 Ask the Wizard: |
Questions about the Lottery |
This article claims that the chances of drawing the same three-digit number two nights in a row are 1 in a million. But since the actual number drawn itself has no significance, the odds are really one in a thousand, right? — Jon from Philadelphia
You're right. The probability of the same sequence of numbers chosen two nights in a row is 1 in 1000. The question that the writer was answering is what is the probability that 1-9-6 is drawn twice is a row, which is indeed one in a million. However, as you note, the pertinent question is what are the odds that any sequence repeats. The answer to that question is (1/10) 3 = 1 in 1000. February 17, 2009
Is it true that state lottery jackpot annuities cease to pay if you die before the annuity runs its term? I heard this was true of New York. This seems outrageous to me. — Alex from Montreal
I checked the New York and California lottery web sites. Both indicated that if the winner dies before all the payments are made, the rest will be paid to the winner's designated heir or estate.
December 8, 2008
Hi, In Australia we have Lotto, where the major cash prize is paid if your six numbers are drawn from a possible 45 numbers (1-45) A lot of people purchase a "Slik Pik," where you get 12 games, each of six, allegedly random, random picks. My friends and I are always amazed that in the 12 games, the same number may appear up to 6 or 7 times out of the 12 games. Surely this is not random!!! My question is what is the expected number of times any number will repeat 6 or 7 times, assuming the selection is random? — Kevin from Perth, Western Australia
The expected number of times any number will appear exactly n times in 12 games is combin (12,n)×(6/45)n×(39/45)n-12. The following table shows the expected number of occurrences from 0 to 12.
| Expected number of repeat numbers |
| Repeats |
Expected |
| 0 | 8.0804888027 |
| 1 | 14.9178254818 |
| 2 | 12.6227754077 |
| 3 | 6.4732181578 |
| 4 | 2.2407293623 |
| 5 | 0.5515641507 |
| 6 | 0.0989986937 |
| 7 | 0.0130547728 |
| 8 | 0.0012552666 |
| 9 | 0.0000858302 |
| 10 | 0.0000039614 |
| 11 | 0.0000001108 |
| 12 | 0.0000000014 |
| Total | 45 |
So, to answer your question, you will see the same number exactly six times about 0.099 times per set of cards, or about once every 10.1 times. The same number appearing exactly seven times will happen 0.0131 times per set of cards, or once every 76.6 times. February 21, 2008
My wife and I bought a $20 raffle ticket for the Indiana lottery. My understanding of this game is that the drawing for the winning prizes (which number 777) will be held on August 16, 2007, regardless of the number of actual tickets sold and with the absolute maximum available number of tickets being 325,000. As of today, only 60,000 tickets have been sold. Would it be a good gamble to buy a few additional tickets? What would our odds of winning a prize be? - David B. from Evansville, IN
According to the Indiana Lottery web site, there is a total of $3,270,000 in prize money given to 325,000 ticket holders. That would make each ticket worth $10.615 each on average, assuming the series sold out. At a cost of $20 each, the return is 50.31%. If only 60,000 tickets were sold then each would be worth $54.50, for a return of 272.50%. The breakeven point is 163,500 tickets sold. If you believe that fewer than that will be sold, then buying tickets becomes a good bet, putting aside tax and the utility of money implications. September 30, 2007
In New York state they have Video Lottery Terminals (VLT’s) at off-track betting spots. You hear the term a machine is approaching its "Set Point" when a VP machine gets "hot" and deals winning hand after winning hand. This would explain why the same machine pays on one day and doesn't want to know you on other days. Also, most of these machines will not allow you to lose a winning dealt hand. Throw it away and it will return the equivalent hand or better.
What are your thoughts on this subject? - Tom from Buffalo, NY
VLT’s are glorified pull-tab games. There is a predetermined pool of outcomes. When you play, the game picks an outcome from the pool at random, and displays the win to the player in the form of a slot machine or video poker game. Since the outcome is predestined, any element of skill is imaginary. For example, if you are dealt a royal flush and throw it away, you’ll get another one on the draw. Usually I say that in gambling the past doesn’t matter, but in this case there is an effect of removal. If you play one time and lose, then it will marginally improve the odds of the remaining game outcomes, until the supply of virtual pull-tabs is exhausted, and I presume the virtual drum is refilled. I believe that your hot and cold swings are just normal luck, and any predestination is imagined.
February 14, 2007
A reader later added the following to this topic.
I have a comment on your February 14 "Ask the Wizard" column (No. 183). It's doesn't really have anything to do with the question you answered. It's just something you might find interesting.
Prior to the passing of Proposition 1A, that allowed to have full class 3 gaming, we had a small installation of VLT style for a couple of years. In our system, which was run by SDG (now part of Bally), the prize pool started with 4 million draws. When the pool was reduced and 2 million remained, the next pool of 4 million was added for a total pool of 6 million draws. When the pool was reduced to 2 million again, the process repeated.
