On which place should you stand in a line, to get a bonus.
$begingroup$
Customers are going inside a store, the first customer whose birthday matches the birthday of someone that has already entered the store will get a bonus discount. Where on the line to stand to get the biggest chance to win a bonus?
probability recreational-mathematics
asked Jan 16 at 23:07
yuner.bekiryuner.bekir
204
$endgroup$
|
show 1 more comment
$begingroup$
Customers are going inside a store, the first customer whose birthday matches the birthday of someone that has already entered the store will get a bonus discount. Where on the line to stand to get the biggest chance to win a bonus?
probability recreational-mathematics
asked Jan 16 at 23:07
yuner.bekiryuner.bekir
204
$endgroup$
$begingroup$
What we can say for sure that the first in the line has no chance for a bonus :)
$endgroup$
– Peter
Jan 16 at 23:11
$begingroup$
Did you try to calculate the probability that the first one, the second one and so one gets the bonus ?
$endgroup$
– Peter
Jan 16 at 23:13
$begingroup$
This question is very poorly specified. What does "first customer" mean? What "line" are you talking about? Please clarify,
$endgroup$
– Rob Arthan
Jan 16 at 23:51
$begingroup$
Also if you're in place number 367 or more, it's impossible to win the bonus (pigeonhole principle).
$endgroup$
– Daniel Schepler
Jan 17 at 0:07
$begingroup$
Intuitively, you don't want to be too near the front (else the chance of getting a duplicate will be too small) nor do you want to be too far back (else somebody will beat you to a match). The expressions for the probabilities are likely to be a bit unwieldy, so I'd just compute them numerically and search for the max.
$endgroup$
– lulu
Jan 17 at 0:17
|
show 1 more comment
$begingroup$
Customers are going inside a store, the first customer whose birthday matches the birthday of someone that has already entered the store will get a bonus discount. Where on the line to stand to get the biggest chance to win a bonus?
probability recreational-mathematics
asked Jan 16 at 23:07
yuner.bekiryuner.bekir
204
$endgroup$
Customers are going inside a store, the first customer whose birthday matches the birthday of someone that has already entered the store will get a bonus discount. Where on the line to stand to get the biggest chance to win a bonus?
probability recreational-mathematics
probability recreational-mathematics
asked Jan 16 at 23:07
yuner.bekiryuner.bekir
204
asked Jan 16 at 23:07
yuner.bekiryuner.bekir
204
asked Jan 16 at 23:07
yuner.bekiryuner.bekir
204
asked Jan 16 at 23:07
yuner.bekiryuner.bekir
204
asked Jan 16 at 23:07
yuner.bekiryuner.bekir
204
204
$begingroup$
What we can say for sure that the first in the line has no chance for a bonus :)
$endgroup$
– Peter
Jan 16 at 23:11
$begingroup$
Did you try to calculate the probability that the first one, the second one and so one gets the bonus ?
$endgroup$
– Peter
Jan 16 at 23:13
$begingroup$
This question is very poorly specified. What does "first customer" mean? What "line" are you talking about? Please clarify,
$endgroup$
– Rob Arthan
Jan 16 at 23:51
$begingroup$
Also if you're in place number 367 or more, it's impossible to win the bonus (pigeonhole principle).
$endgroup$
– Daniel Schepler
Jan 17 at 0:07
$begingroup$
Intuitively, you don't want to be too near the front (else the chance of getting a duplicate will be too small) nor do you want to be too far back (else somebody will beat you to a match). The expressions for the probabilities are likely to be a bit unwieldy, so I'd just compute them numerically and search for the max.
$endgroup$
– lulu
Jan 17 at 0:17
|
show 1 more comment
$begingroup$
What we can say for sure that the first in the line has no chance for a bonus :)
$endgroup$
– Peter
Jan 16 at 23:11
$begingroup$
Did you try to calculate the probability that the first one, the second one and so one gets the bonus ?
