Expected Value

Question

I get to answer 1 question a day that I have a 70% chance of getting that question right. I'm offered prizes for getting a streak of questions right in a row. However, anytime I accept a prize, the next day my streak will start back at 0.

If I get the question right 2 days in a row, I can cash out a $1 prize. 

If I get it right 4 days in a row, I can cash out a $3 prize.

If I get it right 6 days in a row, I can cash out a $6 prize.

If I get it right 8 days in a row, I can cash out a $10 prize.

If I get it right 10 days in a row, I can cash out a $20 prize.

If I get it right 15 days in a row, I can cash out a $50 prize.

If I get it right 20 days in a row, I can cash out a $250 prize.

If I get it right 30 days in a row, I can cash out a $1,000 prize.

If I get it right 50 days in a row, I can cash out a $100,000 prize.

If I get it right 60 days in a row, I can cash out a $1,000,000 prize.

I can only have 1 streak going at a time. There is no cost to attempting this and there is an unlimited amount of streaks I can start. What streak would be the smartest (put the most money in my pocket) to go for?

Answer 1

Hi Chris,

           Interesting question. The binomial distribution gives the probability of k successes in n trials where the probability of a single trial being successful is p is:

 

P(X=k)=_nC_k*(1-p)^n-k*p^k in this problem n=k and p=.7 so P(X=k)=.7^k so the expected amount for each of the games would be Prize*.7^k

 

which is: $ .7203 for the first one (4 days in a row $3 prize) and P(X=4)=.7⁴=.2401. $3*.2401=$.7203 note that this is already smaller than the 2 days in a row that gets you $1. The expected amounts get progressively smaller as the number of days increase.

 

I would just go for 2 days in a row P(X=2)=.7²=.49 which is close to 50/50 odds and is the best bet.

 

Hope this helps

 

Jim

 

 

Answer 2

Go for cxashing out the day before you answer a question incorrectly!