Powerball Lottery: Odds of Winning a Prize
Date: 09/02/2001 at 16:16:57 From: Joshua A. Cippel Subject: Odds of the Powerball Lottery Dr. Math, Hello. In the last few weeks, with all of the attention given to it in the news media, I have become interested in the Powerball lottery. More specifically, the odds of winning. This is not out of any desire to win money, but out of simple mathematical curiosity. As you may know, the object of the game is to choose 5 numbers from 1 to 49, and then the "Powerball" number from 1 to 42. After a number is drawn to become one of the 5 from 1 to 49, it cannot be drawn again. However, any number from 1 to 42 can be drawn as the "Powerball." Of course, the person who picks all 6 numbers correctly wins the jackpot, but one can also win a prize by picking some of the correct numbers. What has me baffled is the statistics of winning. Quoted from www.powerball.com, the odds are: Match 5+ Powerball = 1:80,089,128 Match 5 = 1:1,953,393 Match 4+ Powerball = 1:364,042 Match 4 = 1:8,879 Match 3+ Powerball = 1:8,466 Match 3 = 1:207 Match 2+ Powerball = 1:605 Match 1+ Powerball = 1:118 Match Powerball = 1:74 OVERALL ODDS OF WINNING A PRIZE: 1:35 Trying all of the statistics that I know to manipulate these numbers, I simply cannot determine how these odds are established. I would be very interested if you could help with this puzzle. Thank you.
Date: 09/02/2001 at 21:33:39 From: Doctor Paul Subject: Re: Odds of the Powerball Lottery Let's start with the odds of winning the big prize. To do this, you have to pick all five numbers correctly and you also have to pick the powerball number correctly. The answer depends on whether or not order is important and whether or not the numbers are "replaced" after they are drawn. Clearly, we are working "without replacement" here since a number cannot be drawn twice. That is, once a number has been drawn, it is not placed back into the pool of eligible numbers. In lottery games, the order in which the numbers are chosen is unimportant. All that matters is that you have the same numbers on your card as were drawn by the lottery officials. Since order doesn't matter, we're going to use a combination. We would use a permutation if order mattered. In general, the number of ways to choose r objects from n when order is not important and we are drawing without replacement is given by: C(n,r) = n!/(r! * (n-r)!) The probability of winning the big jackpot is given by the total number of ways to win divided by the possible number of outcomes. The number of possible outcomes (the denominator) is computed as follows: There are C(49,5) = 49!/(5! * (49-5)!) = 1,906,884 ways to pick your five numbers. And there are C(42,1) = 42 ways to pick the powerball (you verify the computation using the formula above). Thus there are 1,906,884 * 42 = 80,089,128 total number of ways that the drawing can occur. To figure out how many ways you can win (the answer is obviously one, but we go through the formalities because we'll need them to do the next part), think of the balls as being partitioned as follows: Among the 49 numbers, 5 are "winners" and 44 are "losers." Similarly, among the 42 powerball numbers, 1 is a "winner" and 41 are "losers." If you want to pick all five numbers correctly and pick the powerball correctly, then what you want to do is pick 5 out of the 5 winners and 0 out of the 44 losers on the non-powerball side, and you want to pick 1 out of 1 winners and 0 out of 41 losers on the powerball side. This makes the numerator: C(5,5) * C(44,0) * C(1,1) * C(41,0) = 1 * 1 * 1 * 1 = 1 Hence the probablity is 1/80,089,128 Now let's look at the odds of the "Match 5." Here you have to get all five numbers right on the non-powerball side but the number on the powerball is not important. Using the tools we developed above, you need to pick 5 of the 5 winners on the non-powerball side and 0 of the 44 losers. But this time, we pick 0 of the 1 winners on the powerball side and 1 of the 41 losers. Hence the numerator is given by: C(5,5) * C(44,0) * C(1,0) * C(41,1) = 1 * 1 * 1 * 41 = 41 The denominator remains the same for all of these problems. The number of possible powerball combinations never changes. Thus the probability of winning a match 5 is: 41/80,089,128, which is pretty close to 1/1,953,393. For the "Match 4 + Powerball" the numerator would be: C(5,4) * C(44,1) * C(1,1) * C(41,0) = 5 * 44 * 1 * 1 = 220 Thus the odds here are 220/80,089,128, which is pretty close to 1/364,042. You should be able to establish the rest of the formulas pretty easily just by following the pattern that has been established. If you have any questions, please write back. The overall odds of winning (1:35) can be verified by adding up all of the individual probabilities of winning: 1/74 + 1/118 + 1/605 + ... + 1/80,089,128 = .028706, which is very close to 1/35. I hope this has helped to take some of the mystery out of the odds. Again - write back if you need further clarification. If you aren't familiar with combinations and permutations in probability, please consult a Probability/Statistics textbook or the Dr. Math FAQ: http://mathforum.org/dr.math/faq/faq.comb.perm.html - Doctor Paul, The Math Forum http://mathforum.org/dr.math/
Date: 11/21/2001 at 14:03:32 From: G Saenger Subject: Question on your powerball odds calcs I read your analysis on the calculation of the Powerball odds, but I have a real problem with the odds of 1:74 that are calculated for selecting the correct powerball number only. I realize you have to take into account selecting 5 incorrect numbers on the non-powerball side, and include that calculation, but it still defies logic: No matter what 5 numbers you select on the non-powerball side (out of the field of 49 numbers), you still select one number on the powerball side, which has a field of 42 numbers. You are therefore selecting 1 number out of a field of 42, and by definition, your odds should be 1:42 of selecting the correct powerball number. It defies logic to think you would have worse odds (1:74) than 1:42 in selecting the correct powerball number. Could you please explain this? Thanks!