$endgroup$
– Peter
Jan 16 at 23:13
$begingroup$
This question is very poorly specified. What does "first customer" mean? What "line" are you talking about? Please clarify,
$endgroup$
– Rob Arthan
Jan 16 at 23:51
$begingroup$
Also if you're in place number 367 or more, it's impossible to win the bonus (pigeonhole principle).
$endgroup$
– Daniel Schepler
Jan 17 at 0:07
$begingroup$
Intuitively, you don't want to be too near the front (else the chance of getting a duplicate will be too small) nor do you want to be too far back (else somebody will beat you to a match). The expressions for the probabilities are likely to be a bit unwieldy, so I'd just compute them numerically and search for the max.
$endgroup$
– lulu
Jan 17 at 0:17
$begingroup$
What we can say for sure that the first in the line has no chance for a bonus :)
$endgroup$
– Peter
Jan 16 at 23:11
$begingroup$
What we can say for sure that the first in the line has no chance for a bonus :)
$endgroup$
– Peter
Jan 16 at 23:11
$begingroup$
Did you try to calculate the probability that the first one, the second one and so one gets the bonus ?
$endgroup$
– Peter
Jan 16 at 23:13
$begingroup$
Did you try to calculate the probability that the first one, the second one and so one gets the bonus ?
$endgroup$
– Peter
Jan 16 at 23:13
$begingroup$
This question is very poorly specified. What does "first customer" mean? What "line" are you talking about? Please clarify,
$endgroup$
– Rob Arthan
Jan 16 at 23:51
$begingroup$
This question is very poorly specified. What does "first customer" mean? What "line" are you talking about? Please clarify,
$endgroup$
– Rob Arthan
Jan 16 at 23:51
$begingroup$
Also if you're in place number 367 or more, it's impossible to win the bonus (pigeonhole principle).
$endgroup$
– Daniel Schepler
Jan 17 at 0:07
$begingroup$
Also if you're in place number 367 or more, it's impossible to win the bonus (pigeonhole principle).
$endgroup$
– Daniel Schepler
Jan 17 at 0:07
$begingroup$
Intuitively, you don't want to be too near the front (else the chance of getting a duplicate will be too small) nor do you want to be too far back (else somebody will beat you to a match). The expressions for the probabilities are likely to be a bit unwieldy, so I'd just compute them numerically and search for the max.
$endgroup$
– lulu
Jan 17 at 0:17
$begingroup$
Intuitively, you don't want to be too near the front (else the chance of getting a duplicate will be too small) nor do you want to be too far back (else somebody will beat you to a match). The expressions for the probabilities are likely to be a bit unwieldy, so I'd just compute them numerically and search for the max.
$endgroup$
– lulu
Jan 17 at 0:17
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
I will use an year with $N=365$ days, every day being equally probable as a birthday for each of the customers in the line.
The $k$.th customer has a chance to get the price equal to $p_k$, say, as a matter of notation. Then $p_k$ is combinatorially obtained by
counting the day configurations $(d_1, dots,d_{k-1},d_k)$ with
different values on the first $(k-1)$ places, and with $d_k$ repeating
one of these $(k-1)$ places, and we divide by the number $N^k$ of all day configurations with $k$ places, so
Corrected version, thanks lulu and Daniel Schepler
$$
begin{aligned}
p_1 &=0 ,\
p_2 &=frac 1N ,\
&qquadtext{ and for }kge 3\
p_k &=
frac 1{N^k}cdot N(N-1)dots(N-k+2)cdot(k-1)
\
&=
left(1-frac 1Nright)
dots
left(1-frac {k-2}Nright)cdot frac {k-1}N .
