Dream Catcher odds: the multiplier chain, solved

The rules and the layout

Evolution's page states a wheel of 54 equal segments: 52 carry a number (1, 2, 5, 10, 20 or 40) and two are multipliers, a silver 2x and a gold 7x. A winning number pays its odds, for example 5 to 1. If the wheel stops on a multiplier, all bets stay in place and the wheel spins again with the next result's odds multiplied by 2 or 7; if another multiplier lands, the multipliers stack and the spinning continues until a number ends the round. Source: Evolution's Dream Catcher page. The page gives no per-number split. We use the usual layout of 23, 15, 7, 4, 2 and 1 segments, which totals the 52 the page states.

Solving the multiplier chain

A round ends only when a number lands, so the question is how big the multiplier is by then. Let M be the multiplier at the end of the round. With probability 52/54 the first spin is a number and M = 1. With probability 1/54 each the wheel stops on 2x or 7x and the process restarts, with M multiplied by 2 or 7. So E[M] = 52/54 + (1/54) x 2 x E[M] + (1/54) x 7 x E[M], which gives E[M] x (1 - 9/54) = 52/54, and E[M] = 52/45 = 1.1556.

The terminal number is independent of the multiplier, and a given number ends the round with probability count/52. A bet on a number n therefore returns count/52 x (n x 1.1556 + 1) of the stake on average.

What the table says

The cheapest bets on a long run are the 10 and the 2 and 1, between 3.4% and 4.7% house edge. The big numbers are the expensive ones: a bet on 40 costs 9.19 per 100 units, nearly three times the edge of a bet on 10. That is a general feature of wheel games. The longer the shot, the bigger the cut taken from its payout.

Multipliers show up on 3.70% of spins. A double chain, two multipliers in a row, happens on 0.14% of spins, about 1 in 730. The advertised 20,000x ceiling needs a 40 plus stacked multipliers, and its probability is tiny. A ceiling is not an average.