Random Answer, Correct! - Part 2

Probability Level 5

John has just registered an account on Brilliant. Each day, starting from today, he logs in, chooses a multiple choice problem, randomly answers it, and logs out. So if he is incorrect on one day, he loses his problem solving streak.

He stops this procedure once he gets a problem incorrect, or once he gets a streak of 10 days. The expected length of John's streak can be written as $\frac{p}{q}$ , where $p$ and $q$ are coprime. Find the last $3$ digits of $p+q$ .

Note: The streak can be 0, in case that John gets the problem he answers on the first day incorrect.

The answer is 101.

1 solution

Shenal Kotuwewatta
Mar 23, 2014

Each day, John has a probability of $\frac{3}{4}$ of losing his streak, and a probability of $\frac{1}{4}$ of maintaining his streak.

So the expected value for his streak is

$\prod _{ n=0 }^{ 9 }{ n{ (\frac { 1 }{ 4 } ) }^{ n } } { (\frac { 3 }{ 4 } ) }=\frac { 349525 }{ 1048576 }$ .

So $p+q=1398101$ and the required answer is $\boxed{101}$ .

Moderator note:

This solution has an incorrect explanation. See Jon Haussmann's comment below.

HOW IS IT TRUE FOR n=10 as when he attains a streak of 10 he doesn't attempt for another day.

Athul Nambolan - 7 years, 2 months ago

I agree with you, since otherwise the total probability would not sum to one. I got 101 on the first try ( $\frac {349525}{1048576} \rightarrow$ 1398101), but after I realized the possibility of this mistake, I "solved" the problem.

A Former Brilliant Member - 6 years, 9 months ago

Thanks. I have updated the answer.

In future, if you spot any errors with a problem, you can “report” it by selecting the “dot dot dot” menu in the lower right corner. You will get a more timely response that way.

Calvin Lin Staff - 6 years, 5 months ago

I get the same answer as the proposer. Let $P(k)$ denote the probability that the streak lasts exactly $k$ days, for $0 \le k \le 10$ . Then $P(k) = (1/4)^k \cdot 3/4$ for $0 \le k \le 9$ , and $P(10) = (1/4)^{10}$ , so the expected value works out to $\sum_{k = 0}^9 \left( k \cdot \left( \frac{1}{4} \right)^k \cdot \frac{3}{4} \right) + 10 \cdot \left( \frac{1}{4} \right)^{10} = \frac{349525}{1048576}.$

Jon Haussmann - 6 years, 5 months ago

Thanks. I was misled by the creator having the wrong formula stated, and based the value off of a calculation submitted in disute. I have verified that this answer is correct, and updated it back to 101.

Calvin Lin Staff - 6 years, 5 months ago

The answer is wrong. No wonder 95% of the people couldn't solve it. Your probabilities don't sum to one and you forgot that the 3/4 only comes in if he got an answer incorrect. However, the question clearly states that he stops if he has a streak of 10 so where did the 3/4 come in then??

Omkar Kamat - 6 years, 5 months ago

See Jon Haussman's comment above.

The numerical answer is correct, though the original explanation is wrong.

Calvin Lin Staff - 6 years, 5 months ago

Random Answer, Correct! - Part 2

The answer is 101.

1 solution

Moderator note:

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