end{aligned}
$$
The maximal value is in the script below obtained for $k=20$, p(20) ~ 0.0323198575
Here is numerically the list of the probabilities for the first customers:
sage: N = 365
sage: for k in [2..40]:
....: pk = binomial(N, k-1)*factorial(k-1)*(k-1) / N^k
....: print "p(%2s) ~ %.10f" % (k, pk.n())
....:
p( 2) ~ 0.0027397260
p( 3) ~ 0.0054644399
p( 4) ~ 0.0081517466
p( 5) ~ 0.0107796612
p( 6) ~ 0.0133269099
p( 7) ~ 0.0157732194
p( 8) ~ 0.0180995893
p( 9) ~ 0.0202885415
p(10) ~ 0.0223243438
p(11) ~ 0.0241932006
p(12) ~ 0.0258834105
p(13) ~ 0.0273854864
p(14) ~ 0.0286922368
p(15) ~ 0.0297988078
p(16) ~ 0.0307026855
p(17) ~ 0.0314036600
p(18) ~ 0.0319037526
p(19) ~ 0.0322071082
p(20) ~ 0.0323198575
p(21) ~ 0.0322499516
p(22) ~ 0.0320069725
p(23) ~ 0.0316019267
p(24) ~ 0.0310470236
p(25) ~ 0.0303554461
p(26) ~ 0.0295411162
p(27) ~ 0.0286184621
p(28) ~ 0.0276021901
p(29) ~ 0.0265070651
p(30) ~ 0.0253477052
p(31) ~ 0.0241383910
p(32) ~ 0.0228928941
p(33) ~ 0.0216243263
p(34) ~ 0.0203450104
p(35) ~ 0.0190663743
p(36) ~ 0.0177988675
p(37) ~ 0.0165519018
p(38) ~ 0.0153338129
p(39) ~ 0.0141518433
p(40) ~ 0.0130121455
sage code was used.
answered Jan 16 at 23:56
dan_fuleadan_fulea
7,1781513
$endgroup$
1
$begingroup$
Not following. Let's take the third in line. For her to win, she needs numbers $1,2$ to be different (a $frac {364}{365}$ probability). Then she must match one of them, a $frac 2{365}$ probability). Thus she wins the prize with probability $frac {364}{365}times frac 2{365}=0.005479452$ which is much greater than your $p(2)$. No?
$endgroup$
– lulu
Jan 17 at 0:09
$begingroup$
Should you be using $N^{underline{k-1}} = (k-1)! binom{N}{k-1}$ instead of $binom{N}{k-1}$?
$endgroup$
– Daniel Schepler
Jan 17 at 0:11
$begingroup$
Note too that $sum p(i)=1$ since somebody has to win, but this is clearly not true for your numbers.
$endgroup$
– lulu
Jan 17 at 0:14
$begingroup$
Can you clarify or correct your post? Unless I am badly misreading the problem, the answer is pretty clearly not the second person in line. In fact, simple calculations show that optimal is $k=20$ for which the probability is $0.032319858$. That seems intuitively solid...as we know from the standard birthday problem, it's in the early 20's that you start to expect to see a duplicate.
$endgroup$
– lulu
Jan 17 at 0:27
1
$begingroup$
With my correction, I calculated $frac{p_{k+1}}{p_k} = frac{N-k}{N} cdot frac{k}{k-1}$ which is greater than 1 if and only if $k^2 < N$; so the maximum would happen at $k = lceil sqrt{N} rceil = 20$.
$endgroup$
– Daniel Schepler
Jan 17 at 0:33
|
show 2 more comments
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
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$begingroup$
I will use an year with $N=365$ days, every day being equally probable as a birthday for each of the customers in the line.
The $k$.th customer has a chance to get the price equal to $p_k$, say, as a matter of notation. Then $p_k$ is combinatorially obtained by
counting the day configurations $(d_1, dots,d_{k-1},d_k)$ with
different values on the first $(k-1)$ places, and with $d_k$ repeating
one of these $(k-1)$ places, and we divide by the number $N^k$ of all day configurations with $k$ places, so
Corrected version, thanks lulu and Daniel Schepler
$$
begin{aligned}
p_1 &=0 ,\
p_2 &=frac 1N ,\
&qquadtext{ and for }kge 3\
p_k &=
frac 1{N^k}cdot N(N-1)dots(N-k+2)cdot(k-1)
\
&=
left(1-frac 1Nright)
dots
left(1-frac {k-2}Nright)cdot frac {k-1}N .
