If the Warriors lose in Portland and the Wizards lose to Indiana, Sacramento would be tied with both for the league's third worst record. If this happens, the teams would have the number of lottery combinations for the 3-4-5 teams pooled and split as evenly as possible. That would work out to each team having the following odds.
Pick No. 1: 12.1% chance
Pick No. 2: 12.6% chance
Pick No. 3: 13.2% chance
Coin flips would decide the order in the event two or more of the teams are pushed out of the top three by No. 4-14 teams who luck into the top of the lotto. The lowest the Kings could pick would by No. 8; however, that would be most unlikely, because it would mean three teams, each with a 6% percent chance or worse, will have snuck into the top three.
MOST LIKELY SCENARIO
The Warriors and Wizards both suck, but I think maybe one of them will win. Don Nelson can add another win to his total, and Wizards host the Pacers, who are 9-31 on the road. Let's say one of them wins. Let's call that team the Wizziors. The Warzards lost. So the Kings would be tied with the Warzards for the league's third worst records. They would split the combos set aside for the 3-4 teams. Here are each team's odds.
Pick No. 1: 13.7% chance
Pick No. 2: 14.1% chance
Pick No. 3: 14.4% chance
As with the other scenario, a coin flip would set up the order in case a usurper infiltrates the righteous top three belonging to the Kings and Warzards. The lowest the Kings could pick would be No. 7.
In a perfect world, the Kings would be in the playoffs and not worrying about this bullsh-t. In a somewhat less perfect but ultimately quite grand world, the Wizziors and Warzards -- oh, sorry, the Warriors and the Wizards -- would both win, leaving the Kings alone in sole possession of the league's third worst record. The Kings lotto hopes would thus look like this.
Pick No. 1: 15.6% chance
Pick No. 2: 15.7% chance
Pick No. 3: 15.7% chance
The lowest the Kings would be able to pick would be Pick No. 6.