Question 533939: Consider a primitive slot machine which, for each quarter inserted, rewards the user with $10 with the probability 1/50. What is the probability of leaving with more money than one started if 100 quarters are inserted? Use the exact binomial distribution
Answer by fcabanski(1391)   (Show Source):
You can put this solution on YOUR website! The key in this problem, as well as any related problems that use approximations such as the normal approximation, is knowing how many of the pulls must result in a "win" in order to walk away with more money.
If that's set up correctly, then it's an easy matter to plug in the numbers into the exact binomial formula, the normal distribution, or Poisson.
For exact binomial, we have to concern ourselves with the number of calculations. That will come up when it's time to find the actual probability.
Examine this experiment.
We don't know how many "win" pulls we need to earn money, so lets call those X. Each X results in a net of $9.75 ($10 won - (minus) the $.25 it cost for the pull.) The value for all the x wins is 9.75x.
We don't know how many "lose" pulls we can have while still staying in the black, so lets call those Y. Each Y results in a $ 25 loss. The value for all the Y losses is .25y.
We now have two equations.
# of Pulls equation: All the pulls together must equal 100. X + Y = 100
Value of all pulls equation: The total value of all the pulls must be positive (greater than 0). That means we walk away with more money than we had when the experiment began.
9.75X + -.25Y > 0
Two equations, two unknowns. Let's solve the # of pulls equation in terms of one of the unknowns (let's use Y), and then substitute that into the value equation so we find out how many X (win) pulls we need to have positive value.
X + Y = 100 (subtract X from both sides) Y = 100 - X
Substitute that into the value equation: 9.75X - .25 (100-x) > 0
Multiply .25 by both items in the ()
9.75X - (.25*100 - .25*X) > 0
Distribute the - (multiply everything inside the () by -1):
9.75X - 25 + .25X >0 Add like terms
10X - 25 > 0 add 25 to both sides.
10x > 25 divide both sides by 10. x > 2.5
So we need more than 2.5 wins in order to walk away with a net gain. That means we must have 3 or more wins, because we can't have fractional pulls. Each pull is 1, there are a whole number number of pulls, and the smallest whole number after 2.5 is one.
Now we perform as many Binomial Distribution calculations, in a series, as needed to find the combined probability of:
3, 4, 5, 6...100 successful pulls. That's way too many. But we can calculate the probability of too few winning pulls (0, 1, or 2), then subtract that from 100. In other words, we find the probability of 2 or less, then subtract that from 100% (1) to find the probability of not 2 or less. That will be three or more.
n - number of opportunities = 100
k - number of wins (0, 1, 2)
p - probability of a win = 1/50 = .02
q - probability of a loss = .98
0 winning pulls (k=0) =
n!/(k!(n-k!))*p^k*q^(n-k) = (100!/(0!*(100-0)!))*.02^0*.98^(100-0) = .1326
+
1 winning pull (k=1) =
n!/(k!(n-k!))*p^k*q^(n-k) = (100!/(1!*(100-1)!))*.02^1*.98^(100-1)= .2706
+
2 winning pulls (k=2) =
n!/(k!(n-k!))*p^k*q^(n-k) = (100!/(2!*(100-2)!))*.02^2*.98^(100-2) = .2734
So the probability we want, which is of 3 or more winning pulls is:
1 - (.1326+.2706+.2734) = approx .3233 or 32.33%
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