end{aligned}
$$
The maximal value is in the script below obtained for $k=20$, p(20) ~ 0.0323198575
Here is numerically the list of the probabilities for the first customers:
sage: N = 365
sage: for k in [2..40]:
....: pk = binomial(N, k-1)*factorial(k-1)*(k-1) / N^k
....: print "p(%2s) ~ %.10f" % (k, pk.n())
....:
p( 2) ~ 0.0027397260
p( 3) ~ 0.0054644399
p( 4) ~ 0.0081517466
p( 5) ~ 0.0107796612
p( 6) ~ 0.0133269099
p( 7) ~ 0.0157732194
p( 8) ~ 0.0180995893
p( 9) ~ 0.0202885415
p(10) ~ 0.0223243438
p(11) ~ 0.0241932006
p(12) ~ 0.0258834105
p(13) ~ 0.0273854864
p(14) ~ 0.0286922368
p(15) ~ 0.0297988078
p(16) ~ 0.0307026855
p(17) ~ 0.0314036600
p(18) ~ 0.0319037526
p(19) ~ 0.0322071082
p(20) ~ 0.0323198575
p(21) ~ 0.0322499516
p(22) ~ 0.0320069725
p(23) ~ 0.0316019267
p(24) ~ 0.0310470236
p(25) ~ 0.0303554461
p(26) ~ 0.0295411162
p(27) ~ 0.0286184621
p(28) ~ 0.0276021901
p(29) ~ 0.0265070651
p(30) ~ 0.0253477052
p(31) ~ 0.0241383910
p(32) ~ 0.0228928941
p(33) ~ 0.0216243263
p(34) ~ 0.0203450104
p(35) ~ 0.0190663743
p(36) ~ 0.0177988675
p(37) ~ 0.0165519018
p(38) ~ 0.0153338129
p(39) ~ 0.0141518433
p(40) ~ 0.0130121455
sage code was used.
answered Jan 16 at 23:56
dan_fuleadan_fulea
7,1781513
$endgroup$
1
$begingroup$
Not following. Let's take the third in line. For her to win, she needs numbers $1,2$ to be different (a $frac {364}{365}$ probability). Then she must match one of them, a $frac 2{365}$ probability). Thus she wins the prize with probability $frac {364}{365}times frac 2{365}=0.005479452$ which is much greater than your $p(2)$. No?
$endgroup$
– lulu
Jan 17 at 0:09
$begingroup$
Should you be using $N^{underline{k-1}} = (k-1)! binom{N}{k-1}$ instead of $binom{N}{k-1}$?
$endgroup$
– Daniel Schepler
Jan 17 at 0:11
$begingroup$
Note too that $sum p(i)=1$ since somebody has to win, but this is clearly not true for your numbers.
$endgroup$
– lulu
Jan 17 at 0:14
$begingroup$
Can you clarify or correct your post? Unless I am badly misreading the problem, the answer is pretty clearly not the second person in line. In fact, simple calculations show that optimal is $k=20$ for which the probability is $0.032319858$. That seems intuitively solid...as we know from the standard birthday problem, it's in the early 20's that you start to expect to see a duplicate.
$endgroup$
– lulu
Jan 17 at 0:27
1
$begingroup$
With my correction, I calculated $frac{p_{k+1}}{p_k} = frac{N-k}{N} cdot frac{k}{k-1}$ which is greater than 1 if and only if $k^2 < N$; so the maximum would happen at $k = lceil sqrt{N} rceil = 20$.
$endgroup$
– Daniel Schepler
Jan 17 at 0:33
|
show 2 more comments
$begingroup$
I will use an year with $N=365$ days, every day being equally probable as a birthday for each of the customers in the line.
