OK, so what is the mathematical likelihood that the two Kneeshaws would win BOTH Grand Prizes??
To be honest, I mis-figured this at least once, as I carelessly forgot to account for the rule that a subscriber could only win ONE prize, and then make the necessary adjustments to handle that.
There were 275 sweepstakes entries, and between them the Kneeshaws had 22 of them (19 for Bobbi – among the highest number – and 3 for Steve). We are drawing two winners, but after a person has won, they must be eliminated from the remaining list for the remainder of the prizes. So we have to account for the fact that Steve could win first, and then (after eliminating his entries,) Bobbi win 2nd, OR that Bobbi could win first, then Steve win second. This is non-technical, but math rules in this case allow us to figure each probability separately and add them.
So, the likelihood that [Steve wins first, then Bobbi] is (3/275)*(19/272) = .00076 (rounded to 5 places).
The chance that [Bobbi wins first, then Steve] is (19/275)*(3/256) = .00081 (rounded to 5 places)
Adding them gives .00157 which works out to roughly 1/637. Expressed as odds, that’s 636 to 1.
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