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:STATEMENT:

Suppose an individual plays a gambling game where it is possible to lose $1.00$, break even, win $3.00$, or win $10.00$ each time she plays. The probability distribution for each outcome is provided by the following table:

Outcome     -$1.00	$0.00   $3.00	   $5.00    
Probability  0.30       0.40     0.20      0.10 

The mean outcome for this game is calculated as follows:

$$ \mu = (-1*.3) + (0*.4) + (3*.2) + (10*0.1) = -0.3 + 0.6 + 0.5 = 0.8$$

In the long run, then, the player can expect to win about 80 cents playing this game -- the odds are in her favor.

In the above gambling example, suppose a woman plays the game five times, with the outcomes $0.00$, -$1.00$, $0.00$, $0.00$, -$1.00$. She might assume, since the true mean of the random variable is $0.80$, that she will win the next few games in order to "make up" for the fact that she has been losing.

Unfortunately for her, this logic has no basis in probability theory. The law of large numbers does not apply for a short string of events, and her chances of winning the next game are no better than if she had won the previous game.


QUESTION

If she plays the game 500 times a day for next 30 days = 15000 games. Wouldn't she will definitely win because it is a large number now?

Original: http://www.stat.yale.edu/Courses/1997-98/101/rvmnvar.htm

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  • $\begingroup$ Suppose the expected value of the game were $\$8.00$ rather than $\$0.80:$ would you suppose she therefore has an 800% chance of winning?? $\endgroup$ – whuber Jul 6 at 14:12
  • $\begingroup$ Question edited $\endgroup$ – Arnuld Jul 6 at 14:15
  • $\begingroup$ en.wikipedia.org/wiki/Gambler's_fallacy $\endgroup$ – Tim Jul 6 at 19:02
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Here are traces of running averages of winnings during four 15,000-play gambling sprees. Nothing is absolutely for sure, but it would be a surprise if the gambler didn't come out ahead (with average winnings near $\$0.80$ shown by blue lines) after many plays of such a favorable game.

enter image description here

R code for figure:

set.seed(2020)
N = 15000; n = 1:N
pay=c(-1,0,3,5); prob=c(.3,.4,.2,.1)
par(mfrow=c(2,2))
for (i in 1:4) {
 w = sample(pay, N, rep=T, p=prob)
 t = cumsum(w); avg = t/n
 plot(n, avg, type="l", lwd=2)
  abline(h=0,col="green2")
  abline(h=.8, col="blue")
}
par(mfrow=c(1,1))
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