Mixing
I have 100 boxes. C of them have a gift. I can open up to 16 boxes. What is the number of C that will give me...
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We start by opening a box. If nothing is in there, we open another one. Once we find a gift, we can stop. Each empty box that was opened is discarded (no revisit).
I can find the number of $C$ that will give probability over $0.5$ by writing a program to try for $C=1, C=2$ .. etc.. , but I can't solve the equation for $C$ to find a more "mathematical" and elegant answer.
My work until now is:
1) Found in 1st box: $P(1) = frac{C}{N}$
2) Found in 2nd box: $P(2) = frac{1-C}{N}cdotfrac{C}{N-1}$
4) Found in 3rd box: $P(3) = frac{1-C}{N}cdotfrac{1-C/}{N-1}cdotfrac{C}{N-2}$
Etc...
Adding them up makes things very complicated to solve for $C$.
Any ideas? Thank you in advance!
probability probability-theory
edited yesterday
amWhy
191k27223437
asked yesterday
XuUserAC
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show 3 more comments
up vote
8
down vote
favorite
Details:
We start by opening a box. If nothing is in there, we open another one. Once we find a gift, we can stop. Each empty box that was opened is discarded (no revisit).
I can find the number of $C$ that will give probability over $0.5$ by writing a program to try for $C=1, C=2$ .. etc.. , but I can't solve the equation for $C$ to find a more "mathematical" and elegant answer.
My work until now is:
1) Found in 1st box: $P(1) = frac{C}{N}$
2) Found in 2nd box: $P(2) = frac{1-C}{N}cdotfrac{C}{N-1}$
4) Found in 3rd box: $P(3) = frac{1-C}{N}cdotfrac{1-C/}{N-1}cdotfrac{C}{N-2}$
Etc...
Adding them up makes things very complicated to solve for $C$.
Any ideas? Thank you in advance!
probability probability-theory
edited yesterday
amWhy
191k27223437
asked yesterday
XuUserAC
442
New contributor
XuUserAC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
There are $binom {100}6$ ways to choose $6$ boxes. How many ways are there to choose $6$ empty boxes?
– lulu
yesterday
Thanks for your response lulu! I do know that, but still it can not be solved for C. Again, this approach will require to try for consecutive values of C in order to find the needed value. Thanks though! :)
– XuUserAC
yesterday
3
Should the question in the title really be "What is the the minimum number of C..."?
– Randall Stewart
21 hours ago
2
This kind of sounds like a software developer’s question about how to rig their game’s loot boxes.
– Josh Detwiler
13 hours ago
1
I wouldnt bother opening them. They are all sweaters.
– Keltari
8 hours ago
|
show 3 more comments
up vote
8
down vote
favorite
up vote
8
down vote
favorite
Details:
We start by opening a box. If nothing is in there, we open another one. Once we find a gift, we can stop. Each empty box that was opened is discarded (no revisit).
I can find the number of $C$ that will give probability over $0.5$ by writing a program to try for $C=1, C=2$ .. etc.. , but I can't solve the equation for $C$ to find a more "mathematical" and elegant answer.
My work until now is:
1) Found in 1st box: $P(1) = frac{C}{N}$
2) Found in 2nd box: $P(2) = frac{1-C}{N}cdotfrac{C}{N-1}$
4) Found in 3rd box: $P(3) = frac{1-C}{N}cdotfrac{1-C/}{N-1}cdotfrac{C}{N-2}$
Etc...
Adding them up makes things very complicated to solve for $C$.
Any ideas? Thank you in advance!
probability probability-theory
edited yesterday
amWhy
191k27223437
asked yesterday
XuUserAC
442
New contributor
XuUserAC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Details:
We start by opening a box. If nothing is in there, we open another one. Once we find a gift, we can stop. Each empty box that was opened is discarded (no revisit).
I can find the number of $C$ that will give probability over $0.5$ by writing a program to try for $C=1, C=2$ .. etc.. , but I can't solve the equation for $C$ to find a more "mathematical" and elegant answer.
