Mixing
Random Dental Floss Odds
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Awhile back I bought 2 identical rolls of dental floss, each with 50 uses, and picked them randomly. Tonight, the one I picked hit the 50 use mark. What is the expected number of uses in the remaining roll?
With a 100000 case brute force, I get an expected answer of 8 uses. Neat, so I have maybe a week.
What is the non-brute force answer?
probability
asked Aug 29 at 5:28
Ed Pegg
9,37932588
add a comment |Â
up vote
20
down vote
favorite
Awhile back I bought 2 identical rolls of dental floss, each with 50 uses, and picked them randomly. Tonight, the one I picked hit the 50 use mark. What is the expected number of uses in the remaining roll?
With a 100000 case brute force, I get an expected answer of 8 uses. Neat, so I have maybe a week.
What is the non-brute force answer?
probability
asked Aug 29 at 5:28
Ed Pegg
9,37932588
5
en.m.wikipedia.org/wiki/Banach%27s_matchbox_problem
– Steve D
Aug 29 at 6:19
3
@SteveD: For a slight difference between the two problems, see this question.
– joriki
Aug 29 at 6:52
1
There is no concrete evidence that flossing is useful for your dental hygiene (I mean, if you have something specific to clean, sure). Some say it is even damaging to your gums. The 8 uses should probably last you about two months, not a week.
– Asaf Karagila♦
Aug 29 at 13:02
@joriki: It's really the same problem though, no? But you set $N=50-1$ to account for stopping on the 50th try, and you add $1$ to the expectation you get because the other roll cannot be empty.
– Steve D
Aug 29 at 14:13
add a comment |Â
up vote
20
down vote
favorite
up vote
20
down vote
favorite
Awhile back I bought 2 identical rolls of dental floss, each with 50 uses, and picked them randomly. Tonight, the one I picked hit the 50 use mark. What is the expected number of uses in the remaining roll?
With a 100000 case brute force, I get an expected answer of 8 uses. Neat, so I have maybe a week.
What is the non-brute force answer?
probability
asked Aug 29 at 5:28
Ed Pegg
9,37932588
Awhile back I bought 2 identical rolls of dental floss, each with 50 uses, and picked them randomly. Tonight, the one I picked hit the 50 use mark. What is the expected number of uses in the remaining roll?
With a 100000 case brute force, I get an expected answer of 8 uses. Neat, so I have maybe a week.
What is the non-brute force answer?
probability
asked Aug 29 at 5:28
Ed Pegg
9,37932588
asked Aug 29 at 5:28
Ed Pegg
9,37932588
asked Aug 29 at 5:28
Ed Pegg
9,37932588
asked Aug 29 at 5:28
Ed Pegg
9,37932588
9,37932588
5
en.m.wikipedia.org/wiki/Banach%27s_matchbox_problem
– Steve D
Aug 29 at 6:19
3
@SteveD: For a slight difference between the two problems, see this question.
– joriki
Aug 29 at 6:52
1
There is no concrete evidence that flossing is useful for your dental hygiene (I mean, if you have something specific to clean, sure). Some say it is even damaging to your gums. The 8 uses should probably last you about two months, not a week.
– Asaf Karagila♦
Aug 29 at 13:02
@joriki: It's really the same problem though, no? But you set $N=50-1$ to account for stopping on the 50th try, and you add $1$ to the expectation you get because the other roll cannot be empty.
– Steve D
Aug 29 at 14:13
add a comment |Â
5
en.m.wikipedia.org/wiki/Banach%27s_matchbox_problem
– Steve D
Aug 29 at 6:19
3
@SteveD: For a slight difference between the two problems, see this question.
– joriki
Aug 29 at 6:52
1
There is no concrete evidence that flossing is useful for your dental hygiene (I mean, if you have something specific to clean, sure). Some say it is even damaging to your gums. The 8 uses should probably last you about two months, not a week.
– Asaf Karagila♦
Aug 29 at 13:02
@joriki: It's really the same problem though, no? But you set $N=50-1$ to account for stopping on the 50th try, and you add $1$ to the expectation you get because the other roll cannot be empty.
