Find the expected value of the additional pickups?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
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$ = $“Double Pickupsâ€Â
Duration of the desired buff$ = 120$s
Average time it takes to proc “Double Pickupsâ€Â$ = (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.
probability statistics proof-verification probability-distributions expected-value
asked Aug 30 at 6:31
Tony Yeung
132
add a comment |Â
up vote
2
down vote
favorite
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$ = $“Double Pickupsâ€Â
Duration of the desired buff$ = 120$s
Average time it takes to proc “Double Pickupsâ€Â$ = (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.
probability statistics proof-verification probability-distributions expected-value
asked Aug 30 at 6:31
Tony Yeung
132
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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$ = $“Double Pickupsâ€Â
Duration of the desired buff$ = 120$s
Average time it takes to proc “Double Pickupsâ€Â$ = (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.
probability statistics proof-verification probability-distributions expected-value
asked Aug 30 at 6:31
Tony Yeung
132
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$ = $“Double Pickupsâ€Â
Duration of the desired buff$ = 120$s
Average time it takes to proc “Double Pickupsâ€Â$ = (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.
probability statistics proof-verification probability-distributions expected-value
probability statistics proof-verification probability-distributions expected-value
asked Aug 30 at 6:31
Tony Yeung
132
asked Aug 30 at 6:31
Tony Yeung
132
asked Aug 30 at 6:31
Tony Yeung
132
asked Aug 30 at 6:31
Tony Yeung
132
asked Aug 30 at 6:31
Tony Yeung
132
132
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
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 $frac16cdotfrac28100=frac7150$. Thus the expected factor by which the pickup rate is multiplied is
begineqnarray*
&½27sum_k=0^5binom5kleft(frac7150right)^kleft(frac143150right)^5-k2^k+frac1527sum_k=0^4binom4kleft(frac7150right)^kleft(frac143150right)^4-k2^k
\
&=&
frac1227left(frac143+2cdot7150right)^5+frac1527left(frac143+2cdot7150right)^4
\
&=&
left(frac157150right)^4left(frac1227cdotfrac157150+frac1527right)
\
&=&
frac418617935489341718750000
\
&approx&
1.225;,
endeqnarray*
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.
answered Aug 30 at 9:08
joriki
167k10180333
Thank you for this. So the quoted solution is right if only one buff can be active at the time. I can follow the part of binomial distribution but I don't understand how to derive the 12s & 15s sector.
– Tony Yeung
Aug 31 at 0:15
@TonyYeung: If you draw a diagram of the $27$-second phases (something like this:|______|______|______|______|______|______) and measure $120$ seconds from one of the buff opportunities, you can cover $4$ entire periods ($4cdot27=108$) with $120-108=12$ seonds left to spare. In those $12$ seconds, the potential buff from which you measured would still be active, as would the intervening four, for a total of five; whereas in the remaining $15$ seconds, only the intervening four would still be active.
– joriki
Aug 31 at 3:08
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
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 $frac16cdotfrac28100=frac7150$. Thus the expected factor by which the pickup rate is multiplied is
begineqnarray*
&½27sum_k=0^5binom5kleft(frac7150right)^kleft(frac143150right)^5-k2^k+frac1527sum_k=0^4binom4kleft(frac7150right)^kleft(frac143150right)^4-k2^k
\
&=&
frac1227left(frac143+2cdot7150right)^5+frac1527left(frac143+2cdot7150right)^4
\
&=&
left(frac157150right)^4left(frac1227cdotfrac157150+frac1527right)
\
&=&
frac418617935489341718750000
\
&approx&
1.225;,
endeqnarray*
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.
answered Aug 30 at 9:08
joriki
167k10180333
Thank you for this. So the quoted solution is right if only one buff can be active at the time. I can follow the part of binomial distribution but I don't understand how to derive the 12s & 15s sector.
– Tony Yeung
Aug 31 at 0:15
@TonyYeung: If you draw a diagram of the $27$-second phases (something like this:|______|______|______|______|______|______) and measure $120$ seconds from one of the buff opportunities, you can cover $4$ entire periods ($4cdot27=108$) with $120-108=12$ seonds left to spare. In those $12$ seconds, the potential buff from which you measured would still be active, as would the intervening four, for a total of five; whereas in the remaining $15$ seconds, only the intervening four would still be active.
– joriki
Aug 31 at 3:08
add a comment |Â
up vote
0
down vote
accepted
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 $frac16cdotfrac28100=frac7150$. Thus the expected factor by which the pickup rate is multiplied is
begineqnarray*
&½27sum_k=0^5binom5kleft(frac7150right)^kleft(frac143150right)^5-k2^k+frac1527sum_k=0^4binom4kleft(frac7150right)^kleft(frac143150right)^4-k2^k
\
&=&
frac1227left(frac143+2cdot7150right)^5+frac1527left(frac143+2cdot7150right)^4
\
&=&
left(frac157150right)^4left(frac1227cdotfrac157150+frac1527right)
\
&=&
frac418617935489341718750000
\
&approx&
1.225;,
endeqnarray*
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.
answered Aug 30 at 9:08
joriki
167k10180333
Thank you for this. So the quoted solution is right if only one buff can be active at the time. I can follow the part of binomial distribution but I don't understand how to derive the 12s & 15s sector.