My work until now is:
1) Found in 1st box: $P(1) = frac{C}{N}$
2) Found in 2nd box: $P(2) = frac{1-C}{N}cdotfrac{C}{N-1}$
4) Found in 3rd box: $P(3) = frac{1-C}{N}cdotfrac{1-C/}{N-1}cdotfrac{C}{N-2}$
Etc...
Adding them up makes things very complicated to solve for $C$.
Any ideas? Thank you in advance!
probability probability-theory
probability probability-theory
edited yesterday
amWhy
191k27223437
asked yesterday
XuUserAC
442
New contributor
XuUserAC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
amWhy
191k27223437
asked yesterday
XuUserAC
442
New contributor
XuUserAC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
amWhy
191k27223437
edited yesterday
amWhy
191k27223437
edited yesterday
amWhy
191k27223437
191k27223437
asked yesterday
XuUserAC
442
New contributor
XuUserAC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
XuUserAC
442
asked yesterday
XuUserAC
442
442
New contributor
XuUserAC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
XuUserAC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
XuUserAC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
There are $binom {100}6$ ways to choose $6$ boxes. How many ways are there to choose $6$ empty boxes?
– lulu
yesterday
Thanks for your response lulu! I do know that, but still it can not be solved for C. Again, this approach will require to try for consecutive values of C in order to find the needed value. Thanks though! :)
– XuUserAC
yesterday
3
Should the question in the title really be "What is the the minimum number of C..."?
– Randall Stewart
21 hours ago
2
This kind of sounds like a software developer’s question about how to rig their game’s loot boxes.
– Josh Detwiler
13 hours ago
1
I wouldnt bother opening them. They are all sweaters.
– Keltari
8 hours ago
|
show 3 more comments
There are $binom {100}6$ ways to choose $6$ boxes. How many ways are there to choose $6$ empty boxes?
– lulu
yesterday
Thanks for your response lulu! I do know that, but still it can not be solved for C. Again, this approach will require to try for consecutive values of C in order to find the needed value. Thanks though! :)
– XuUserAC
yesterday
3
Should the question in the title really be "What is the the minimum number of C..."?
– Randall Stewart
21 hours ago
2
This kind of sounds like a software developer’s question about how to rig their game’s loot boxes.
– Josh Detwiler
13 hours ago
1
I wouldnt bother opening them. They are all sweaters.
– Keltari
8 hours ago
There are $binom {100}6$ ways to choose $6$ boxes. How many ways are there to choose $6$ empty boxes?
– lulu
yesterday
There are $binom {100}6$ ways to choose $6$ boxes. How many ways are there to choose $6$ empty boxes?
– lulu
yesterday
Thanks for your response lulu! I do know that, but still it can not be solved for C. Again, this approach will require to try for consecutive values of C in order to find the needed value. Thanks though! :)
– XuUserAC
yesterday
Thanks for your response lulu! I do know that, but still it can not be solved for C. Again, this approach will require to try for consecutive values of C in order to find the needed value. Thanks though! :)
– XuUserAC
yesterday
3
3
Should the question in the title really be "What is the the minimum number of C..."?
– Randall Stewart
21 hours ago
Should the question in the title really be "What is the the minimum number of C..."?
– Randall Stewart
21 hours ago
2
2
This kind of sounds like a software developer’s question about how to rig their game’s loot boxes.
– Josh Detwiler
13 hours ago
This kind of sounds like a software developer’s question about how to rig their game’s loot boxes.
– Josh Detwiler
13 hours ago
1
1
I wouldnt bother opening them. They are all sweaters.
– Keltari
8 hours ago
I wouldnt bother opening them. They are all sweaters.
– Keltari
8 hours ago
|
show 3 more comments
4 Answers
4
active
oldest
votes
up vote
18
down vote
Start with a rough estimate: If the box contents were independent, the probability of losing would be $(1-C/100)^{16}$. Equating this to $0.5$ gives us $Capprox 4.2$.
Hence, we boldly check the cases $C=4$:
$C=4$ leads to a losing probability of $$frac{96choose 16}{100choose 16}=frac{96!84!}{80!100!}=frac{84cdot 83cdot 82cdot 81}{100cdot 99cdot 98cdot 97}approx 0.492$$
so a winning probability slightly above $frac12$. A close look reveals that $C=3$ leads to a winning probability below $frac12$, so the correct answer is $C=4$.