– Steve D
Aug 29 at 14:13
5
5
en.m.wikipedia.org/wiki/Banach%27s_matchbox_problem
– Steve D
Aug 29 at 6:19
en.m.wikipedia.org/wiki/Banach%27s_matchbox_problem
– Steve D
Aug 29 at 6:19
3
3
@SteveD: For a slight difference between the two problems, see this question.
– joriki
Aug 29 at 6:52
@SteveD: For a slight difference between the two problems, see this question.
– joriki
Aug 29 at 6:52
1
1
There is no concrete evidence that flossing is useful for your dental hygiene (I mean, if you have something specific to clean, sure). Some say it is even damaging to your gums. The 8 uses should probably last you about two months, not a week.
– Asaf Karagila♦
Aug 29 at 13:02
There is no concrete evidence that flossing is useful for your dental hygiene (I mean, if you have something specific to clean, sure). Some say it is even damaging to your gums. The 8 uses should probably last you about two months, not a week.
– Asaf Karagila♦
Aug 29 at 13:02
@joriki: It's really the same problem though, no? But you set $N=50-1$ to account for stopping on the 50th try, and you add $1$ to the expectation you get because the other roll cannot be empty.
– Steve D
Aug 29 at 14:13
@joriki: It's really the same problem though, no? But you set $N=50-1$ to account for stopping on the 50th try, and you add $1$ to the expectation you get because the other roll cannot be empty.
– Steve D
Aug 29 at 14:13
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
12
down vote
I understand “hit the $50$-use mark†to mean that you can tell when the $50$-th use has used up the roll, and you don't need to pick it one more time to find that it's empty.
The probability that the other roll has $0lt kle 50$ uses left is
$$
mathsf P(K=k)=2^-(100-k)binom99-k49,
$$
since in that case you chose the now empty roll any $49$ times out of $99-k$ times and then once tonight.
I assume that you're implying that you haven't hit the $50$-use mark on the other roll yet, so we want the conditional probabilities $mathsf P(K=kmid Kgt0)$. Fortunately, we have $mathsf P(Kgt0)$ by symmetry, so for $kgt0$ we have
$$
mathsf P(K=kmid Kgt0)=fracmathsf P(K=kland Kgt0)mathsf P(Kgt0)=fracmathsf P(K=k)mathsf P(Kgt0)=2mathsf P(K=k);.
$$
Thus
begineqnarray*
mathsf E[Kmid Kgt0]
&=&
sum_k=1^50k,mathsf P(K=kmid Kgt0)
\
&=&
sum_k=1^50kcdot2cdot2^-(100-k)binom99-k49
\
&=&
frac31528545170488810417128905392539614081257132168796771975168
\
&approx&
7.96;,
endeqnarray*
in agreement with your simulation.
Alternatively, you could get the factor of $2$ by saying that the last use emptied either of the rolls, so you don't get a factor of $frac12$ for the last use, and then using $2^-(99-k)$ for the probability of a certain pattern of choices of that or the other roll.
answered Aug 29 at 6:22
joriki
167k10180333
Can you explain why the exponent is not $-(99-k)$?
– Steve D
Aug 29 at 6:25
@SteveD: I will in a bit, but I need to fix an error first :-)
– joriki
Aug 29 at 6:26
@SteveD: Actually, it turned out the error yielded a factor of $2$ :-) Does that resolve your question?
– joriki
Aug 29 at 6:36
@SteveD: I added another paragraph about another way of arriving at $2^-(99-k)$ -- perhaps this is what you had in mind?
– joriki
Aug 29 at 6:40
2
Yes, thanks! See my comment above for the history of this nice problem :)
– Steve D
Aug 29 at 6:41
add a comment |Â
up vote
11
down vote
Model your floss use as a random walk in the plane starting at $(0,0)$ and taking a unit step to the right when you pick from roll A, and a unit step up when you pick from roll B.
The first time you use up one of the rolls, there are $k$ uses remaining in the other roll, for some $k$ between $1$ and $50$. There are two mutually exclusive possibilities: you must have arrived at the point $(50, 50-k)$ by taking a step to the right of the point $(49, 50-k)$, or you arrived at the point $(50-k, 50)$ by taking a step up from the point $(50-k, 49)$.