The $k$.th customer has a chance to get the price equal to $p_k$, say, as a matter of notation. Then $p_k$ is combinatorially obtained by
counting the day configurations $(d_1, dots,d_{k-1},d_k)$ with
different values on the first $(k-1)$ places, and with $d_k$ repeating
one of these $(k-1)$ places, and we divide by the number $N^k$ of all day configurations with $k$ places, so
Corrected version, thanks lulu and Daniel Schepler
$$
begin{aligned}
p_1 &=0 ,\
p_2 &=frac 1N ,\
&qquadtext{ and for }kge 3\
p_k &=
frac 1{N^k}cdot N(N-1)dots(N-k+2)cdot(k-1)
\
&=
left(1-frac 1Nright)
dots
left(1-frac {k-2}Nright)cdot frac {k-1}N .
end{aligned}
$$
The maximal value is in the script below obtained for $k=20$, p(20) ~ 0.0323198575
Here is numerically the list of the probabilities for the first customers:
sage: N = 365
sage: for k in [2..40]:
....: pk = binomial(N, k-1)*factorial(k-1)*(k-1) / N^k
....: print "p(%2s) ~ %.10f" % (k, pk.n())
....:
p( 2) ~ 0.0027397260
p( 3) ~ 0.0054644399
p( 4) ~ 0.0081517466
p( 5) ~ 0.0107796612
p( 6) ~ 0.0133269099
p( 7) ~ 0.0157732194
p( 8) ~ 0.0180995893
p( 9) ~ 0.0202885415
p(10) ~ 0.0223243438
p(11) ~ 0.0241932006
p(12) ~ 0.0258834105
p(13) ~ 0.0273854864
p(14) ~ 0.0286922368
p(15) ~ 0.0297988078
p(16) ~ 0.0307026855
p(17) ~ 0.0314036600
p(18) ~ 0.0319037526
p(19) ~ 0.0322071082
p(20) ~ 0.0323198575
p(21) ~ 0.0322499516
p(22) ~ 0.0320069725
p(23) ~ 0.0316019267
p(24) ~ 0.0310470236
p(25) ~ 0.0303554461
p(26) ~ 0.0295411162
p(27) ~ 0.0286184621
p(28) ~ 0.0276021901
p(29) ~ 0.0265070651
p(30) ~ 0.0253477052
p(31) ~ 0.0241383910
p(32) ~ 0.0228928941
p(33) ~ 0.0216243263
p(34) ~ 0.0203450104
p(35) ~ 0.0190663743
p(36) ~ 0.0177988675
p(37) ~ 0.0165519018
p(38) ~ 0.0153338129
p(39) ~ 0.0141518433
p(40) ~ 0.0130121455
sage code was used.
answered Jan 16 at 23:56
dan_fuleadan_fulea
7,1781513
$endgroup$
1
$begingroup$
Not following. Let's take the third in line. For her to win, she needs numbers $1,2$ to be different (a $frac {364}{365}$ probability). Then she must match one of them, a $frac 2{365}$ probability). Thus she wins the prize with probability $frac {364}{365}times frac 2{365}=0.005479452$ which is much greater than your $p(2)$. No?
$endgroup$
– lulu
Jan 17 at 0:09
$begingroup$
Should you be using $N^{underline{k-1}} = (k-1)! binom{N}{k-1}$ instead of $binom{N}{k-1}$?
$endgroup$
– Daniel Schepler
Jan 17 at 0:11
$begingroup$
Note too that $sum p(i)=1$ since somebody has to win, but this is clearly not true for your numbers.
$endgroup$
– lulu
Jan 17 at 0:14
$begingroup$
Can you clarify or correct your post? Unless I am badly misreading the problem, the answer is pretty clearly not the second person in line. In fact, simple calculations show that optimal is $k=20$ for which the probability is $0.032319858$. That seems intuitively solid...as we know from the standard birthday problem, it's in the early 20's that you start to expect to see a duplicate.
$endgroup$
– lulu
Jan 17 at 0:27
1
$begingroup$
With my correction, I calculated $frac{p_{k+1}}{p_k} = frac{N-k}{N} cdot frac{k}{k-1}$ which is greater than 1 if and only if $k^2 < N$; so the maximum would happen at $k = lceil sqrt{N} rceil = 20$.
$endgroup$
– Daniel Schepler
Jan 17 at 0:33
|
show 2 more comments
$begingroup$
I will use an year with $N=365$ days, every day being equally probable as a birthday for each of the customers in the line.