Note that the "true" breaking point is thus between $3$ and $4$, not between $4$ and $5$ as the rough estimate suggested - the box contents are not independent after all (namely, if you find a - rare - gift, the probability of finding a gift in another box falls dramatically).
answered yesterday
Hagen von Eitzen
273k21266493
add a comment |
up vote
5
down vote
As pointed out in the comments, finding the chance of not getting a gift is rather easier, though the patterns involved assist with the computation. Suppose we had six, rather than sixteen, to choose. We have $$binom {100}{6}=frac {100!}{6!94!}=frac {100cdot 99 cdot 98cdot 97cdot 96cdot 95}{6!}$$ ways of choosing six boxes, and $$binom {100-C}{6}=frac {(100-C)cdot (99-C) cdot (98-C)cdot (97-C)cdot (96-C)cdot (95-C)}{6!}$$ ways of choosing six empty ones, so the probability of an empty box is $$p=frac {(100-C)cdot (99-C) cdot (98-C)cdot (97-C)cdot (96-C)cdot (95-C)}{100cdot 99 cdot 98cdot 97cdot 96cdot 95}$$
Now setting this equal to $0.5$ we get a sextic for $C$. The numerator is monotone in $C$ so we know that trial can work. Can we do better? Well if we take $q=frac {98-C}{98}$ we can estimate the probability as $p=q^6$, and that gives us a potential starting place for trial to reduce the amount of effort involved.
[I see there is another solution which works with a simpler, but slightly different, estimate]
answered yesterday
Mark Bennet
79.6k978177
Given the symmetry, isn't $q=frac {97.5-C}{97.5}$ an even better approximation? Then $C approx 97.5(1-q)$ . Also, might as well develop the full expression for 16, not 6.
– smci
6 hours ago
@smci Yes indeed. At the time I was writing this there was not much in the way of answer and discussion, so I didn't give a direct answer to the question which was being asked, but rather an indication of how to go about it, so that the person who asked the question would have to think about it a little more. Also there was an original focus on integers. But both your suggestions are good.
– Mark Bennet
2 hours ago
And since there were 16 boxes not 6, the arithmetic midpoint would be 92.5
– smci
2 hours ago
add a comment |
up vote
1
down vote
As with a lot of binomial problems, the easiest way to calculate the probability of success from N tries is to start by calculating the probability of N failures and subtracting the answer from 1.
The probability of opening 16 empty boxes (and thus failing to find a prize) in this case is:
$frac{100-Cchoose 16}{100choose 16}
= frac{(100-C)!}{16!(84-C)!}frac{16!84!}{100!}
= frac{(100-C)!84!}{100!(84-C)!}
= frac{84×83×...×(85-C)}{100×99×...×(101-C)}
= frac{84}{100}×frac{83}{99}×...×frac{85-C}{101-C}
$
At this point we can proceed by trial and error multiplying by one term at a time.
For C=1 we get $frac{84}{100}$ which is clearly $>frac{1}{2}$
For C=2, $frac{84}{100}×frac{83}{99}=frac{6972}{9900} approx 0.704$
For C=3, $frac{6972}{9900}×frac{82}{98} approx 0.589$
For C=4, $0.589...×frac{81}{97} approx 0.492$
So the minimum C for which the probability of losing drops below 0.5 (and thus the winning probability is above 0.5) is 4.
answered 10 hours ago
IanF1
1,314812
add a comment |
up vote
-1
down vote
An alternative approach is to work in base 10 logarithms. Chance of failing on 16 tries is
$displaystyle f(C) = left(frac{100-C}{100}right) times
left(frac{99-C}{99}right) times
left(frac{98-C}{98}right) times cdots times
left(frac{85-C}{85}right).
$
Assume that you've written a computer program that calculates
$;log_{10}n;$ for $nin{30, 31, cdots, 100}.$
Then it becomes a simple matter to calculate
$;g(C) = log_{10}f(C).$
answered 15 hours ago
user2661923
411112
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
18
down vote
Start with a rough estimate: If the box contents were independent, the probability of losing would be $(1-C/100)^{16}$. Equating this to $0.5$ gives us $Capprox 4.2$.
