Jill likes to play Blackjack. She heard about a winning strategy to ensure she does not lose. The strategy is: choose a starting amount to bet, and then double your bet if you lose. Keep doubling your bet until you win. When you win a hand, go back to the starting bet. Hoping for only a modest win, she chose a table that had a minimum bet requirement of only $3. If she brings $300 to play with, betting $3 for the first hand, how many hands in a row can she afford to lose before running out of money (so she wouldn’t have enough to bet double on the next hand)? Use an appropriate formula and Algebra, not guessing and checking or making a table, etc. Show work.
I still make a chart or table to establish some kind of pattern for my formulas or equations.
consider consecutive losses
first bet = 3
total losses = 3
second bet = 6
total losses = 9
third bet = 12
total losses = 9+12 = 21
looks like a geometric series where
a = 3, r = 2 and we want to know Sum(n) < 300
Sum(n) = a(rn - 1)/(r - 1)
300 = 3(2^n - 1)/(2-1)
2^n - 1 = 100
2^n = 101
n = log 101/log2 = 6.65
So after 6 losses she lost 3(2^6 - 1)/2-1) = 189
after 7 losses she lost 3(2^7 - 1)/2-1) = 381
she can play 6 times in a row, expecting to lose each time , running out on the 7th turn.
To solve this problem, we can use a formula to find the number of consecutive losses Jill can afford before running out of money.
Let's break down the information given:
- Jill starts with $300.
- She bets $3 for the first hand.
- After a loss, she doubles her bet for the next hand.
- Jill wants to know the number of consecutive losses she can afford before she runs out of money.
Let's assume Jill loses x consecutive hands before winning.
In terms of bets, the first bet is $3. After x consecutive losses, Jill's bet will be doubled each time. So, the second bet will be $3 x 2 = $6, the third bet will be $6 x 2 = $12, and so on.
The amount of money Jill spends on bets can be calculated using a geometric series. The sum of a geometric series is given by the formula:
S = a * (1 - r^n) / (1 - r)
Where:
S = sum of the series
a = first term
r = common ratio between terms
n = number of terms
In this case, the first term (a) is $3, the common ratio (r) is 2 (since she doubles her bet every time), and the number of terms (n) is x (the number of consecutive losses).
We want to find the maximum number of consecutive losses (x) before Jill runs out of money, which means the sum of the series should be less than or equal to her starting amount of $300.
So, we can set up the following inequality:
a * (1 - r^n) / (1 - r) ≤ $300
Plugging in the given values:
$3 * (1 - 2^n) / (1 - 2) ≤ $300
Simplifying the equation:
3 * (1 - 2^n) / -1 ≤ $300
Multiplying both sides by -1 (to change the direction of the inequality):
3 * (2^n - 1) ≥ $300
Now we have an inequality that we can solve for n.
First, divide both sides of the inequality by 3:
2^n - 1 ≥ $100
Next, add 1 to both sides:
2^n ≥ $101
To solve for n, we can take the logarithm (base 2) of both sides:
log₂(2^n) ≥ log₂($101)
n ≥ log₂($101)
Using a calculator, we can determine that log₂($101) is approximately 6.673.
Therefore, Jill can afford to lose a maximum of 6 consecutive hands before running out of money.