Date: 11/21/2001 at 14:40:10 From: Doctor Paul Subject: Re: Question on your powerball odds calcs Perhaps it would help to recall what is meant by the probability of an event. The probability that an event occurs is given by: the total number of ways the event can occur -------------------------------------------- the total number of possible outcomes In the case above (picking only the powerball number correctly), you need to count the number of ways to draw only the powerball. This means you pick the powerball number correctly but it also means that you don't pick any of the 5 numbers from the non-powerball side correctly. As mentioned above, it is helpful to think of the numbers as being partitioned in the following way: Among the 49 non powerball numbers, there are 5 "winners" and 44 "losers." Similarly, among the 42 powerball numbers, there are 1 "winner" and 41 "losers." In order to draw only the powerball, you have to pick: 0 of the 5 winners on the non-powerball side 5 of the 44 losers on the non-powerball side 1 of the 1 winners on the powerball side 0 of the 41 losers on the powerball side In your argument above, you seem to be making the assumption that you can pick any 5 numbers from the 49 available on the non-powerball side as long as you pick the powerball number correctly. This is in fact not the case. What if one of the five you chose on the powerball side was a winner? Then you wouldn't be in the scenario of picking only the powerball anymore. You would be in the "Match 1 + Powerball" scenario. Using the method explained in the link above, the number of ways to pick: 0 of the 5 winners on the non powerball side 5 of the 44 losers on the non powerball side 1 of the 1 winners on the powerball side 0 of the 41 losers on the powerball side is: C(5,0) * C(44,5) * C(1,1) * C(41,0) = 1 * 1086008 * 1 * 1 = 1086008 So the odds of picking only the powerball correctly are: 1086008 / 80089128 = .013559992812 which is very close to 1 / 74 = .013513513514 Now - if we wanted to know the odds of getting the powerball correct (and we didn't care about what happened on the other side) then we would just lump all 49 non-powerball numbers together and pick any five of them. This would give: C(49,5) * C(1,1) * C(41,0) = 1906884 * 1 * 1 = 1906884 ways of picking our numbers in such a way that guarantees that we get the powerball number correct. The probability of this happening is: 1906884 / 80089128 = .02380952381 and this is exactly 1/42. So the odds of picking the powerball number correctly are in fact 1/42 (as you mentioned should intuitavely be the case). But you have to realize that the odds of picking only the powerball correctly are in fact lower because in picking only the powerball correctly, you must guarantee that you don't pick any of the non-powerball numbers correctly. Finally, notice that 1/42 is the sum of the probabilities given above that are associated with winning the powerball. Here is the chart from above: Match 5+ Powerball = 1:80,089,128 Match 5 = 1:1,953,393 Match 4+ Powerball = 1:364,042 Match 4 = 1:8,879 Match 3+ Powerball = 1:8,466 Match 3 = 1:207 Match 2+ Powerball = 1:605 Match 1+ Powerball = 1:118 Match Powerball = 1:74 If you sum all of the ones that mean correctly picking the powerball, the answer will be 1/42: 1/74 + 1/118 + 1/605 + 1/8466 + 1/364042 + 1/80089128 = .023761861306 and 1/42 = .02380952381 These numbers are not exactly the same because the probabilities listed in the chart above are rounded off. For example - notice that when we computed the "Match Powerball" probability above, we got a number that was close to 1/74 but it wasn't exactly 1/74. I hope this helps. Please write back if you'd like to talk about this some more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/
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