Quote from: Bayes on December 30, 2014, 12:14:24 PM
Example 2
In an even chance there are 4 equally likely outcomes taking 2 decisions at a time, i.e., BB, PP, PB, BP. What is the chance of at least one repeat in 4 decisions (note that this equates to 8 hands, not 4, because each outcome consists of two hands)
N = 4 and r = 4, so
[math]P(repeat)=\frac{4^4(4-4)!-4!}{4^4(4-4)!} = 0.906[/math]
or a little over 90%.
You can find a scientific calculator here.
BB
PP
PB lose 1 bet
P? now what
Eirescott posted many years ago, to place the P bet after the PP, and PB outcomes, because if the result is P it then becomes ambiguous therefore a no-bet scenario for which to recoup the prior lost bet.
Either way you will lose more hand per shoe than win, despite the 90% expectation figure.
What is the formula if you include Eirescotts adaptation?