For the first option there are $49 + (50-k)choose 49$ ways to reach the penultimate point, followed by a step to the right; each path has probability $(1/2)^99-kcdot(1/2)$, giving an overall probability of $99-kchoose 49(1/2)^100-k$. Since same result holds for the second option, the probability that the remaining roll has $k$ uses is twice this last number:
$$p_k:=99-kchoose 49left(frac12right)^99-k.tag1$$
To compute $E(K)$, the expected number of uses in the remaining roll, you can evaluate the summation $sum k p_k$, as in @joriki's approach. Alternatively you can use the following device to obtain a closed form expression: Rearrange (1) into the recursion relation
$$
2[50-k]p_k = [100-(k+1)]p_k+1,tag2
$$
valid for all $k=1,ldots,50$. Sum both sides of (2) over all $k$:
$$
2[50 - E(K)] = 100(1-p_1) - (E(K)-p_1).tag3
$$
Finally rearrange (3) to obtain
$$E(K)=99p_1= 9998choose49left(frac12right)^98approx7.96.$$
answered Aug 29 at 7:22
grand_chat
18.3k11122
Your calculation of the expectation is very nice! :-)
– joriki
Aug 29 at 19:10
1
@joriki Thanks! I just now fixed an error in (3) which fortunately leads to the same result :-)
– grand_chat
Aug 29 at 19:38
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
I understand “hit the $50$-use mark†to mean that you can tell when the $50$-th use has used up the roll, and you don't need to pick it one more time to find that it's empty.
The probability that the other roll has $0lt kle 50$ uses left is
$$
mathsf P(K=k)=2^-(100-k)binom99-k49,
$$
since in that case you chose the now empty roll any $49$ times out of $99-k$ times and then once tonight.
I assume that you're implying that you haven't hit the $50$-use mark on the other roll yet, so we want the conditional probabilities $mathsf P(K=kmid Kgt0)$. Fortunately, we have $mathsf P(Kgt0)$ by symmetry, so for $kgt0$ we have
$$
mathsf P(K=kmid Kgt0)=fracmathsf P(K=kland Kgt0)mathsf P(Kgt0)=fracmathsf P(K=k)mathsf P(Kgt0)=2mathsf P(K=k);.
$$
Thus
begineqnarray*
mathsf E[Kmid Kgt0]
&=&
sum_k=1^50k,mathsf P(K=kmid Kgt0)
\
&=&
sum_k=1^50kcdot2cdot2^-(100-k)binom99-k49
\
&=&
frac31528545170488810417128905392539614081257132168796771975168
\
&approx&
7.96;,
endeqnarray*
in agreement with your simulation.
Alternatively, you could get the factor of $2$ by saying that the last use emptied either of the rolls, so you don't get a factor of $frac12$ for the last use, and then using $2^-(99-k)$ for the probability of a certain pattern of choices of that or the other roll.
answered Aug 29 at 6:22
joriki
167k10180333
Can you explain why the exponent is not $-(99-k)$?
– Steve D
Aug 29 at 6:25
@SteveD: I will in a bit, but I need to fix an error first :-)
– joriki
Aug 29 at 6:26
@SteveD: Actually, it turned out the error yielded a factor of $2$ :-) Does that resolve your question?
– joriki
Aug 29 at 6:36
@SteveD: I added another paragraph about another way of arriving at $2^-(99-k)$ -- perhaps this is what you had in mind?
– joriki
Aug 29 at 6:40
2
Yes, thanks! See my comment above for the history of this nice problem :)
– Steve D
Aug 29 at 6:41
add a comment |Â
up vote
12
down vote
I understand “hit the $50$-use mark†to mean that you can tell when the $50$-th use has used up the roll, and you don't need to pick it one more time to find that it's empty.
The probability that the other roll has $0lt kle 50$ uses left is
$$
mathsf P(K=k)=2^-(100-k)binom99-k49,
$$
since in that case you chose the now empty roll any $49$ times out of $99-k$ times and then once tonight.
