Re: Risk/probability interactions with the public -Reply

[CHARLES BLUE <BLUE.CHARLES@epamail.epa.gov>]

Some people just have too much time on their hands.)

>>> "Dukelow, James S Jr" <jim.dukelow@pnl.gov> 07/29/98 03:42pm
>>>


On 29 July 1998 Bob Giansiracusa wrote:

> On the other hand, if one interprets the statement of the odds 
> "really" being 1 in 80 million (for a $250,000,000 win), the expected 
> return on the $1 investment is $ 250/80 = $3.125 -- also not a bad 
> investment.
>

Well, not quite.

A single Powerball ticket's probability of matching the six numbers is 

        p = 1.249E-8   or one chance in 80 million

You have to match 5 white balls numbered 1 through 49 and match the
red 
ball, chosen from red balls numbered 1 through 42. 

However, if there are 100 million tickets sold for this drawing, we can 
use a binomial distribution with  N = 1.0E8 and p = 1.249E-8  or a 
Poisson distribution with  lamda = N*p = 1.249  to calculate the 
probabilities of 0, 1, 2, ... winners in the drawing.  If there are two 
or more winners, then the $137 million immediate payout will be split 
among the winners.  The $250 million dollars is the annuitized payout 
of $10M per year for 25 years, so if you are calculating the expected 
value of your $1 ticket, you should use either the immediate payout 
(less taxes, of course) or the net present value of the annuitized 
payout (less taxes, of course). 

With N = 100 million tickets, using the binomial distribution we have

        P(No winners) = (1 - 1.249E-8)^N = 0.2868
        P(1 winner) = P(No winners)*N*p/(1-p) = 0.3582
        P(2 winners) = P(1 winner)*((N-1)/2)*p/(1-p) = 0.2237
        P(3 winners) = P(2 winners)*((N-2)/3)*p/(1-p) = 0.09313
        P(4 winners) = P(3 winners)*((N-3)/4)*p/(1-p) = 0.02908
        P(5 winners) = P(4 winners)*((N-4)/5)*p/(1-p) = 0.00726
        P(6 or more winners) = 0.00183

Now, if you hold a winning ticket, it is not the case that there were 
no winners.  Therefore we need to calculate conditional probabilities.

        P(1 winner | you won ) = 0.3582/(1 - P(no winners))
                               = 0.3582 * 1.4021 = 0.5022
        P(2 winners | you won ) = 0.2237 * 1.4021 = 0.3136
        P(3 winners | you won ) = 0.09313 * 1.4021 = 0.13058
        P(4 winners | you won ) = 0.02908 * 1.4021 = 0.04077
        P(5 winners | you won ) = 0.00726 * 1.4021 = 0.01019
        P(6 or more winners | you won ) = 0.00183 * 1.4021 = 0.00257

So, assuming 100 million tickets sold, your expected winnings on your 
$1 investment (from winning or sharing the grand prize only) would be 

  E[winnings] = 1.249E-8 * $1.37E8 * ( 0.5022 + 0.3136/2 + 0.13058/3
                              + 0.04077/4 + 0.01019/5 + 0.00257/6 )
              = 1.249E-8 * $1.37E8 * 0.71525
              = $1.225

This ignores your expected winnings from prizes less than the grand 
prize, but also ignores the taxes you will have to pay on the winnings.  
If more than 100 million tickets are sold, your expected winnings will 
be smaller (and vice versa). 

This is, however, better expected winnings than I expected.

One other thing about communicating probabilities to the public.  This 
morning on National Public Radio, we were told that the chance of 
winning was 1 in 80 million, which was "less than your probability of 
being hit by lightning 14 times in a single year".  The number of 
deaths per year in the U.S. caused by lightning was between 80 and 100 
for the years 1986, 1987, and 1988.  I assume there is an opening at 
NPR for a probability consultant, or, perhaps, an English language 
consultant. 

Best regards.

Jim Dukelow
Pacific Northwest National Laboratory
Richland, WA

The comments are mine and have not been reviewed and/or approved by
my 
management or by the U.S. Department of Energy.