`Poisson(k) = (e^lambda lambda^k)/(k!)` 
Lottery Power PicksProbability of Winning Calculator  



The purpose of this page is to provide an interactive interface to determine the probability that someone will win a particular lottery drawing based on the number of tickets sold. It relies on the Poisson Distribution to calculate these probabilities, but uses 2 distinct methods. The first is the famed Poisson Formula; and the second is a more lengthy binomial distribution. Both methods generate very similar results, and the differences are displayed in graph P3 below.
Five similar charts are displayed. Graph P1 is a Pie Chart showing the breakdown of winners. It's results change depending on whether the "Only" or "or More" radio button is checked. Graph P2 shows the probabilities of individual ("Only") occurrences. Graph P3 is the difference between the closed form Poisson Formula and the brute force binomial expansion. Graphs P4 and P5 display the cumulative breakdown of the distributions: where P4 contains the total amounts expected, and P5 contains the corresponding probabilities (these are synonymous with the "or More" option results).
Although this page is targeted at lotteries, it can also be used as a tool to analyze any Poisson Distribution problem: (the Birthday Problem, Traffic Flow, Congestion, Stacking, Defect Analysis, Disease Spread, etc). To utilize these, simply change the Select Game Pulldown to "Other", and then enter the appropriate information.
Poisson Formulas
Let: 
`\Poisson = ` Poisson Event Probability
`N = ` Total Number of Combinations `S = ` Number Selected `p = 1/N ` `lambda = S/N ` 
Then: 
`Poisson(k)_{k=0}^{\infty} = (e^lambda lambda^k)/(k!)` 


Lastly: 
`Poisson(k) ~= \P(k)` 
Binomial Formulas
Let: 
`\T^(Cum) = ` Cumulative Number Selected `\P^(Cum) = ` Cumulative Probability `\P = ` Event Probability `N = ` Total Number of Combinations `S = ` Number Selected `p = 1/N ` 
Then: 
`\T^(Cum)(0) = \sum_{i=1}^{i lt S) ((S),(i+1)) * p^(i) ` `\T^(Cum)(1) = N  \T^(Cum)(0) ` `\T^(Cum)(k)_{k=2}^{\infty} = \sum_{i=1}^{(k+i1)>S) ((S),(k+i1)) * ((k+i2),(i1)) * p^(k+i2) ` 
And: 
`\P^(Cum)(k)_{k=0}^{\infty} = \T^(Cum)(k) * p ` `\P(0) = \P^(Cum)(0)` `\P(k)_{k=1}^{\infty} = \P^(Cum)(k)  \P^(Cum)(k+1)` 
Notes
The classic Birthday Problem asks: What is the probability that two people in a group will have the same birthday?
The generally accepted answer is stated in terms of 50% and 99% probabilities. To be 50% sure that there will be a duplicate birthday, we need a group of 23 people. To be 99% sure, we need a group of 57 people.
Using this Poisson Calculator, we learn that we will be slightly more than 50% sure after the 21st person is examined. You can verify this by setting the pulldown Menu to "Other", entering 2 in the "Number of Winners" field, and clicking the "or More" radio button. When done, we will see that the expected abount of duplicate people will be 0.52. Which means that when the 21st person is examined, there will be a 52% chance that this person will have the same birthday as one of the others.
Further, after 28 people are examined, we find that we will have 1 person sharing the same birthday as another.
Links
Birthday problem
The Birthday Problem by Becky Schmoyer
The Birthday Problem  from Wolfram MathWorld
Probability of the Same Birthday within a Group
Assume that 180 cars per hour pass through a certain point on a highway during the morning rush hour. Due to planned construction, it is estimated that congestion will occur when more than 5 cars pass through this the point in any one minute. What is the probability of congestion occurring?
To answer this, we note that 3 cars will pass trough the point every minute. Using this Poisson Calculator, we enter 1 in the "Number of Combinations" field (representing 1 minute), and then enter 3 in the "Ticket Sales" field (i.e. 3 cars). Next, we enter 6 in the "Number of Winners" field, and check the "or More" radio button.
Now, we can see the answer that the Probability of "more than 5 cars" is equal to 8.382%.
Links
Automobile Congestion (see example on page 10)
The Poisson Probability Distribution (see example 4)
It has been observed that 720 vehicles pass through a certain point every hour.
Calculate the probability that exactly 5 vehicles pass through this point in 5 seconds.
To answer this, we note that there are 3600 seconds in an hour. So, we enter: 3600 in the "Number of Combinations" field; and, 3600 (720 * 5 seconds) in the "Ticket Sales" field. Lastly, we enter 5 in the "Number of Winners" field, and check the "Only" radio button.
The answer is found to be 0.307% or 0.00307.
Links
Distributions of Traffic Flow
When Scale checked, all data is scaled such that the number of combinations approaches 1,000; otherwise, the raw values are used for calculations.
This is important because the larger the number of combinations becomes, the closer the binomial distributation values approach those of the closed form Poisson formula.
Lambda is the Poisson Distribution Average, computed by dividing the Number of Items by the Number selected
Note: it can only be entered when the pulldown selection is set to Other
Only and or More determine what is shown in the Probability and Expected Amount fields. Only means the exact matches, whereas or More represents the cumulative total.
Note, when Number of Winners is set to 0 and or More is clicked, the Probability will always be 100%. But, we do not show this. Instead we display the Only value as we believe it is more meaningful.