A gambler decides to play roulette. His initial investment is $200. He bets $100 each time and will quit when one of the following events happens:
i.He loses his money, within less than 8 bets.
ii. He's got $500, within less than 8 bets.
iii. He has played 8 times ending up with any amount.
a. What is the number of ways the betting can ocur?
b. What is the most likely length of his play (number of bets) within the above constraints?
c. Provide the probabilities of each of the three events (i, ii, iii).
Remark: Assume betting on even chances i.e. w/l=1
d. (bonus question) Would your answers change if the w/l odds were 18/19?
(In reply to computer assisted solution
I verified parts a,b,c by hand. I also found the expected number of turns: 39/8 and the expected amount of money at termination: unsurprisingly $200.
One really cool result I happened upon:
The game cannot end in one turn. After that it can end on an even number of turns at $0 and an odd number of turns at $500.
Observe the probabilities of ending in 1,2,3 etc turns with denominators as powers of 2:
0/2, 1/4, 1/8, 2/16, 3/32, 5/64, 8/128, 13/256, 21/512 (assuming 9th turn)
Anyone care to prove the numerators are the Fibonacci sequence?
Posted by Jer
on 2016-05-20 21:46:01