Find the expected value of the additional pickups? I find a probability problem in a computer game. The problem is as following.
The player has a passive ability that has $28$% chance every $27$ second to grant random buff. There are 6 types of buff and they have equal chance to occur. The desired buff lasts $120$ second and it doubles($\times2$) the pickups of the player during the period. The desired buff effect can be piled and the player earns $\times2^n$ pickups for n numbers of the desired buff. What is the expected value of the additional pickups?
Someone has given the following solution:
Chance of random buff$ = 28$%
Time between trying buff$ = 27$s
Number of buff types$ = 6$
Desired buff$ = $“Double Pickups”
Duration of the desired buff$ = 120$s
Average time it takes to proc “Double Pickups”$ = (27$s$ \times 1/0.28) \times 6 = 578$s$ = 9.64 $min
How often is the “Double Pickups” active $ = 120$s$ / 578$s$ = 20.76$% of the time
How much pickups does one gain over all $ = 1 + (1 \times 0.2) = 1.2 = 120$%$ = 20$% more pickups overall
I don't think this is the right solution. May someone tell me if it is correct or not? What is the correct answer? Please explain.
 A: From the solution that you quote (which is incorrect, as you suspected), I gather that by “the expected value of the additional pickups” you mean the expected relative increase in the pickup rate due to the buff. I'll assume that the pickups form a time-invariant process (e.g. a Poisson process) that is independent of the rhythm of granting buffs.
Then we can obtain the expected relative increase by averaging over time. In any $27$-second period, there are $12$ seconds during which up to $5$ buffs could be active and $15$ seconds during which up to $4$ buffs could be active. Each of the desired buffs is active with probability $\frac16\cdot\frac{28}{100}=\frac7{150}$. Thus the expected factor by which the pickup rate is multiplied is
\begin{eqnarray*}
&&\frac{12}{27}\sum_{k=0}^5\binom5k\left(\frac7{150}\right)^k\left(\frac{143}{150}\right)^{5-k}2^k+\frac{15}{27}\sum_{k=0}^4\binom4k\left(\frac7{150}\right)^k\left(\frac{143}{150}\right)^{4-k}2^k
\\
&=&
\frac{12}{27}\left(\frac{143+2\cdot7}{150}\right)^5+\frac{15}{27}\left(\frac{143+2\cdot7}{150}\right)^4
\\
&=&
\left(\frac{157}{150}\right)^4\left(\frac{12}{27}\cdot\frac{157}{150}+\frac{15}{27}\right)
\\
&=&
\frac{418617935489}{341718750000}
\\
&\approx&
1.225\;,
\end{eqnarray*}
so the relative increase is about $22.5\%$, slightly higher than the result of the calculation that doesn't take into account multiple active buffs. The difference is small because the probability for more than one buff to be active simultaneously is low.
