In a casino game based on the standard gambler's ruin problem, the gambler and the dealer EACH start with 20 tokens.?

.45 = (1 -(p/q)^20)(1 -(p/q)^40)
.45 = 1/(1 +p/q)^20
.45(p/q)^20 = .55
(p/q)^20 = .55/.45 = 11/9
p/q = (11/9)^(1/20) = 1.01008
This is based on probability of ending penniless.
1/(1 +1.01008) = .4975
49.75% is the gambler's chance of winning an event.

Note that 49.75% plus 50.25% shown in the other result = 100%. Since they have equal starting amounts, the one with the advantage will more often bankrupt thd other.

https://en.wikipedia.org/wiki/Gambler%27s_ruin#Unfair_coin_flipping

p = ?
q = ?
n1 = 20
n2 = 20

We know that P1 = 0.45

0.45 = (1 - (p/q)^20) / (1 - (p/q)^40)
0.45 = (1 - (p/q)^20) / ((1 - (p/q)^20) * (1 + (p/q)^20))
0.45 = 1 / (1 + (p/q)^20)

p/q = k

0.45 = 1 / (1 + k^20)
9/20 = 1 / (1 + k^20)
20/9 = 1 + k^20
20/9 - 1 = k^20
11/9 = k^20
(11/9)^(1/20) = k
1.0100840394552964420805360402189... = k
1.0100840394552964420805360402189... = p/q
q * 1.0100840394552964420805360402189... = p

p + q = 1
q * 1.0100840394552964420805360402189... + q = 1
q * (1.0100840394552964420805360402189 + 1) = 1
q * (2.0100840394552964420805360402189) = 1
q = 1 / 2.0100840394552964420805360402189
q = 0.49749163735014057665957153582118

1 - q = p
1 - 0.49749163735014057665957153582118... = p
p = 0.50250836264985942334042846417882....

The chances of the player winning any single time is 50.2508%, or 0.5025