can debate this among yourselves.

We've said all we have to up to this point.

JADE,

I think the discrepancy you're seeing is just due to the fundamental problem with the lottery industry referring to probability as odds.

probability = chances for / total chances (or x in y) odds = chances for : chances against (or x to y)

So for Powerplay, what the lottery calls "odds" is really the probability: 2x - 1 in 2 (or 5 in 10) 3x - 1 in 3.33 (or 3 in 10) 4x - 1 in 10 5x - 1 in 10

The equivalent *actual* odds for Powerplay: 2x - 5 to 5 (or 1:1, aka "50:50") 3x - 3 to 7 (or 1:2.33) 4x - 1 to 9 5x - 1 to 9 Which you have confirmed, and your assessment is technically correct based on proper terminology.

Have a Happy Easter!

P.S. Don't mind LottoMetro's arrogant condescending remark from above: "There's only been 26 draws of Power Play and you already believe it's rigged/bogus/biased....despite me providing statistical evidence to the contrary." LottoMetro can't help acting like a jerk if anyone dares question him. He has even admitted this himself in a previous post: I admit my social skills are lacking. But since LottoMetro lacks that sensitivity chip, he's the last one to be evaluating wherther or not he's being offensive.