1. Consider a simple gambling game. Each time you bet $10. If you lose (with pro
ID: 3126852 • Letter: 1
Question
1. Consider a simple gambling game. Each time you bet $10. If you lose (with probability 1 p), you lose the $10 you bet; if you win(with probability p), you get back your $10 and win extra $10 dollars. The games are assumed to be independent from one time to another. Suppose you play the game repeatedly. Define a random variable N to be the number of games until your first winning. (a) Give the distribution of N. (b) Let L be the total amount of dollars you have lost until your first winning. Write L as a function of N. (c) Find the the expected loss E(L) until your first winning.
Explanation / Answer
If N is the number of games to be played to have the first game to be won, then the first N - 1 games have to be lost.
So,
The probability distribution = (1-p)N - 1 * p [ this is an example of geometric districbution ]
L = amount of dollars lost in N-1 games.
Since, each lost games cost $10,
L = -10(N-1) [ negative sign indicates the loss. It can be dropped if L is assumed to be inherently negative ]
If the net loss is required, then : Lnet = -10(N-1) + 10
c)
Expected loss : Expected number of failures before first success = ( 1 - p ) / p
Thus,
Expected loss = 10 (1-p) / p
Hope this helps. Ask if you have doubts.
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