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Eastern Time (GMT-5:00) # Bose/Einstein Problem applied to a lottery example - read this to feel "smart"Prev TopicNext Topic

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• Sugar Land
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August 28, 2019
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Bose/Einstein is a famous probability problem that was originally posed between Albert Einstein and another scientist named Satyendra Bose. The analysis has broad application, not just the physics problem it was originally addressing.

Suppose we have "n" lottery balls in a pool and we pick one at random "k" times, replacing the selected ball each time. Then, when the order of selection is considered distinct, there are n^k possible outcomes, where "^" is "to the power of".

However, what if the order of selection does not matter, just the number of selections of each ball and how frequently they occur? For the sake of argument, say there are 10 balls with enumerated (0,1,2,...,9) and we select 5 balls. We consider the outcome 1-1-3-3-5 the same as 1-5-1-3-3, i.e. the distribution of balls and their frequencies are the same. How do we count the number of possible outcomes? This is the Bose-Einstein problem.

Consider a physical experiment where we have "walls" represented by "|" pictorially, which in turn represent the boundaries of "boxes", each "box" containing common balls - e.g. "3", "5" or so on. Again, there are "n" boxes and "k" selections. Represent a given selected ball as "*". For a given "box", all the "*"s are indistinguishable.

As an example, we might have the following for 10 balls (corresponding to "boxes") and 15 random selections from the ball pool (with replacement) represented by "*". A random outcome could be as follows. The space between each two dividers "|" represents each box.

||****|*|**||***|*||***|*| which is equivalent to 0 count of ball 0, 4 count of ball 1, 1 count of ball 2, 2 count of ball 3, 0 count of ball 4, 3 count of ball 5, 1 count of ball 6, 0 count of ball 7, 3 count of ball 8, and 1 count of ball 9. The total number of boxes is 10 and the total selections is 15.

So, in general, the outer dividers "|" are fixed (i.e. first and last positions in the sequence), while the inner dividers can move around. Specifically, there are n-1 "|"s between the outer "|"s and k "*"s, for a total of n-1+k. Then the number of possible arrangements of the "|"s is (n-1+k)!/[(n-1)!k!] , or C(n-1+k,n-1). Let's also denote k_1, k_2, k_3,...k_n as the number of counts of "*"s in each box. Then, we have the bounding condition that k_1 + k_2 + k_3 + ... k_n = k.

As an illustrative example using the Powerball lottery, let's correlate the "boxes" to each tier of prizes, including no prize. These are (w/o Powerplay): \$0,\$4,\$7,\$100,\$50k,\$1M, and \$Jackpot. So there are a total of 7 possibilities. Let's say there are 20 people in an office (corresponding to the "*"s) and each purchases a Quickpick. What are the number of possible prize tier assortments across the 20 players, where only the counts for each prize tier matter, not which employee in the office gets a certain tier? Answer: n=7 and k=20, Result = C(n-1+k,n-1) = C(7-1+20,7-1) = C(26,6) = 230230.

What if, in the above example, we wanted to count the number of assortments where no player loses, i.e. result of tier \$0. Now, effectively, n=6, while k remains as 20. Therefore, Result = C(6-1+20,6-1) = C(25,5) = 53100.

These calculations just reflect the number of unordered assortments, not the probabilities. The probabilities of each assortment, which can be calculated in any and all cases, will vary widely.

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