I assume that you're implying that you haven't hit the $50$-use mark on the other roll yet, so we want the conditional probabilities $mathsf P(K=kmid Kgt0)$. Fortunately, we have $mathsf P(Kgt0)$ by symmetry, so for $kgt0$ we have
$$
mathsf P(K=kmid Kgt0)=fracmathsf P(K=kland Kgt0)mathsf P(Kgt0)=fracmathsf P(K=k)mathsf P(Kgt0)=2mathsf P(K=k);.
$$
Thus
begineqnarray*
mathsf E[Kmid Kgt0]
&=&
sum_k=1^50k,mathsf P(K=kmid Kgt0)
\
&=&
sum_k=1^50kcdot2cdot2^-(100-k)binom99-k49
\
&=&
frac31528545170488810417128905392539614081257132168796771975168
\
&approx&
7.96;,
endeqnarray*
in agreement with your simulation.
Alternatively, you could get the factor of $2$ by saying that the last use emptied either of the rolls, so you don't get a factor of $frac12$ for the last use, and then using $2^-(99-k)$ for the probability of a certain pattern of choices of that or the other roll.
answered Aug 29 at 6:22
joriki
167k10180333
Can you explain why the exponent is not $-(99-k)$?
– Steve D
Aug 29 at 6:25
@SteveD: I will in a bit, but I need to fix an error first :-)
– joriki
Aug 29 at 6:26
@SteveD: Actually, it turned out the error yielded a factor of $2$ :-) Does that resolve your question?
– joriki
Aug 29 at 6:36
@SteveD: I added another paragraph about another way of arriving at $2^-(99-k)$ -- perhaps this is what you had in mind?
– joriki
Aug 29 at 6:40
2
Yes, thanks! See my comment above for the history of this nice problem :)
– Steve D
Aug 29 at 6:41
add a comment |Â
up vote
12
down vote
up vote
12
down vote
I understand “hit the $50$-use mark†to mean that you can tell when the $50$-th use has used up the roll, and you don't need to pick it one more time to find that it's empty.
The probability that the other roll has $0lt kle 50$ uses left is
$$
mathsf P(K=k)=2^-(100-k)binom99-k49,
$$
since in that case you chose the now empty roll any $49$ times out of $99-k$ times and then once tonight.
I assume that you're implying that you haven't hit the $50$-use mark on the other roll yet, so we want the conditional probabilities $mathsf P(K=kmid Kgt0)$. Fortunately, we have $mathsf P(Kgt0)$ by symmetry, so for $kgt0$ we have
$$
mathsf P(K=kmid Kgt0)=fracmathsf P(K=kland Kgt0)mathsf P(Kgt0)=fracmathsf P(K=k)mathsf P(Kgt0)=2mathsf P(K=k);.
$$
Thus
begineqnarray*
mathsf E[Kmid Kgt0]
&=&
sum_k=1^50k,mathsf P(K=kmid Kgt0)
\
&=&
sum_k=1^50kcdot2cdot2^-(100-k)binom99-k49
\
&=&
frac31528545170488810417128905392539614081257132168796771975168
\
&approx&
7.96;,
endeqnarray*
in agreement with your simulation.
Alternatively, you could get the factor of $2$ by saying that the last use emptied either of the rolls, so you don't get a factor of $frac12$ for the last use, and then using $2^-(99-k)$ for the probability of a certain pattern of choices of that or the other roll.
answered Aug 29 at 6:22
joriki
167k10180333
I understand “hit the $50$-use mark†to mean that you can tell when the $50$-th use has used up the roll, and you don't need to pick it one more time to find that it's empty.
The probability that the other roll has $0lt kle 50$ uses left is
$$
mathsf P(K=k)=2^-(100-k)binom99-k49,
$$
since in that case you chose the now empty roll any $49$ times out of $99-k$ times and then once tonight.
I assume that you're implying that you haven't hit the $50$-use mark on the other roll yet, so we want the conditional probabilities $mathsf P(K=kmid Kgt0)$. Fortunately, we have $mathsf P(Kgt0)$ by symmetry, so for $kgt0$ we have
$$
mathsf P(K=kmid Kgt0)=fracmathsf P(K=kland Kgt0)mathsf P(Kgt0)=fracmathsf P(K=k)mathsf P(Kgt0)=2mathsf P(K=k);.
$$
Thus
begineqnarray*
mathsf E[Kmid Kgt0]
&=&
sum_k=1^50k,mathsf P(K=kmid Kgt0)
\
&=&
sum_k=1^50kcdot2cdot2^-(100-k)binom99-k49
\
&=&
frac31528545170488810417128905392539614081257132168796771975168
\
&approx&
7.96;,
endeqnarray*
in agreement with your simulation.
