calculate chance of winning this progressive jackpot

Step 1: Probability
Numbers 1 through 8 are arranged in random order.
House draws 8 numbers.
To win, the order of the drawn 8 numbers have to match the original arrangement.

Step 2: Return
jackpot = 20000
8! = 1/40320
return = (1/40320)*20000=0.496031746
rtp = return/bet size



But the game rules say: The more you bet, the greater the chance to win!
Does this mean that it's really impossible to calculate the chance of winning the jackpot?
Does the house manipulate the probability of winning the jackpot by not randomly drawing the 8 numbers?

Quote: bumblingfool

But the game rules say: The more you bet, the greater the chance to win!
Does this mean that it's really impossible to calculate the chance of winning the jackpot?
Does the house manipulate the probability of winning the jackpot by not randomly drawing the 8 numbers?

I don't understand the specifics of your game, but I imagine that if I bet 1 unit, I'd get one chance at the jackpot, whereas if I bet 2 units, I'd get two chances at the jackpot.

I would further imagine that would happen with a second (third, fourth...) number draw for additional units bet... just like buying a second (third, fourth...) lotto ticket.

edit: It could just as easily be that there is a second (third, fourth...) jackpot winning permutation, and additional units "buy a pay".

Perhaps "the more you bet" means "the more permutations of eight numbers you select".

Quote: ThatDonGuy

Perhaps "the more you bet" means "the more permutations of eight numbers you select".

You only get one permutation of 8 numbers regardless of how much you bet. So if the draw was really random, bet size would not affect probability of winning the jackpot.