Hence, we boldly check the cases $C=4$:
$C=4$ leads to a losing probability of $$frac{96choose 16}{100choose 16}=frac{96!84!}{80!100!}=frac{84cdot 83cdot 82cdot 81}{100cdot 99cdot 98cdot 97}approx 0.492$$
so a winning probability slightly above $frac12$. A close look reveals that $C=3$ leads to a winning probability below $frac12$, so the correct answer is $C=4$.
Note that the "true" breaking point is thus between $3$ and $4$, not between $4$ and $5$ as the rough estimate suggested - the box contents are not independent after all (namely, if you find a - rare - gift, the probability of finding a gift in another box falls dramatically).
answered yesterday
Hagen von Eitzen
273k21266493
add a comment |
up vote
18
down vote
Start with a rough estimate: If the box contents were independent, the probability of losing would be $(1-C/100)^{16}$. Equating this to $0.5$ gives us $Capprox 4.2$.
Hence, we boldly check the cases $C=4$:
$C=4$ leads to a losing probability of $$frac{96choose 16}{100choose 16}=frac{96!84!}{80!100!}=frac{84cdot 83cdot 82cdot 81}{100cdot 99cdot 98cdot 97}approx 0.492$$
so a winning probability slightly above $frac12$. A close look reveals that $C=3$ leads to a winning probability below $frac12$, so the correct answer is $C=4$.
Note that the "true" breaking point is thus between $3$ and $4$, not between $4$ and $5$ as the rough estimate suggested - the box contents are not independent after all (namely, if you find a - rare - gift, the probability of finding a gift in another box falls dramatically).
answered yesterday
Hagen von Eitzen
273k21266493
add a comment |
up vote
18
down vote
up vote
18
down vote
Start with a rough estimate: If the box contents were independent, the probability of losing would be $(1-C/100)^{16}$. Equating this to $0.5$ gives us $Capprox 4.2$.
Hence, we boldly check the cases $C=4$:
$C=4$ leads to a losing probability of $$frac{96choose 16}{100choose 16}=frac{96!84!}{80!100!}=frac{84cdot 83cdot 82cdot 81}{100cdot 99cdot 98cdot 97}approx 0.492$$
so a winning probability slightly above $frac12$. A close look reveals that $C=3$ leads to a winning probability below $frac12$, so the correct answer is $C=4$.
Note that the "true" breaking point is thus between $3$ and $4$, not between $4$ and $5$ as the rough estimate suggested - the box contents are not independent after all (namely, if you find a - rare - gift, the probability of finding a gift in another box falls dramatically).
answered yesterday
Hagen von Eitzen
273k21266493
Start with a rough estimate: If the box contents were independent, the probability of losing would be $(1-C/100)^{16}$. Equating this to $0.5$ gives us $Capprox 4.2$.
Hence, we boldly check the cases $C=4$:
$C=4$ leads to a losing probability of $$frac{96choose 16}{100choose 16}=frac{96!84!}{80!100!}=frac{84cdot 83cdot 82cdot 81}{100cdot 99cdot 98cdot 97}approx 0.492$$
so a winning probability slightly above $frac12$. A close look reveals that $C=3$ leads to a winning probability below $frac12$, so the correct answer is $C=4$.
Note that the "true" breaking point is thus between $3$ and $4$, not between $4$ and $5$ as the rough estimate suggested - the box contents are not independent after all (namely, if you find a - rare - gift, the probability of finding a gift in another box falls dramatically).
answered yesterday
Hagen von Eitzen
273k21266493
answered yesterday
Hagen von Eitzen
273k21266493
answered yesterday
Hagen von Eitzen
273k21266493
answered yesterday
Hagen von Eitzen
273k21266493
273k21266493
add a comment |
add a comment |
up vote
5
down vote
As pointed out in the comments, finding the chance of not getting a gift is rather easier, though the patterns involved assist with the computation. Suppose we had six, rather than sixteen, to choose. We have $$binom {100}{6}=frac {100!}{6!94!}=frac {100cdot 99 cdot 98cdot 97cdot 96cdot 95}{6!}$$ ways of choosing six boxes, and $$binom {100-C}{6}=frac {(100-C)cdot (99-C) cdot (98-C)cdot (97-C)cdot (96-C)cdot (95-C)}{6!}$$ ways of choosing six empty ones, so the probability of an empty box is $$p=frac {(100-C)cdot (99-C) cdot (98-C)cdot (97-C)cdot (96-C)cdot (95-C)}{100cdot 99 cdot 98cdot 97cdot 96cdot 95}$$
Now setting this equal to $0.5$ we get a sextic for $C$. The numerator is monotone in $C$ so we know that trial can work. Can we do better? Well if we take $q=frac {98-C}{98}$ we can estimate the probability as $p=q^6$, and that gives us a potential starting place for trial to reduce the amount of effort involved.