Alternatively, you could get the factor of $2$ by saying that the last use emptied either of the rolls, so you don't get a factor of $frac12$ for the last use, and then using $2^-(99-k)$ for the probability of a certain pattern of choices of that or the other roll.
answered Aug 29 at 6:22
joriki
167k10180333
edited Aug 29 at 6:35
answered Aug 29 at 6:22
joriki
167k10180333
answered Aug 29 at 6:22
joriki
167k10180333
answered Aug 29 at 6:22
joriki
167k10180333
167k10180333
Can you explain why the exponent is not $-(99-k)$?
– Steve D
Aug 29 at 6:25
@SteveD: I will in a bit, but I need to fix an error first :-)
– joriki
Aug 29 at 6:26
@SteveD: Actually, it turned out the error yielded a factor of $2$ :-) Does that resolve your question?
– joriki
Aug 29 at 6:36
@SteveD: I added another paragraph about another way of arriving at $2^-(99-k)$ -- perhaps this is what you had in mind?
– joriki
Aug 29 at 6:40
2
Yes, thanks! See my comment above for the history of this nice problem :)
– Steve D
Aug 29 at 6:41
add a comment |Â
Can you explain why the exponent is not $-(99-k)$?
– Steve D
Aug 29 at 6:25
@SteveD: I will in a bit, but I need to fix an error first :-)
– joriki
Aug 29 at 6:26
@SteveD: Actually, it turned out the error yielded a factor of $2$ :-) Does that resolve your question?
– joriki
Aug 29 at 6:36
@SteveD: I added another paragraph about another way of arriving at $2^-(99-k)$ -- perhaps this is what you had in mind?
– joriki
Aug 29 at 6:40
2
Yes, thanks! See my comment above for the history of this nice problem :)
– Steve D
Aug 29 at 6:41
Can you explain why the exponent is not $-(99-k)$?
– Steve D
Aug 29 at 6:25
Can you explain why the exponent is not $-(99-k)$?
– Steve D
Aug 29 at 6:25
@SteveD: I will in a bit, but I need to fix an error first :-)
– joriki
Aug 29 at 6:26
@SteveD: I will in a bit, but I need to fix an error first :-)
– joriki
Aug 29 at 6:26
@SteveD: Actually, it turned out the error yielded a factor of $2$ :-) Does that resolve your question?
– joriki
Aug 29 at 6:36
@SteveD: Actually, it turned out the error yielded a factor of $2$ :-) Does that resolve your question?
– joriki
Aug 29 at 6:36
@SteveD: I added another paragraph about another way of arriving at $2^-(99-k)$ -- perhaps this is what you had in mind?
– joriki
Aug 29 at 6:40
@SteveD: I added another paragraph about another way of arriving at $2^-(99-k)$ -- perhaps this is what you had in mind?
– joriki
Aug 29 at 6:40
2
2
Yes, thanks! See my comment above for the history of this nice problem :)
– Steve D
Aug 29 at 6:41
Yes, thanks! See my comment above for the history of this nice problem :)
– Steve D
Aug 29 at 6:41
add a comment |Â
up vote
11
down vote
Model your floss use as a random walk in the plane starting at $(0,0)$ and taking a unit step to the right when you pick from roll A, and a unit step up when you pick from roll B.
The first time you use up one of the rolls, there are $k$ uses remaining in the other roll, for some $k$ between $1$ and $50$. There are two mutually exclusive possibilities: you must have arrived at the point $(50, 50-k)$ by taking a step to the right of the point $(49, 50-k)$, or you arrived at the point $(50-k, 50)$ by taking a step up from the point $(50-k, 49)$.
