calculus
posted by Nic on .
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)/(21)
2^n  1 = 100
2^n = 101
n = log 101/log2 = 6.65
So after 6 losses she lost 3(2^6  1)/21) = 189
after 7 losses she lost 3(2^7  1)/21) = 381
she can play 6 times in a row, expecting to lose each time , running out on the 7th turn.