Let X be the random variable for the number of draw attempts needed to get the blue counter.

For any X = x, Amount spent = $1 per draw × xdraws
= $x

Total amount of money he can be expected to spend

= E(X) (basically the probabilities for each possible number of attempts × amount spent for them)

= ∑ xP(X = x)

Since they are drawn with replacement, the probability of getting a blue counter in a draw/trial = 2/7 and the trials are all independent of each other.


We can use this formula for P(X = x)

= p (1 - p)ˣ⁻¹, where p = 2/7 and x - 1 = number of failed attempts (i.e he draws a red counter)


E(X) = ∑ xP(X = x) , from getting it on the 1st try, all the way up to an infinite number of tries.


E(X) = (1)P(X = 1) + 2 P(X = 2) + 3 P(X = 3) + 4 P(X = 4) + ...


E(X) = 1(2/7)(5/7)¹⁻¹ + 2(2/7)(5/7)²⁻¹ + 3(2/7)(5/7)³⁻¹ + 4(2/7)(5/7)⁴-¹ + ....


= 2/7 + 2(2/7)(5/7) + 3(2/7)(5/7)² + 4(2/7)(5/7)³ ....


= 2/7 (1 + 2(5/7) + 3(5/7)² + 4(5/7)³ + ...)


= 2/7 (1/(1 - 5/7)²)

(Using the approximation given in the question. Here, x = 5/7. Note that |5/7| = 5/7 < 1)


= 2/7 (1/(2/7)²)

= 1/(2/7)

= 7/2 or 3.5


So, he should he receiving $3.50 when he wins, since he is expected to spent $3.50.

(Recall that E(X) is basically a weighted average of the amount expected to be spent , after taking into account all the possible outcomes and their probabilities.

The concept is similar to finding the relative atomic mass of an element with different isotopes in chemistry)


This way, it is fair as he will be making no loss.

Edited. Hope it helps and do comment if you have any questions about the working.

thank you so much! rlly appreciate it:)

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