The $k$.th customer has a chance to get the price equal to $p_k$, say, as a matter of notation. Then $p_k$ is combinatorially obtained by
counting the day configurations $(d_1, dots,d_{k-1},d_k)$ with
different values on the first $(k-1)$ places, and with $d_k$ repeating
one of these $(k-1)$ places, and we divide by the number $N^k$ of all day configurations with $k$ places, so
Corrected version, thanks lulu and Daniel Schepler
$$
begin{aligned}
p_1 &=0 ,\
p_2 &=frac 1N ,\
&qquadtext{ and for }kge 3\
p_k &=
frac 1{N^k}cdot N(N-1)dots(N-k+2)cdot(k-1)
\
&=
left(1-frac 1Nright)
dots
left(1-frac {k-2}Nright)cdot frac {k-1}N .
end{aligned}
$$
The maximal value is in the script below obtained for $k=20$, p(20) ~ 0.0323198575
Here is numerically the list of the probabilities for the first customers:
sage: N = 365
sage: for k in [2..40]:
....: pk = binomial(N, k-1)*factorial(k-1)*(k-1) / N^k
....: print "p(%2s) ~ %.10f" % (k, pk.n())
....:
p( 2) ~ 0.0027397260
p( 3) ~ 0.0054644399
p( 4) ~ 0.0081517466
p( 5) ~ 0.0107796612
p( 6) ~ 0.0133269099
p( 7) ~ 0.0157732194
p( 8) ~ 0.0180995893
p( 9) ~ 0.0202885415
p(10) ~ 0.0223243438
p(11) ~ 0.0241932006
p(12) ~ 0.0258834105
p(13) ~ 0.0273854864
p(14) ~ 0.0286922368
p(15) ~ 0.0297988078
p(16) ~ 0.0307026855
p(17) ~ 0.0314036600
p(18) ~ 0.0319037526
p(19) ~ 0.0322071082
p(20) ~ 0.0323198575
p(21) ~ 0.0322499516
p(22) ~ 0.0320069725
p(23) ~ 0.0316019267
p(24) ~ 0.0310470236
p(25) ~ 0.0303554461
p(26) ~ 0.0295411162
p(27) ~ 0.0286184621
p(28) ~ 0.0276021901
p(29) ~ 0.0265070651
p(30) ~ 0.0253477052
p(31) ~ 0.0241383910
p(32) ~ 0.0228928941
p(33) ~ 0.0216243263
p(34) ~ 0.0203450104
p(35) ~ 0.0190663743
p(36) ~ 0.0177988675
p(37) ~ 0.0165519018
p(38) ~ 0.0153338129
p(39) ~ 0.0141518433
p(40) ~ 0.0130121455
sage code was used.
answered Jan 16 at 23:56
dan_fuleadan_fulea
7,1781513
$endgroup$
I will use an year with $N=365$ days, every day being equally probable as a birthday for each of the customers in the line.
The $k$.th customer has a chance to get the price equal to $p_k$, say, as a matter of notation. Then $p_k$ is combinatorially obtained by
counting the day configurations $(d_1, dots,d_{k-1},d_k)$ with
different values on the first $(k-1)$ places, and with $d_k$ repeating
one of these $(k-1)$ places, and we divide by the number $N^k$ of all day configurations with $k$ places, so
Corrected version, thanks lulu and Daniel Schepler
$$
begin{aligned}
p_1 &=0 ,\
p_2 &=frac 1N ,\
&qquadtext{ and for }kge 3\
p_k &=
frac 1{N^k}cdot N(N-1)dots(N-k+2)cdot(k-1)
\
&=
left(1-frac 1Nright)
dots
left(1-frac {k-2}Nright)cdot frac {k-1}N .
