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1 vote (12.5%)
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pcket5s
pcket5s
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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.

link to my google spreadsheet:

My Spreadsheet

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?


Thanks for all comments!
I like to play blackjack. I'm not addicted to gambling. I'm addicted to sitting in a semi-circle.
Ayecarumba
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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
pcket5s
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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
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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.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Ayecarumba
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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
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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.
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Wizard
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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%.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Ayecarumba
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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
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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
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January 5th, 2011 at 2:17:01 AM permalink
Quote: pcket5s

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.



Looks like a bad bet. Just heard 2 people won. One in Idaho and the other in Washington. Sorry.
pcket5s
pcket5s
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January 5th, 2011 at 5:51:25 AM 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.



Wiz thanks for taking the time to look at this problem. Sorry I didn't get back before the drawing but was busy with other commitments. I came up with the 403M tickets by plotting the last 5 estimated number of tickets vs the jackpot amount. Then adding a polynomial trendline in the second order to the points. I used the polynomial trendline because it has the best R sqaure value, and made sense by showing a big increase as the lottery drawing amount went up.


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%.




Two people did win the jackpot. I took at look at the distribution from the megamillions website and in fact there was a large number of 2 + 1 winners as a member predicted earlier in the other lottery thread. I'm not sure if this is due to the numbers being from Lost or just an anomaly, but nevertheless I thought it was interesting. Using the weighed average formula, I came up with an estimated number of tickets being sold as approx 221,200,000. The 2 + 1 category would indicate approximately 352,600,000 tickets were sold. I way over estimated the number of tickets sold, but am glad I did because it saved me 20 bucks as I didn't make the bet. Thanks to all members who aided in this discussion.
I like to play blackjack. I'm not addicted to gambling. I'm addicted to sitting in a semi-circle.
miplet
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January 5th, 2011 at 8:47:20 AM permalink
There were $229,421,186 in tickets sold according to http://lottoreport.com/mmsales.htm just scroll to where the charts are.
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Ibeatyouraces
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January 10th, 2011 at 9:11:39 AM permalink
deleted
DUHHIIIIIIIII HEARD THAT!
Alan
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June 17th, 2011 at 8:48:57 AM permalink
So did you take the bet or pass?
FleaStiff
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June 17th, 2011 at 8:59:56 AM permalink
Quote: Alan

So did you take the bet or pass?

I'm curious about that also, but I tend to think that he took (and therefore lost) the bet because he was so specific as to the terms and that means he and his friend had probably finalized the details of the bet.

I think it was a poor bet because the purchase of additional tickets is often stimulated by media stories and the relatively slow news days in December coupled with a year end need for "feel good" features about one year ending and another one starting meant that staggering lottery amounts would get good publicity.
Ayecarumba
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June 17th, 2011 at 10:54:34 AM permalink
Quote: pcket5s

I way over estimated the number of tickets sold, but am glad I did because it saved me 20 bucks as I didn't make the bet. Thanks to all members who aided in this discussion.



He didn't take the action.
Simplicity is the ultimate sophistication - Leonardo da Vinci
Alan
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June 17th, 2011 at 11:29:18 AM permalink
Thanks, I missed that one. Good for him.
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