## Poll

 Yes!!! 1 vote (12.5%) No... 7 votes (87.5%)

8 members have voted

pcket5s
Joined: Jan 3, 2011
• Posts: 22
January 4th, 2011 at 1:43:04 PM permalink
Hi all first time poster long time lurker. Got a question for you math / gambling wiz's out there. My roommate and I were having a discussion last night about the Mega Millions lottery. He believes that more than two people will win the lottery in tonight's drawing, mainly because no one has won in a long time and two because a lot of people will be playing it because it is so high. I differed and said guessed only one person or no one will win and told him the only validity in his reasoning is that the more people that play it the more likely of some one winning, as you all here obviously know.

So he says I will bet you straight up- \$20 bucks that two or more people win. So basically I win \$20 if no one or one person wins, he wins \$20 if two or more people win. I tried to do a little statistical analysis to determine if this was a good bet. I offer the following and welcome any criticism because my probability skills are weak.

I first looked at the last five mega million drawings to try to determine the avg. number of tickets bought in respect to the jackpot amount. I estimated the number of tickets won from the number of winning tickets per category - multiplied by the odds of winning. I then used a weighted average to aid in determining the actual number of tickets bought. The weighted average is the number of winning tickets divided by the sum of all winning tickets multiplied by the estimated number of winning tickets. I then sumed each of the weighted average to give me my final value.

I then took this number and divided by the jackpot prize for this week to give me a ratio.

I plotted all five ratios and used a polynomial trendline in the second order to generate the equation to estimate the amount of tickets that would be sold for the upcoming drawing.

I came up with 403,722,000, dividing that by the odds of winning the Mega Millions I come up with an estimated number of winners of 2.3. So no bet for me. Thoughts?

I like to play blackjack. I'm not addicted to gambling. I'm addicted to sitting in a semi-circle.
Ayecarumba
Joined: Nov 17, 2009
• Posts: 6763
January 4th, 2011 at 2:09:19 PM permalink
Is the proposal for "one ticket" or "one person"? Even if one ticket wins, it is often part of a pool of tickets purchased by a group. Perhaps another reason to not take the action.
Simplicity is the ultimate sophistication - Leonardo da Vinci
pcket5s
Joined: Jan 3, 2011
• Posts: 22
January 4th, 2011 at 2:12:17 PM permalink
Sorry I shoulda been more clear. The proposal is for winning tickets (i.e. - the number that is listed next to jackpot on the megamillion website tomorrow morning).
I like to play blackjack. I'm not addicted to gambling. I'm addicted to sitting in a semi-circle.
Wizard
Joined: Oct 14, 2009
• Posts: 21756
January 4th, 2011 at 3:56:22 PM permalink
I looked at the spreadsheet but couldn't follow how you got the 403,722,000 tickets will be sold. Can't we just do the jackpot growth divided by contribution per ticket? The odds of winning are 1 in 1 in 175,711,536. That would result in an expected 2.29764 winners. Using the Poisson distribution, the probability of 2 or more winners is 66.9%.

I addressed split jackpots in a recent Ask the Wizard, but it doesn't indicate how much jackpot size induces sales.
It's not whether you win or lose; it's whether or not you had a good bet.
Ayecarumba
Joined: Nov 17, 2009
• Posts: 6763
January 4th, 2011 at 4:23:33 PM permalink
Quote: Wizard

I looked at the spreadsheet but couldn't follow how you got the 403,722,000 tickets will be sold. Can't we just do the jackpot growth divided by contribution per ticket? The odds of winning are 1 in 1 in 175,711,536. That would result in an expected 2.29764 winners. Using the Poisson distribution, the probability of 2 or more winners is 66.9%.

I addressed split jackpots in a recent Ask the Wizard, but it doesn't indicate how much jackpot size induces sales.

It is possible, (but unlikely) that less tickets will be sold (bad weather, system malfunction, player ruin, birds falling from the sky, zombie attack), reducing the chance of a split jackpot.
Simplicity is the ultimate sophistication - Leonardo da Vinci
ahiromu
Joined: Jan 15, 2010
• Posts: 2101
January 4th, 2011 at 6:00:37 PM permalink
I think December 31st is an anomaly and is greatly skewing your data. You would need hundreds, maybe even thousands, to come up with a proper estimate for tickets sold vs. jackpot value - although I agree that it is exponential in some way. I'd take the bet.
Its - Possessive; It's - "It is" / "It has"; There - Location; Their - Possessive; They're - "They are"
Wizard
Joined: Oct 14, 2009
• Posts: 21756
January 4th, 2011 at 6:43:22 PM permalink
I have looked into this some more.

First lottoreport.com shows the jackpot grew from \$242M to \$355M, or by \$113M.

According to Durango Bill you can roughly get the number of tickets sold by multiplying the jackpot growth by 3. So that would be 339M tickets sold.

The probability of winning is 1 in 175,711,536. So there should be 1.93 winners.

The probability of 0 is 14.53%.
The probability of 1 is 28.03%.
The probability of 2 or more is 57.44%.
It's not whether you win or lose; it's whether or not you had a good bet.
Ayecarumba
Joined: Nov 17, 2009
• Posts: 6763
January 4th, 2011 at 8:00:49 PM permalink
Quote: Wizard

I have looked into this some more.

First lottoreport.com shows the jackpot grew from \$242M to \$355M, or by \$113M.

According to Durango Bill you can roughly get the number of tickets sold by multiplying the jackpot growth by 3. So that would be 339M tickets sold.

The probability of winning is 1 in 175,711,536. So there should be 1.93 winners.

The probability of 0 is 14.53%.
The probability of 1 is 28.03%.
The probability of 2 or more is 57.44%.

Hmm, the Calfornia Mega Millions website states that 53.9 cents of every dollar goes to prizes. Here are some figures from the last 2 drawings:

Draw 12/24 12/31
Mega Prize 200M 242M
Increase in Mega vs. prior draw 32M 42M
Pct. Chng. 19.05% 21.00%
Prizes Awarded \$2,445,053 \$3,763,734
Awards vs. Chg in Mega 7.64% 8.96%

So assuming the trend holds, the expected payout for tonight's drawing is:
Draw 12/24 12/31 1/4
Mega Prize 200M 242M 355M
Increase in Mega vs. prior draw 32M 42M 113M
Pct. Chng. 19.05% 21.00% 46.7%
Prizes Awarded (assuming no Mega) \$2,445,053 \$3,763,734 \$10,170,000
Awards vs. Chg in Mega 7.64% 8.96% 9% (assumption)

\$113M+\$10.2M = \$123.2M in prizes. If this is 53.9% of the new money, there were approximately 228.5M tickets sold since the last draw. Odds of hitting the Mega are 1 in 175,711,536, so it is possible that every combination has been sold.
Simplicity is the ultimate sophistication - Leonardo da Vinci
DeMango
Joined: Feb 2, 2010
• Posts: 2452
January 5th, 2011 at 1:18:06 AM permalink
Does anyone else here think that the lottery is the best invention for taxing the poor? Positive expected value? Please! Double digit million to one odds against, on winning with a single ticket? I would rather find a way to put a dent in Steve Wynn's bottom line!
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
Yoyomama
Joined: Oct 11, 2010