[I see there is another solution which works with a simpler, but slightly different, estimate]
answered yesterday
Mark Bennet
79.6k978177
Given the symmetry, isn't $q=frac {97.5-C}{97.5}$ an even better approximation? Then $C approx 97.5(1-q)$ . Also, might as well develop the full expression for 16, not 6.
– smci
6 hours ago
@smci Yes indeed. At the time I was writing this there was not much in the way of answer and discussion, so I didn't give a direct answer to the question which was being asked, but rather an indication of how to go about it, so that the person who asked the question would have to think about it a little more. Also there was an original focus on integers. But both your suggestions are good.
– Mark Bennet
2 hours ago
And since there were 16 boxes not 6, the arithmetic midpoint would be 92.5
– smci
2 hours ago
add a comment |
up vote
5
down vote
As pointed out in the comments, finding the chance of not getting a gift is rather easier, though the patterns involved assist with the computation. Suppose we had six, rather than sixteen, to choose. We have $$binom {100}{6}=frac {100!}{6!94!}=frac {100cdot 99 cdot 98cdot 97cdot 96cdot 95}{6!}$$ ways of choosing six boxes, and $$binom {100-C}{6}=frac {(100-C)cdot (99-C) cdot (98-C)cdot (97-C)cdot (96-C)cdot (95-C)}{6!}$$ ways of choosing six empty ones, so the probability of an empty box is $$p=frac {(100-C)cdot (99-C) cdot (98-C)cdot (97-C)cdot (96-C)cdot (95-C)}{100cdot 99 cdot 98cdot 97cdot 96cdot 95}$$
Now setting this equal to $0.5$ we get a sextic for $C$. The numerator is monotone in $C$ so we know that trial can work. Can we do better? Well if we take $q=frac {98-C}{98}$ we can estimate the probability as $p=q^6$, and that gives us a potential starting place for trial to reduce the amount of effort involved.
[I see there is another solution which works with a simpler, but slightly different, estimate]
answered yesterday
Mark Bennet
79.6k978177
Given the symmetry, isn't $q=frac {97.5-C}{97.5}$ an even better approximation? Then $C approx 97.5(1-q)$ . Also, might as well develop the full expression for 16, not 6.
– smci
6 hours ago
@smci Yes indeed. At the time I was writing this there was not much in the way of answer and discussion, so I didn't give a direct answer to the question which was being asked, but rather an indication of how to go about it, so that the person who asked the question would have to think about it a little more. Also there was an original focus on integers. But both your suggestions are good.
– Mark Bennet
2 hours ago
And since there were 16 boxes not 6, the arithmetic midpoint would be 92.5
– smci
2 hours ago
add a comment |
up vote
5
down vote
up vote
5
down vote
As pointed out in the comments, finding the chance of not getting a gift is rather easier, though the patterns involved assist with the computation. Suppose we had six, rather than sixteen, to choose. We have $$binom {100}{6}=frac {100!}{6!94!}=frac {100cdot 99 cdot 98cdot 97cdot 96cdot 95}{6!}$$ ways of choosing six boxes, and $$binom {100-C}{6}=frac {(100-C)cdot (99-C) cdot (98-C)cdot (97-C)cdot (96-C)cdot (95-C)}{6!}$$ ways of choosing six empty ones, so the probability of an empty box is $$p=frac {(100-C)cdot (99-C) cdot (98-C)cdot (97-C)cdot (96-C)cdot (95-C)}{100cdot 99 cdot 98cdot 97cdot 96cdot 95}$$
Now setting this equal to $0.5$ we get a sextic for $C$. The numerator is monotone in $C$ so we know that trial can work. Can we do better? Well if we take $q=frac {98-C}{98}$ we can estimate the probability as $p=q^6$, and that gives us a potential starting place for trial to reduce the amount of effort involved.
