Probabilities? (Correct mathematical solution)?
At the roulette table
Manuela, Markus, Pedro and Pauline visit a casino at their holiday resort.
At the roulette table you can bet on the following events, among others:
Rouge: all red numbers
Noir: all black numbers
Column 34: the first longitudinal row (1;4;7;..; 34) Column 35: the second longitudinal row (2; 5; 8; …; 35) Column 36: the third longitudinal row (3; 6; 9; …; 36)
Manuela always chooses a Colonne, Markus always chooses Noir.
Pedro wants to bet in such a way that Markus's probability of winning changes as little as possible when Pedro is known to win. State how Pedro should place his chip. Explain why there is no betting option that results in no change in Markus' probability of winning.
-Pauline wants to bet on column 35 and a color. Indicate which of the two colors she should bet on to maximize the probability of winning.
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Part 1)
If Pedro sits on a single red number and wins, the probability is that Markus also wins zero.
If Pedro puts on a single black number and wins, the probability is that Markus also wins equal 1.
If Pedro sits on a red and a black number and wins, the probability that Markus also wins equal to 1/2.
The latter corresponds approximately to the probability that Markus wins independently of Pedro :
1/2 ~ 18/37
For Pedro, there are many other ways to put chips with similar results, e.g. on n red and at the same time on n black numbers.
The assertion “Because there is no possibility of setting for which there is no change in the probability of winning from Markus.” is disgusting.
Put Pedro on all numbers from 0-36, Pedro always wins, and the likelihood that then Markus wins remains at 18/37.
Pedro makes a loss with this strategy, but it’s not in demand. If you link the term “profit probability” with the event of achieving financial gain, the probability 18/37 is never exactly achieved. Proof:
Be Em the event “Markus wins”
Be Ep the event “Pedro wins”
For the conditional probability:
(I): p (Em under condition Ep) = p(Em and Ep) / p(Ep)
This probability should be 18/37:
(I): p(Em and Ep) / p(Ep) = 18/37
It shall apply:
p(Ep) = n/37, p(Em and Ep) = m/37 with n,m = {1,2,…,36}
Use values:
(I): (m/37) / (n/37) = (18/37)
This follows:
(I): n/m = 37/18
Because of n and m <= 36, there is no solution to the equation.
Part 2)
Column 35 has 4 red and 8 black numbers. The probability of winning is increased if you additionally rely on all red numbers.