end{aligned}
$$
The maximal value is in the script below obtained for $k=20$, p(20) ~ 0.0323198575
Here is numerically the list of the probabilities for the first customers:
sage: N = 365
sage: for k in [2..40]:
....: pk = binomial(N, k-1)*factorial(k-1)*(k-1) / N^k
....: print "p(%2s) ~ %.10f" % (k, pk.n())
....:
p( 2) ~ 0.0027397260
p( 3) ~ 0.0054644399
p( 4) ~ 0.0081517466
p( 5) ~ 0.0107796612
p( 6) ~ 0.0133269099
p( 7) ~ 0.0157732194
p( 8) ~ 0.0180995893
p( 9) ~ 0.0202885415
p(10) ~ 0.0223243438
p(11) ~ 0.0241932006
p(12) ~ 0.0258834105
p(13) ~ 0.0273854864
p(14) ~ 0.0286922368
p(15) ~ 0.0297988078
p(16) ~ 0.0307026855
p(17) ~ 0.0314036600
p(18) ~ 0.0319037526
p(19) ~ 0.0322071082
p(20) ~ 0.0323198575
p(21) ~ 0.0322499516
p(22) ~ 0.0320069725
p(23) ~ 0.0316019267
p(24) ~ 0.0310470236
p(25) ~ 0.0303554461
p(26) ~ 0.0295411162
p(27) ~ 0.0286184621
p(28) ~ 0.0276021901
p(29) ~ 0.0265070651
p(30) ~ 0.0253477052
p(31) ~ 0.0241383910
p(32) ~ 0.0228928941
p(33) ~ 0.0216243263
p(34) ~ 0.0203450104
p(35) ~ 0.0190663743
p(36) ~ 0.0177988675
p(37) ~ 0.0165519018
p(38) ~ 0.0153338129
p(39) ~ 0.0141518433
p(40) ~ 0.0130121455
sage code was used.
answered Jan 16 at 23:56
dan_fuleadan_fulea
7,1781513
edited Jan 17 at 1:05
answered Jan 16 at 23:56
dan_fuleadan_fulea
7,1781513
answered Jan 16 at 23:56
dan_fuleadan_fulea
7,1781513
answered Jan 16 at 23:56
dan_fuleadan_fulea
7,1781513
7,1781513
1
$begingroup$
Not following. Let's take the third in line. For her to win, she needs numbers $1,2$ to be different (a $frac {364}{365}$ probability). Then she must match one of them, a $frac 2{365}$ probability). Thus she wins the prize with probability $frac {364}{365}times frac 2{365}=0.005479452$ which is much greater than your $p(2)$. No?
$endgroup$
– lulu
Jan 17 at 0:09
$begingroup$
Should you be using $N^{underline{k-1}} = (k-1)! binom{N}{k-1}$ instead of $binom{N}{k-1}$?
$endgroup$
– Daniel Schepler
Jan 17 at 0:11
$begingroup$
Note too that $sum p(i)=1$ since somebody has to win, but this is clearly not true for your numbers.
$endgroup$
– lulu
Jan 17 at 0:14
$begingroup$
Can you clarify or correct your post? Unless I am badly misreading the problem, the answer is pretty clearly not the second person in line. In fact, simple calculations show that optimal is $k=20$ for which the probability is $0.032319858$. That seems intuitively solid...as we know from the standard birthday problem, it's in the early 20's that you start to expect to see a duplicate.
$endgroup$
– lulu
Jan 17 at 0:27
1
$begingroup$
With my correction, I calculated $frac{p_{k+1}}{p_k} = frac{N-k}{N} cdot frac{k}{k-1}$ which is greater than 1 if and only if $k^2 < N$; so the maximum would happen at $k = lceil sqrt{N} rceil = 20$.
$endgroup$
– Daniel Schepler
Jan 17 at 0:33
|
show 2 more comments
1
$begingroup$
Not following. Let's take the third in line. For her to win, she needs numbers $1,2$ to be different (a $frac {364}{365}$ probability). Then she must match one of them, a $frac 2{365}$ probability). Thus she wins the prize with probability $frac {364}{365}times frac 2{365}=0.005479452$ which is much greater than your $p(2)$. No?
$endgroup$
– lulu
Jan 17 at 0:09
$begingroup$
Should you be using $N^{underline{k-1}} = (k-1)! binom{N}{k-1}$ instead of $binom{N}{k-1}$?
$endgroup$
– Daniel Schepler
Jan 17 at 0:11
$begingroup$
Note too that $sum p(i)=1$ since somebody has to win, but this is clearly not true for your numbers.
