Ask Question
12 February, 17:50

A coin is tossed until the first time a head turns up. If this occurs on the n th toss and n is odd you win 2n/n, but if n is even then you lose 2n/n. Then if your expected winnings exist they are given by the convergent series.

+4
Answers (1)
  1. 12 February, 18:11
    0
    This is a series of variable harmonics that converges to log (2).

    Step-by-step explanation:

    Let's ignore for a moment that n is even or odd, and look at the expected value for any n. Let X be profit, which can be negative if we lose. The wait is given using (and assuming the coin is valid)

    E[X]=∑n=1∞ ((-1) n+1*2∧n/n) ⋅1/2∧n=∑n=1∞ (-1) n+1*1/n.

    This is a series of variable harmonics that converges to log (2). However, expectation exists only if it absolutely converges! Looking at

    ∑n=1∞∣ (-1) n+1*2∧n/n∣*1/2n=∑n=1∞1/n

    we notice that a number of harmonics diverge, therefore, in fact, there is no expectation of X.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A coin is tossed until the first time a head turns up. If this occurs on the n th toss and n is odd you win 2n/n, but if n is even then you ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers