Wouldn't P be 2 since either place can win. Even if they are the same we still look at them and each one can win even though we aren't increasing the payout.
Nah. I hear ya, but that is why there is an "n" in the top right too. "n" places to win on, but only "P" gets paid... so if all "n" places turns out to be the identical number, only "P" times will the jackpot be paid. In this case, it's 2 minus 1. Out of the 2 spots, only 1 gets paid at any given time, even if both spots have the same winning hex.
I spent over 35 minutes figuring out a formula that is scalable. Feel free to play with the numbers, you'll see it works for all base numbers and paid places!! (I'm pretty proud of that)
Try these numbers:
base 10 [0 - 9], 2 spots, 1 paid = 19:100 (1 in 5.26)
base 8, 2 spots, 1 paid = 15:64 (1 in 4.27)
base 16, 3 spots, 1 paid = 46:4096 (1 in 89)
base 2 [binary], 3 spots, 1 paid = 4:8 (1 in 2)
base 16, 2, 2 = 32:256 (1 in 8 for single payout, 1 in 256 for double payout)
It works. The formula. I do think it's easier to just pay out for any one instance.