[I see there is another solution which works with a simpler, but slightly different, estimate]
answered yesterday
Mark Bennet
79.6k978177
As pointed out in the comments, finding the chance of not getting a gift is rather easier, though the patterns involved assist with the computation. Suppose we had six, rather than sixteen, to choose. We have $$binom {100}{6}=frac {100!}{6!94!}=frac {100cdot 99 cdot 98cdot 97cdot 96cdot 95}{6!}$$ ways of choosing six boxes, and $$binom {100-C}{6}=frac {(100-C)cdot (99-C) cdot (98-C)cdot (97-C)cdot (96-C)cdot (95-C)}{6!}$$ ways of choosing six empty ones, so the probability of an empty box is $$p=frac {(100-C)cdot (99-C) cdot (98-C)cdot (97-C)cdot (96-C)cdot (95-C)}{100cdot 99 cdot 98cdot 97cdot 96cdot 95}$$
Now setting this equal to $0.5$ we get a sextic for $C$. The numerator is monotone in $C$ so we know that trial can work. Can we do better? Well if we take $q=frac {98-C}{98}$ we can estimate the probability as $p=q^6$, and that gives us a potential starting place for trial to reduce the amount of effort involved.
[I see there is another solution which works with a simpler, but slightly different, estimate]
answered yesterday
Mark Bennet
79.6k978177
edited 10 hours ago
answered yesterday
Mark Bennet
79.6k978177
answered yesterday
Mark Bennet
79.6k978177
answered yesterday
Mark Bennet
79.6k978177
79.6k978177
Given the symmetry, isn't $q=frac {97.5-C}{97.5}$ an even better approximation? Then $C approx 97.5(1-q)$ . Also, might as well develop the full expression for 16, not 6.
– smci
6 hours ago
@smci Yes indeed. At the time I was writing this there was not much in the way of answer and discussion, so I didn't give a direct answer to the question which was being asked, but rather an indication of how to go about it, so that the person who asked the question would have to think about it a little more. Also there was an original focus on integers. But both your suggestions are good.
– Mark Bennet
2 hours ago
And since there were 16 boxes not 6, the arithmetic midpoint would be 92.5
– smci
2 hours ago
add a comment |
Given the symmetry, isn't $q=frac {97.5-C}{97.5}$ an even better approximation? Then $C approx 97.5(1-q)$ . Also, might as well develop the full expression for 16, not 6.
– smci
6 hours ago
@smci Yes indeed. At the time I was writing this there was not much in the way of answer and discussion, so I didn't give a direct answer to the question which was being asked, but rather an indication of how to go about it, so that the person who asked the question would have to think about it a little more. Also there was an original focus on integers. But both your suggestions are good.
– Mark Bennet
2 hours ago
And since there were 16 boxes not 6, the arithmetic midpoint would be 92.5
– smci
2 hours ago
Given the symmetry, isn't $q=frac {97.5-C}{97.5}$ an even better approximation? Then $C approx 97.5(1-q)$ . Also, might as well develop the full expression for 16, not 6.
– smci
6 hours ago
Given the symmetry, isn't $q=frac {97.5-C}{97.5}$ an even better approximation? Then $C approx 97.5(1-q)$ . Also, might as well develop the full expression for 16, not 6.
– smci
6 hours ago
@smci Yes indeed. At the time I was writing this there was not much in the way of answer and discussion, so I didn't give a direct answer to the question which was being asked, but rather an indication of how to go about it, so that the person who asked the question would have to think about it a little more. Also there was an original focus on integers. But both your suggestions are good.
– Mark Bennet
2 hours ago
@smci Yes indeed. At the time I was writing this there was not much in the way of answer and discussion, so I didn't give a direct answer to the question which was being asked, but rather an indication of how to go about it, so that the person who asked the question would have to think about it a little more. Also there was an original focus on integers. But both your suggestions are good.