– Tony Yeung
Aug 31 at 0:15
@TonyYeung: If you draw a diagram of the $27$-second phases (something like this:|______|______|______|______|______|______) and measure $120$ seconds from one of the buff opportunities, you can cover $4$ entire periods ($4cdot27=108$) with $120-108=12$ seonds left to spare. In those $12$ seconds, the potential buff from which you measured would still be active, as would the intervening four, for a total of five; whereas in the remaining $15$ seconds, only the intervening four would still be active.
– joriki
Aug 31 at 3:08
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
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 $frac16cdotfrac28100=frac7150$. Thus the expected factor by which the pickup rate is multiplied is
begineqnarray*
&½27sum_k=0^5binom5kleft(frac7150right)^kleft(frac143150right)^5-k2^k+frac1527sum_k=0^4binom4kleft(frac7150right)^kleft(frac143150right)^4-k2^k
\
&=&
frac1227left(frac143+2cdot7150right)^5+frac1527left(frac143+2cdot7150right)^4
\
&=&
left(frac157150right)^4left(frac1227cdotfrac157150+frac1527right)
\
&=&
frac418617935489341718750000
\
&approx&
1.225;,
endeqnarray*
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.
answered Aug 30 at 9:08
joriki
167k10180333
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 $frac16cdotfrac28100=frac7150$. Thus the expected factor by which the pickup rate is multiplied is
begineqnarray*
&½27sum_k=0^5binom5kleft(frac7150right)^kleft(frac143150right)^5-k2^k+frac1527sum_k=0^4binom4kleft(frac7150right)^kleft(frac143150right)^4-k2^k
\
&=&
frac1227left(frac143+2cdot7150right)^5+frac1527left(frac143+2cdot7150right)^4
\
&=&
left(frac157150right)^4left(frac1227cdotfrac157150+frac1527right)
\
&=&
frac418617935489341718750000
\
&approx&
1.225;,
endeqnarray*
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.
answered Aug 30 at 9:08
joriki
167k10180333
answered Aug 30 at 9:08
joriki
167k10180333
answered Aug 30 at 9:08
joriki
167k10180333
answered Aug 30 at 9:08
joriki
167k10180333
167k10180333
Thank you for this. So the quoted solution is right if only one buff can be active at the time. I can follow the part of binomial distribution but I don't understand how to derive the 12s & 15s sector.
– Tony Yeung
Aug 31 at 0:15
@TonyYeung: If you draw a diagram of the $27$-second phases (something like this:|______|______|______|______|______|______) and measure $120$ seconds from one of the buff opportunities, you can cover $4$ entire periods ($4cdot27=108$) with $120-108=12$ seonds left to spare. In those $12$ seconds, the potential buff from which you measured would still be active, as would the intervening four, for a total of five; whereas in the remaining $15$ seconds, only the intervening four would still be active.
– joriki
Aug 31 at 3:08
add a comment |Â
Thank you for this. So the quoted solution is right if only one buff can be active at the time. I can follow the part of binomial distribution but I don't understand how to derive the 12s & 15s sector.
– Tony Yeung
Aug 31 at 0:15
@TonyYeung: If you draw a diagram of the $27$-second phases (something like this:|______|______|______|______|______|______) and measure $120$ seconds from one of the buff opportunities, you can cover $4$ entire periods ($4cdot27=108$) with $120-108=12$ seonds left to spare. In those $12$ seconds, the potential buff from which you measured would still be active, as would the intervening four, for a total of five; whereas in the remaining $15$ seconds, only the intervening four would still be active.
– joriki
Aug 31 at 3:08
Thank you for this. So the quoted solution is right if only one buff can be active at the time. I can follow the part of binomial distribution but I don't understand how to derive the 12s & 15s sector.
– Tony Yeung
Aug 31 at 0:15
Thank you for this. So the quoted solution is right if only one buff can be active at the time. I can follow the part of binomial distribution but I don't understand how to derive the 12s & 15s sector.
– Tony Yeung
Aug 31 at 0:15
@TonyYeung: If you draw a diagram of the $27$-second phases (something like this:
|______|______|______|______|______|______) and measure $120$ seconds from one of the buff opportunities, you can cover $4$ entire periods ($4cdot27=108$) with $120-108=12$ seonds left to spare. In those $12$ seconds, the potential buff from which you measured would still be active, as would the intervening four, for a total of five; whereas in the remaining $15$ seconds, only the intervening four would still be active.– joriki
Aug 31 at 3:08
@TonyYeung: If you draw a diagram of the $27$-second phases (something like this:
|______|______|______|______|______|______) and measure $120$ seconds from one of the buff opportunities, you can cover $4$ entire periods ($4cdot27=108$) with $120-108=12$ seonds left to spare. In those $12$ seconds, the potential buff from which you measured would still be active, as would the intervening four, for a total of five; whereas in the remaining $15$ seconds, only the intervening four would still be active.– joriki
Aug 31 at 3:08
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2899202%2ffind-the-expected-value-of-the-additional-pickups%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