For the first option there are $49 + (50-k)choose 49$ ways to reach the penultimate point, followed by a step to the right; each path has probability $(1/2)^99-kcdot(1/2)$, giving an overall probability of $99-kchoose 49(1/2)^100-k$. Since same result holds for the second option, the probability that the remaining roll has $k$ uses is twice this last number:
$$p_k:=99-kchoose 49left(frac12right)^99-k.tag1$$
To compute $E(K)$, the expected number of uses in the remaining roll, you can evaluate the summation $sum k p_k$, as in @joriki's approach. Alternatively you can use the following device to obtain a closed form expression: Rearrange (1) into the recursion relation
$$
2[50-k]p_k = [100-(k+1)]p_k+1,tag2
$$
valid for all $k=1,ldots,50$. Sum both sides of (2) over all $k$:
$$
2[50 - E(K)] = 100(1-p_1) - (E(K)-p_1).tag3
$$
Finally rearrange (3) to obtain
$$E(K)=99p_1= 9998choose49left(frac12right)^98approx7.96.$$
answered Aug 29 at 7:22
grand_chat
18.3k11122
Your calculation of the expectation is very nice! :-)
– joriki
Aug 29 at 19:10
1
@joriki Thanks! I just now fixed an error in (3) which fortunately leads to the same result :-)
– grand_chat
Aug 29 at 19:38
add a comment |Â
up vote
11
down vote
Model your floss use as a random walk in the plane starting at $(0,0)$ and taking a unit step to the right when you pick from roll A, and a unit step up when you pick from roll B.
The first time you use up one of the rolls, there are $k$ uses remaining in the other roll, for some $k$ between $1$ and $50$. There are two mutually exclusive possibilities: you must have arrived at the point $(50, 50-k)$ by taking a step to the right of the point $(49, 50-k)$, or you arrived at the point $(50-k, 50)$ by taking a step up from the point $(50-k, 49)$.
For the first option there are $49 + (50-k)choose 49$ ways to reach the penultimate point, followed by a step to the right; each path has probability $(1/2)^99-kcdot(1/2)$, giving an overall probability of $99-kchoose 49(1/2)^100-k$. Since same result holds for the second option, the probability that the remaining roll has $k$ uses is twice this last number:
$$p_k:=99-kchoose 49left(frac12right)^99-k.tag1$$
To compute $E(K)$, the expected number of uses in the remaining roll, you can evaluate the summation $sum k p_k$, as in @joriki's approach. Alternatively you can use the following device to obtain a closed form expression: Rearrange (1) into the recursion relation
$$
2[50-k]p_k = [100-(k+1)]p_k+1,tag2
$$
valid for all $k=1,ldots,50$. Sum both sides of (2) over all $k$:
$$
2[50 - E(K)] = 100(1-p_1) - (E(K)-p_1).tag3
$$
Finally rearrange (3) to obtain
$$E(K)=99p_1= 9998choose49left(frac12right)^98approx7.96.$$
answered Aug 29 at 7:22
grand_chat
18.3k11122
Your calculation of the expectation is very nice! :-)
– joriki
Aug 29 at 19:10
1
@joriki Thanks! I just now fixed an error in (3) which fortunately leads to the same result :-)
– grand_chat
Aug 29 at 19:38
add a comment |Â
up vote
11
down vote
up vote
11
down vote
Model your floss use as a random walk in the plane starting at $(0,0)$ and taking a unit step to the right when you pick from roll A, and a unit step up when you pick from roll B.
The first time you use up one of the rolls, there are $k$ uses remaining in the other roll, for some $k$ between $1$ and $50$. There are two mutually exclusive possibilities: you must have arrived at the point $(50, 50-k)$ by taking a step to the right of the point $(49, 50-k)$, or you arrived at the point $(50-k, 50)$ by taking a step up from the point $(50-k, 49)$.
