Another quiz - this time staying on lottery topic!Prev TopicNext Topic

• Orange71

  

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  Jun 8, 2025, 7:39 pm

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  Suppose we have a 6/54 lottery. In a given lottery draw 6 balls are randomly picked in the range of 1 to 54. What is the probability that, in 50 draws, at least 1 ball is not drawn?
• Orange71

  

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  Jun 15, 2025, 12:19 pm

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  To make it more interesting, solve for the probability that at least m  balls are not drawn out of n 6/54 lottery games where  1 ≤ m ≤ 48 and n ≥ 9.
• Tucker Black

  

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  Jun 19, 2025, 12:43 pm

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  I'm stuck. It seems easy but my formula doesn't match a simulation. My formula said 0.139078 but my simulation said 0.1399.
• Orange71

  

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  Jun 21, 2025, 11:53 am

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  In response to Tucker Black

  Quote: Originally posted by Tucker Black on Jun 19, 2025

  I'm stuck. It seems easy but my formula doesn't match a simulation. My formula said 0.139078 but my simulation said 0.1399.

  To eight significant digits I get 0.14015016. 

  To derive the solution, we need to apply the "Inclusion, Exclusion" Principle. In general, only mutually exclusive events can be added to arrive at a total probability. If events are not mutually exclusive, their intersections must be subtracted from the total. However, with combination of events adding or subtracting a particular number of combinations may then need to be compensated by (alternately) adding or subtracting higher levels of combinations.

  Let's say in this case we start with the probability of a specific ball not being drawn in 50 lottery games. There are 54 balls, and the probability of this event for each is the same. We could at this point add these together for a provisional total. However, these are not mutually exclusive events, so we need to subtract out event combinations, such as neither the balls 8 and 10 being drawn. There are 54C2 combinations of two balls. However, we are only getting started. There are combinations of m = 3, 4, 5, ..., 48 balls. Each case is alternately added or subtracted to the total to eventually compute the total probability. Each probability sum for m+1 balls is lower magnitude than than that for m balls. Why stop at 48? The game is 6 balls drawn out of 54. So those most extremely improbable event is the same 6 balls being picked 50 times. Therefore at most 48 balls will not be drawn.

  

  If we want the probability of "at least 2 balls not drawn" it would be:

  

  Note that "exactly one ball" and "at least two balls" are mutually exclusive, so we can add them to total P of "at least one ball". The same logic can be extrapolated to "exactly m balls" or "at least m balls". For P(1), i.e. exactly 1, I get 0.13075679.
• Il2c

  

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  Jul 14, 2025, 5:13 pm

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  Well.

  Let's just round it up to 0.2

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• Il2c

  

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  Jul 14, 2025, 5:21 pm

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  In response to Orange71

  Quote: Originally posted by Orange71 on Jun 15, 2025

  To make it more interesting, solve for the probability that at least m  balls are not drawn out of n 6/54 lottery games where  1 ≤ m ≤ 48 and n ≥ 9.

  This would basically be giving you the winning lottery numbers.

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