Answer:
the number of games that you should play to at least break even with probability 0.99 is 30
Step-by-step explanation:
Given the data in the question;
we represent the total number of games by n
and among the n number of games, k is the number of wins.
so
P( at least break even ) = P( k × $5 ≥ $2( n -k ) )
P( k × $5 ≥ $2( n -k ) ) = P( 5K ≥ 2n - 2k ) = P( 7k ≥ 2n )
so
P( at least break even ) = P( 7k ≥ 2n ) ≥ 0.99
⇒ P( p' ≥ (2/7) ) ≥ 0.99
p' ~ Normal( Mean = 0.5, standard deviation = √( (0.5×0.5) / n )
so
P( Z ≥ ((2/7) - 0.5) / √( 0.25/n ) ≥ 0.99
⇒ ((2/7) - 0.5) / √( 0.25/n ) ≥ -2.326
⇒ (0.2857-0.5) / (0.5/√n) ≥ -2.326
⇒ -0.4286 × √n ≥ -2.326
√n ≥ -2.326 / -0.4286
√n ≥ 5.42697
n ≥ (5.42697)²
n ≥ 29.35 since its sample size
It approximately becomes 30
Therefore, the number of games that you should play to at least break even with probability 0.99 is 30