For the first option there are $49 + (50-k)choose 49$ ways to reach the penultimate point, followed by a step to the right; each path has probability $(1/2)^99-kcdot(1/2)$, giving an overall probability of $99-kchoose 49(1/2)^100-k$. Since same result holds for the second option, the probability that the remaining roll has $k$ uses is twice this last number:
$$p_k:=99-kchoose 49left(frac12right)^99-k.tag1$$
To compute $E(K)$, the expected number of uses in the remaining roll, you can evaluate the summation $sum k p_k$, as in @joriki's approach. Alternatively you can use the following device to obtain a closed form expression: Rearrange (1) into the recursion relation
$$
2[50-k]p_k = [100-(k+1)]p_k+1,tag2
$$
valid for all $k=1,ldots,50$. Sum both sides of (2) over all $k$:
$$
2[50 - E(K)] = 100(1-p_1) - (E(K)-p_1).tag3
$$
Finally rearrange (3) to obtain
$$E(K)=99p_1= 9998choose49left(frac12right)^98approx7.96.$$
answered Aug 29 at 7:22
grand_chat
18.3k11122
Model your floss use as a random walk in the plane starting at $(0,0)$ and taking a unit step to the right when you pick from roll A, and a unit step up when you pick from roll B.
The first time you use up one of the rolls, there are $k$ uses remaining in the other roll, for some $k$ between $1$ and $50$. There are two mutually exclusive possibilities: you must have arrived at the point $(50, 50-k)$ by taking a step to the right of the point $(49, 50-k)$, or you arrived at the point $(50-k, 50)$ by taking a step up from the point $(50-k, 49)$.
For the first option there are $49 + (50-k)choose 49$ ways to reach the penultimate point, followed by a step to the right; each path has probability $(1/2)^99-kcdot(1/2)$, giving an overall probability of $99-kchoose 49(1/2)^100-k$. Since same result holds for the second option, the probability that the remaining roll has $k$ uses is twice this last number:
$$p_k:=99-kchoose 49left(frac12right)^99-k.tag1$$
To compute $E(K)$, the expected number of uses in the remaining roll, you can evaluate the summation $sum k p_k$, as in @joriki's approach. Alternatively you can use the following device to obtain a closed form expression: Rearrange (1) into the recursion relation
$$
2[50-k]p_k = [100-(k+1)]p_k+1,tag2
$$
valid for all $k=1,ldots,50$. Sum both sides of (2) over all $k$:
$$
2[50 - E(K)] = 100(1-p_1) - (E(K)-p_1).tag3
$$
Finally rearrange (3) to obtain
$$E(K)=99p_1= 9998choose49left(frac12right)^98approx7.96.$$
answered Aug 29 at 7:22
grand_chat
18.3k11122
edited Aug 29 at 19:37
answered Aug 29 at 7:22
grand_chat
18.3k11122
answered Aug 29 at 7:22
grand_chat
18.3k11122
answered Aug 29 at 7:22
grand_chat
18.3k11122
18.3k11122
Your calculation of the expectation is very nice! :-)
– joriki
Aug 29 at 19:10
1
@joriki Thanks! I just now fixed an error in (3) which fortunately leads to the same result :-)
– grand_chat
Aug 29 at 19:38
add a comment |Â
Your calculation of the expectation is very nice! :-)
– joriki
Aug 29 at 19:10
1
@joriki Thanks! I just now fixed an error in (3) which fortunately leads to the same result :-)
– grand_chat
Aug 29 at 19:38
Your calculation of the expectation is very nice! :-)
– joriki
Aug 29 at 19:10
Your calculation of the expectation is very nice! :-)
– joriki
Aug 29 at 19:10
1
1
@joriki Thanks! I just now fixed an error in (3) which fortunately leads to the same result :-)
– grand_chat
Aug 29 at 19:38
@joriki Thanks! I just now fixed an error in (3) which fortunately leads to the same result :-)
– grand_chat
Aug 29 at 19:38
add a comment |Â
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5
en.m.wikipedia.org/wiki/Banach%27s_matchbox_problem
– Steve D
Aug 29 at 6:19
3
@SteveD: For a slight difference between the two problems, see this question.
– joriki
Aug 29 at 6:52
1
There is no concrete evidence that flossing is useful for your dental hygiene (I mean, if you have something specific to clean, sure). Some say it is even damaging to your gums. The 8 uses should probably last you about two months, not a week.
– Asaf Karagila♦
Aug 29 at 13:02
@joriki: It's really the same problem though, no? But you set $N=50-1$ to account for stopping on the 50th try, and you add $1$ to the expectation you get because the other roll cannot be empty.
– Steve D
Aug 29 at 14:13