Wizard, could you please describe the equivalent odds of the California SuperLotto Plus (1 in 41.4 Million), in terms of number of consecutive times of rolling 7 or 11? I heard it somewhere before. Most people cannot comprehend the lottery odds. But, the rolling of dice -- they can relate. – Tim from Belmont, CA
Let n be your answer. The probability of rolling a 7 or 11 is 8/36. To solve for n:
(8/36)n = 1/41,400,000
log((8/36)n) = log(1/41,400,000)
n × log(8/36) = log(1/41,400,000)
n = log(1/41,400,000)/log(8/36)
n = -7.617 / -0.65321
n = 11.6608
So there you go, the probability of hitting the SuperLotto is the same as rolling a seven or eleven 11.66 times in a row. For those who can't comprehend a partial throw I would rephrase as the probability falls between 11 and 12 consecutive rolls.
September 22, 2006
Is the powerplay option in the Powerball lottery a "sucker's bet"?
Powerplay multiplies non-grand prizes from 2x - 5x. The Powerball site has listed testimonials from its winners that say "powerplay is the only way to go". I'm thinking it's a sucker's bet. – John from Morrisville, NC
The lottery is always a sucker bet! Briefly, the return of the Powerplay option is 49.28%. The return from a Powerball ticket alone is so much worse that it would be better to buy x/2 tickets with the Powerplay option than x without it. I added details about this option to my lottery section if you want more information.
September 13, 2006
They say the probability of winning the Powerball lottery is 1 in 147,107,962. In the recent drawing for a $340 million jackpot the local media said the number of tickets sold was 105,000,000. My questions are if you win what is the probability you will have to share the jackpot, and how much does this reduce the expected value? - Mitch F.
from Hopkins, MN
First let's confirm that probability. The player must match 5 regular numbers from a pool of 55, and one Power Ball from a pool of 42. The probability of winning would be 1 in combin(55,5)*42 = 1 in 146,107,962. So I agree with your probability. I like to use the Poisson distribution for questions such as yours. The mean number of winnings would be 105,000,000/146,107,962 = 0.71865. The general formula for the probability of n winners, with a mean of m, is e-m*mn/n!. In this case the mean is 0.71865 so the probability of zero is e-0.71865*0.718650/0! = 0.48741. So the probability of at least one winner is 1-0.48741 = 0.51259. So, 0.71865 winners will have 0.51259 of a jackpot to split up. That is 0.51259/0.71865 = 0.71327 jackpots per winner. So jackpot sharing reduces your expected win to 71.327% of the jackpot amount, or a reduction of 28.673%.
Oct. 26, 2005
Concerning your answer about the Italian lottery (see
Sep 11, 2005
column), you showed that the probability of 53
not being picked in two years is 1 in 1,707,460. You should
have followed up with the probability that any one (or more)
of the 90 numbers would be missed during the same two-year
period; I think that's what the person asking really wanted
to know. Also, you could have explained (once again) why 53
is no more likely to be picked in the next lottery than any
other number, despite the unlikely past situation.
The probability that any number would not be hit
in a two-year period could be closely approximated as
90*(1/ 1,707,460) = 1 in 18,972. The actual probability
would be slightly less than that because I double-counted
two numbers being missed, which is very negligible. Of
course, the past does not matter in the lottery and every
drawing has the same probability of picking the 53.
Sept.
18, 2005
I'm assuming you're aware, but if not, in the Italian
lottery, there is a twice-weekly drawing of 50 out of number
1 through 90 (five numbers from each of 10 cities). For
roughly 2 years, the number 53 has not shown up, leading up
to a "number 53 frenzy", to the point where people have
committed suicide after betting everything they have on what
they were sure would be a corrective! So I got to
thinking - what are the probabilities that the 53
would not come up for two years? (link
for more information) - Andrew from
Hollywood
I did some research and six numbers are chosen
each drawing. In any given drawing the probability of the
number 53 not making an appearance is combin(89,6)/combin(90,6)
= 93.333%. In two years there would be 208 drawings. So
the probability of 53 not occurring in a specific two
year period would be 0.93333208 =
0.000000585665, or 1 in 1,707,460. Sept.
11, 2005
If I buy two quick-pick lottery tickets what is the
probability I get the same number on both cards. Assume a
6/49 lottery.
The probability of correctly picking 6
numbers out of 49 is 1 in combin(49,6)
= 1 in 13,983,816. This is also the probability of your
two tickets matching. March 10,
2005
Dear Wiz: The horserace track that I attend is
introducing video lottery machines. Can you tell me anything
about them? Are they the same as slots? Any info you can
give would be helpful and appreciated. Thanks, Mike
S.
Another Mike S., what are the odds? Lots of
racetracks permit what is called "class 2" gaming, which
must be lottery or bingo based. The way to offer slots
under this rule is to have a lottery or bingo game going
on behind the scenes and the outcome is displayed in the
form of a slot machine win. For example if the lottery
game determines that you win 20 times your bet it will
display whatever slot machine symbols pay 20. It is a
clever illusion. April 4,
2004
I am a student from a very poor home please I will
like you to help me with the week draws thanks... - Frank
from Benin City, Nigeria
I take it you mean you want lottery numbers.
