Feb 8, 2015, 9:30 am

The powerball jackpot has reached $450M annuity, taking an unprecedented 19 draws to do so. (The highest previous number of draws required by Powerball was 16.)

Based on the cash value of the previous jackpot, $256.8M and the cash value of this one, $304.1M, this means that the lottery is estimating that they will have sales of around $148M, which translates into roughly 74 million tickets.

If these figures prove accurate, the Poisson distribution for randomly picked numbers suggest the following probabilities for numbers of winners.

k, number of winners | p(m,k) |

0 | 65.57% |

1 | 27.68% |

2 | 5.84% |

3 | 0.82% |

4 | 0.09% |

5 | 0.01% |

As things stand, the most probable outcome is another rollover. These probabilities are subject to change if the sales figures increase, and do not take into account situations in which people skew the distribution by doing things like playing their birthdates or other things which reduce the randomness of ticket choices.

Feb 8, 2015, 1:34 pm

According to lottoreport ticket sales for the last drawing equaled that amount so that is a conservative estimation since sales usually increase every drawing until it is won. I wouldn't be surprised if the jackpot guesstimation is a half billion dollars before Wednesday's drawing.

Feb 8, 2015, 5:12 pm

RJOH: I would not be surprised either; however these probabilities, which certainly have limits, are based on what the current situation is.

Feb 8, 2015, 9:54 pm

This is the first time the Powerball jackpot rolled over 20 times in a row. In the past it has rolled over 19 times in a row and it reached 18 consecutive rollovers twice before.

Feb 9, 2015, 1:36 pm

In response to winoneday

**Quote:** Originally posted by winoneday on Feb 8, 2015

This is the first time the Powerball jackpot rolled over 20 times in a row. In the past it has rolled over 19 times in a row and it reached 18 consecutive rollovers twice before.

The word is they want it to roll over even more so in April they will change the matrix.

Feb 10, 2015, 9:13 pm

An update:

The lottery is now predicting a $327.7M cash jackpot, $485M annuity.

This suggests they are predicting $221M in sales, or roughly 110.5M tickets to be sold.

The Poisson distribution for randomized tickets thus predicts the following probabilities for numbers of winners:

k, number of winners | p(m,k) |

0 | 53.11% |

1 | 33.61% |

2 | 10.63% |

3 | 2.24% |

4 | 0.35% |

5 | 0.04% |

The single most probable outcome is still another rollover, and at least until the ticket sales rise further, as I expect they will, the probability of another rollover is greater than the probability that there will be any number of winners greater than zero.

Feb 10, 2015, 10:09 pm

In response to Prob988

**Quote:** Originally posted by Prob988 on Feb 10, 2015

An update:

The lottery is now predicting a $327.7M cash jackpot, $485M annuity.

This suggests they are predicting $221M in sales, or roughly 110.5M tickets to be sold.

The Poisson distribution for randomized tickets thus predicts the following probabilities for numbers of winners:

k, number of winners | p(m,k) |

0 | 53.11% |

1 | 33.61% |

2 | 10.63% |

3 | 2.24% |

4 | 0.35% |

5 | 0.04% |

The single most probable outcome is still another rollover, and at least until the ticket sales rise further, as I expect they will, the probability of another rollover is greater than the probability that there will be any number of winners greater than zero.

so more than 5 winners gives a lower than 0.04% chance of winning... so that means that my 12 winner idea is pretty far fetched then hmm? (at least statistically speaking)

Feb 11, 2015, 1:20 pm

In response to maximumfun

**Quote:** Originally posted by maximumfun on Feb 10, 2015

so more than 5 winners gives a lower than 0.04% chance of winning... so that means that my 12 winner idea is pretty far fetched then hmm? (at least statistically speaking)

Did you really need to see the math to know that?

The probability for each additional winner is a bigger change than the previous one. As an extremely rough guess, having 12 winners would be at least 20 to 30 million times less likely than 5 winners if it was all based on random probability.

Of course, it's not entirely random, and a lot of people play birthday numbers and a few very common combinations (bet slip patterns, consecutive numbers, common multiples and fortune cookie numbers). Let's figure there are 175 of those combinations, since that conveniently works out to an actual chance of 12 or more winners being 1 in a million. More likely than the chance that you'll win, but still very unlikely.

Feb 11, 2015, 1:50 pm

In response to maximumfun

**Quote:** Originally posted by maximumfun on Feb 10, 2015

so more than 5 winners gives a lower than 0.04% chance of winning... so that means that my 12 winner idea is pretty far fetched then hmm? (at least statistically speaking)

The probability for 12 independent winning tickes for a * randomized set * is not zero, but it is on the order of 1 in 230 billion, much smaller than the odds of actually winning the lottery.

However, if it happens that 50 people all play the same numbers because they shows up in 50 fortune cookies that have the same number because the fortune cookie maker does not randomize, or where the drawn numbers are something like 1, 2, 3, 4, 5 and powerball 6, your scenario is more likely.

Feb 11, 2015, 2:00 pm

In response to RJOh

**Quote:** Originally posted by RJOh on Feb 8, 2015

According to lottoreport ticket sales for the last drawing equaled that amount so that is a conservative estimation since sales usually increase every drawing until it is won. I wouldn't be surprised if the jackpot guesstimation is a half billion dollars before Wednesday's drawing.

"I wouldn't be surprised if the jackpot guesstimation is a half billion dollars before Wednesday's drawing."

Today's headline about PowerBall jackpot climbing to a half billion dollars didn't surprise me, I expected it.

Feb 14, 2015, 7:35 am

For the drawing that actually occurred, with a 564.1M annuity jackpot, the Poisson probabilities for actual numbers of winners, based on a $379M sales figure was as follows:

k, number of winners | p(m,k) |

0 | 33.90% |

1 | 36.67% |

2 | 19.83% |

3 | 7.15% |

4 | 1.93% |

5 | 0.42% |

6 | 0.08% |

7 | 0.01% |

The most probable outcome was a single winner, followed by another rollover.

It's possible I'll see you in another couple of months, but not probable.

Feb 14, 2015, 2:31 pm

In response to KY Floyd

**Quote:** Originally posted by KY Floyd on Feb 11, 2015

Did you really need to see the math to know that?

The probability for each additional winner is a bigger change than the previous one. As an extremely rough guess, having 12 winners would be at least 20 to 30 million times less likely than 5 winners if it was all based on random probability.

Of course, it's not entirely random, and a lot of people play birthday numbers and a few very common combinations (bet slip patterns, consecutive numbers, common multiples and fortune cookie numbers). Let's figure there are 175 of those combinations, since that conveniently works out to an actual chance of 12 or more winners being 1 in a million. More likely than the chance that you'll win, but still very unlikely.

No I did not need see the math to realize the possibility was slim, but it was also slim that a winner would be from Puerto Rico. It would have been quite something to see 12 winners. A gal can hope.

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