– Mark Bennet
2 hours ago
And since there were 16 boxes not 6, the arithmetic midpoint would be 92.5
– smci
2 hours ago
And since there were 16 boxes not 6, the arithmetic midpoint would be 92.5
– smci
2 hours ago
add a comment |
up vote
1
down vote
As with a lot of binomial problems, the easiest way to calculate the probability of success from N tries is to start by calculating the probability of N failures and subtracting the answer from 1.
The probability of opening 16 empty boxes (and thus failing to find a prize) in this case is:
$frac{100-Cchoose 16}{100choose 16}
= frac{(100-C)!}{16!(84-C)!}frac{16!84!}{100!}
= frac{(100-C)!84!}{100!(84-C)!}
= frac{84×83×...×(85-C)}{100×99×...×(101-C)}
= frac{84}{100}×frac{83}{99}×...×frac{85-C}{101-C}
$
At this point we can proceed by trial and error multiplying by one term at a time.
For C=1 we get $frac{84}{100}$ which is clearly $>frac{1}{2}$
For C=2, $frac{84}{100}×frac{83}{99}=frac{6972}{9900} approx 0.704$
For C=3, $frac{6972}{9900}×frac{82}{98} approx 0.589$
For C=4, $0.589...×frac{81}{97} approx 0.492$
So the minimum C for which the probability of losing drops below 0.5 (and thus the winning probability is above 0.5) is 4.
answered 10 hours ago
IanF1
1,314812
add a comment |
up vote
1
down vote
As with a lot of binomial problems, the easiest way to calculate the probability of success from N tries is to start by calculating the probability of N failures and subtracting the answer from 1.
The probability of opening 16 empty boxes (and thus failing to find a prize) in this case is:
$frac{100-Cchoose 16}{100choose 16}
= frac{(100-C)!}{16!(84-C)!}frac{16!84!}{100!}
= frac{(100-C)!84!}{100!(84-C)!}
= frac{84×83×...×(85-C)}{100×99×...×(101-C)}
= frac{84}{100}×frac{83}{99}×...×frac{85-C}{101-C}
$
At this point we can proceed by trial and error multiplying by one term at a time.
For C=1 we get $frac{84}{100}$ which is clearly $>frac{1}{2}$
For C=2, $frac{84}{100}×frac{83}{99}=frac{6972}{9900} approx 0.704$
For C=3, $frac{6972}{9900}×frac{82}{98} approx 0.589$
For C=4, $0.589...×frac{81}{97} approx 0.492$
So the minimum C for which the probability of losing drops below 0.5 (and thus the winning probability is above 0.5) is 4.
answered 10 hours ago
IanF1
1,314812
add a comment |
up vote
1
down vote
up vote
1
down vote
As with a lot of binomial problems, the easiest way to calculate the probability of success from N tries is to start by calculating the probability of N failures and subtracting the answer from 1.
The probability of opening 16 empty boxes (and thus failing to find a prize) in this case is:
$frac{100-Cchoose 16}{100choose 16}
= frac{(100-C)!}{16!(84-C)!}frac{16!84!}{100!}
= frac{(100-C)!84!}{100!(84-C)!}
= frac{84×83×...×(85-C)}{100×99×...×(101-C)}
= frac{84}{100}×frac{83}{99}×...×frac{85-C}{101-C}
$
At this point we can proceed by trial and error multiplying by one term at a time.
For C=1 we get $frac{84}{100}$ which is clearly $>frac{1}{2}$
For C=2, $frac{84}{100}×frac{83}{99}=frac{6972}{9900} approx 0.704$
For C=3, $frac{6972}{9900}×frac{82}{98} approx 0.589$
For C=4, $0.589...×frac{81}{97} approx 0.492$
So the minimum C for which the probability of losing drops below 0.5 (and thus the winning probability is above 0.5) is 4.
answered 10 hours ago
IanF1
1,314812
As with a lot of binomial problems, the easiest way to calculate the probability of success from N tries is to start by calculating the probability of N failures and subtracting the answer from 1.