Sorry but I can't do any better than you can. However, I
would recommend you not play at all, especially if you
are very poor. There seem to be a lot of former generals
and dictators over there trying to wire me 17 million
dollars, maybe one of them will give you a scholarship.
Feb. 20, 2003
In your lottery probability chart for the MD lotto
game you make no allowance for the probability of a split
jackpot. What effect does this possibility have on expected
value? - Bob Schottler from Falls Church, Virginia
No, I didn't account for splitting the jackpot.
That definitely does depress the value, the more people
that play the more it reduces the expected return. I
didn't have enough information about number of players
when I wrote that article to properly factor that in.
June 28, 2002
I have noticed something over the years watching our
Quinto lottery drawing here in Washington State. It is a 52
'card' game that draws 5. I have noticed that the vast
majority of the time, three suits are drawn. From poker
numbers you find that the chance of getting just one suit (a
flush) is 5148 out of about 2.6 million. What are the odds
of 2,3, or all 4 suits coming up? - Kevin Larson from
Tacoma, Washington
Let's define f(x,y) as the probability of
getting x of one suit and y of another. This function is
not limited to two terms.
With two arguments f(x,y)=
combin(13,x)*combin(13,y)*12/combin(52,5).
With three arguments f(x,y,z)=
combin(13,x)*combin(13,y)*combin(13,z)*12/combin(52,5).
With four arguments
f(w,x,y,z)=combin(13,w)*combin(13,x)*combin(13,y)*combin(13,z)*4/combin(52,5).
The probability of all four suits is
COMBIN(13,1)3*COMBIN(13,2)*4/combin(52,5) =
26.37%.
The probability of three suits is
COMBIN(13,3)*COMBIN(13,1)2*12 +
COMBIN(13,1)*COMBIN(13,2)^2*12/combin(52,5) = 58.84%
The probability of two suits is
COMBIN(13,3)*COMBIN(13,2)*12 +
COMBIN(13,4)*COMBIN(13,1)*12/combin(52,5) = 14.59%
The probability of one suit (including straight and
royal flushes) is 4*combin(13,5)/combin(52,5) =
0.20%.
So, three suits are the most frequent outcome.
Oct. 30, 2001
I have been playing lotteries and sweepstakes now
for two months straight, will I ever hit a jackpot? and
when? - Mavis of Mattapan, U.S.
The short answer is no, you will never win.
The odds of winning the usual 6/49 lottery are 1 in
13,983,816. You would have to play the game
ln(.5)/ln(1-1/combin(49,6)) = 9,692,842 times to have a
50/50 chance of winning at least once. Assuming you
bought 100 lottery tickets a day this would take 265.6
years. To have a 90% chance of winning it would take
882.2 years. July 30,
2000
This is my second question to you, this time the
subject is state lotteries. I'm sure you've heard of a group
of "investors" which used to wait for a jackpot to reach a
certain level, at which point they would purchase tickets
with every combination of numbers possible. This would
assure them a share of the prize.
A group of mathematicians decided they would put their
abilities to use by playing the lottery. They would monitor
the various state lotteries and when the net payout of a
given lottery reached a certain amount they would proceed to
purchase one of each possible combination of numbers.
Assuming the cost of a ticket is $1 how high must the
jackpot get for this to be a profitable venture? -
Ted
A factor in the answer is the total number of
tickets sold to other players. In the event more than one
player hits the jackpot it will have to be shared. Let's
call the number of possible combinations n, the total
number of other tickets sold t, the rate of return on the
small prizes r (in the case of the Big Game r=0.179612),
and j be the jackpot size. For this to be a break-even
venture j*n/(n+t) + r*n - n=0. This works out to
j=(1-r)*(n+t). April 29,
2000
Given that a lottery has 10mm potential
combinations, what are the odds that someone will win with
90% confidence given that 10mm tickets are sold. Clearly it
would not be 100% since some tickets would be duplicates. I
am less interested in the answer than in the methodolgy used
to solve it. - Scott Kinder of New York, New York
Lets try to rephrase the question. Assuming
the lottery has 10 million combinations and all players
choose their numbers randomly (allowing for duplicates)
how many tickets would the lottery need to sell so that
the probability of at least one person winning is 90%?
Lets let p be the probability of winning and n be the
number of tickets sold. The probability of 1 person
losing is 1-p. The probability of all n people losing is
(1-p)n. The probability of at least one winner
is 1 - (1-p)n. So we need to set this equal to
.9 and solve for n.
.9 = 1 - (1-p)n
.1 = (1-p)n
ln(.1) = ln((1-p)n)
ln(.1) = n*ln(1-p)
n = ln(.1)/ln(1-p)
n = ln(.1)/ln(.9999999)
n = 23,025,850.
So the lottery would need to sell 23,025,850 tickets
for the probability of at least one winner to be 90%. In
case you were wondering if the lottery sold exactly ten
million tickets the probability of at least one winner
would be 63.2%. March 4,
2000
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