The probability of opening 16 empty boxes (and thus failing to find a prize) in this case is:
$frac{100-Cchoose 16}{100choose 16}
= frac{(100-C)!}{16!(84-C)!}frac{16!84!}{100!}
= frac{(100-C)!84!}{100!(84-C)!}
= frac{84×83×...×(85-C)}{100×99×...×(101-C)}
= frac{84}{100}×frac{83}{99}×...×frac{85-C}{101-C}
$
At this point we can proceed by trial and error multiplying by one term at a time.
For C=1 we get $frac{84}{100}$ which is clearly $>frac{1}{2}$
For C=2, $frac{84}{100}×frac{83}{99}=frac{6972}{9900} approx 0.704$
For C=3, $frac{6972}{9900}×frac{82}{98} approx 0.589$
For C=4, $0.589...×frac{81}{97} approx 0.492$
So the minimum C for which the probability of losing drops below 0.5 (and thus the winning probability is above 0.5) is 4.
answered 10 hours ago
IanF1
1,314812
answered 10 hours ago
IanF1
1,314812
answered 10 hours ago
IanF1
1,314812
answered 10 hours ago
IanF1
1,314812
1,314812
add a comment |
add a comment |
up vote
-1
down vote
An alternative approach is to work in base 10 logarithms. Chance of failing on 16 tries is
$displaystyle f(C) = left(frac{100-C}{100}right) times
left(frac{99-C}{99}right) times
left(frac{98-C}{98}right) times cdots times
left(frac{85-C}{85}right).
$
Assume that you've written a computer program that calculates
$;log_{10}n;$ for $nin{30, 31, cdots, 100}.$
Then it becomes a simple matter to calculate
$;g(C) = log_{10}f(C).$
answered 15 hours ago
user2661923
411112
add a comment |
up vote
-1
down vote
An alternative approach is to work in base 10 logarithms. Chance of failing on 16 tries is
$displaystyle f(C) = left(frac{100-C}{100}right) times
left(frac{99-C}{99}right) times
left(frac{98-C}{98}right) times cdots times
left(frac{85-C}{85}right).
$
Assume that you've written a computer program that calculates
$;log_{10}n;$ for $nin{30, 31, cdots, 100}.$
Then it becomes a simple matter to calculate
$;g(C) = log_{10}f(C).$
answered 15 hours ago
user2661923
411112
add a comment |
up vote
-1
down vote
up vote
-1
down vote
An alternative approach is to work in base 10 logarithms. Chance of failing on 16 tries is
$displaystyle f(C) = left(frac{100-C}{100}right) times
left(frac{99-C}{99}right) times
left(frac{98-C}{98}right) times cdots times
left(frac{85-C}{85}right).
$
Assume that you've written a computer program that calculates
$;log_{10}n;$ for $nin{30, 31, cdots, 100}.$
Then it becomes a simple matter to calculate
$;g(C) = log_{10}f(C).$
answered 15 hours ago
user2661923
411112
An alternative approach is to work in base 10 logarithms. Chance of failing on 16 tries is
$displaystyle f(C) = left(frac{100-C}{100}right) times
left(frac{99-C}{99}right) times
left(frac{98-C}{98}right) times cdots times
left(frac{85-C}{85}right).
$
Assume that you've written a computer program that calculates
$;log_{10}n;$ for $nin{30, 31, cdots, 100}.$
Then it becomes a simple matter to calculate
$;g(C) = log_{10}f(C).$
answered 15 hours ago
user2661923
411112
answered 15 hours ago
user2661923
411112
answered 15 hours ago
user2661923
411112
answered 15 hours ago
user2661923
411112
411112
add a comment |
add a comment |
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There are $binom {100}6$ ways to choose $6$ boxes. How many ways are there to choose $6$ empty boxes?
– lulu
yesterday
Thanks for your response lulu! I do know that, but still it can not be solved for C. Again, this approach will require to try for consecutive values of C in order to find the needed value. Thanks though! :)
– XuUserAC
yesterday
3
Should the question in the title really be "What is the the minimum number of C..."?
– Randall Stewart
21 hours ago
2
This kind of sounds like a software developer’s question about how to rig their game’s loot boxes.
– Josh Detwiler
13 hours ago
1
I wouldnt bother opening them. They are all sweaters.
– Keltari
8 hours ago