$endgroup$
– lulu
Jan 17 at 0:14
$begingroup$
Can you clarify or correct your post? Unless I am badly misreading the problem, the answer is pretty clearly not the second person in line. In fact, simple calculations show that optimal is $k=20$ for which the probability is $0.032319858$. That seems intuitively solid...as we know from the standard birthday problem, it's in the early 20's that you start to expect to see a duplicate.
$endgroup$
– lulu
Jan 17 at 0:27
1
$begingroup$
With my correction, I calculated $frac{p_{k+1}}{p_k} = frac{N-k}{N} cdot frac{k}{k-1}$ which is greater than 1 if and only if $k^2 < N$; so the maximum would happen at $k = lceil sqrt{N} rceil = 20$.
$endgroup$
– Daniel Schepler
Jan 17 at 0:33
1
1
$begingroup$
Not following. Let's take the third in line. For her to win, she needs numbers $1,2$ to be different (a $frac {364}{365}$ probability). Then she must match one of them, a $frac 2{365}$ probability). Thus she wins the prize with probability $frac {364}{365}times frac 2{365}=0.005479452$ which is much greater than your $p(2)$. No?
$endgroup$
– lulu
Jan 17 at 0:09
$begingroup$
Not following. Let's take the third in line. For her to win, she needs numbers $1,2$ to be different (a $frac {364}{365}$ probability). Then she must match one of them, a $frac 2{365}$ probability). Thus she wins the prize with probability $frac {364}{365}times frac 2{365}=0.005479452$ which is much greater than your $p(2)$. No?
$endgroup$
– lulu
Jan 17 at 0:09
$begingroup$
Should you be using $N^{underline{k-1}} = (k-1)! binom{N}{k-1}$ instead of $binom{N}{k-1}$?
$endgroup$
– Daniel Schepler
Jan 17 at 0:11
$begingroup$
Should you be using $N^{underline{k-1}} = (k-1)! binom{N}{k-1}$ instead of $binom{N}{k-1}$?
$endgroup$
– Daniel Schepler
Jan 17 at 0:11
$begingroup$
Note too that $sum p(i)=1$ since somebody has to win, but this is clearly not true for your numbers.
$endgroup$
– lulu
Jan 17 at 0:14
$begingroup$
Note too that $sum p(i)=1$ since somebody has to win, but this is clearly not true for your numbers.
$endgroup$
– lulu
Jan 17 at 0:14
$begingroup$
Can you clarify or correct your post? Unless I am badly misreading the problem, the answer is pretty clearly not the second person in line. In fact, simple calculations show that optimal is $k=20$ for which the probability is $0.032319858$. That seems intuitively solid...as we know from the standard birthday problem, it's in the early 20's that you start to expect to see a duplicate.
$endgroup$
– lulu
Jan 17 at 0:27
$begingroup$
Can you clarify or correct your post? Unless I am badly misreading the problem, the answer is pretty clearly not the second person in line. In fact, simple calculations show that optimal is $k=20$ for which the probability is $0.032319858$. That seems intuitively solid...as we know from the standard birthday problem, it's in the early 20's that you start to expect to see a duplicate.
$endgroup$
– lulu
Jan 17 at 0:27
1
1
$begingroup$
With my correction, I calculated $frac{p_{k+1}}{p_k} = frac{N-k}{N} cdot frac{k}{k-1}$ which is greater than 1 if and only if $k^2 < N$; so the maximum would happen at $k = lceil sqrt{N} rceil = 20$.
$endgroup$
– Daniel Schepler
Jan 17 at 0:33
$begingroup$
With my correction, I calculated $frac{p_{k+1}}{p_k} = frac{N-k}{N} cdot frac{k}{k-1}$ which is greater than 1 if and only if $k^2 < N$; so the maximum would happen at $k = lceil sqrt{N} rceil = 20$.
$endgroup$
– Daniel Schepler
Jan 17 at 0:33
|
show 2 more comments
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$begingroup$
What we can say for sure that the first in the line has no chance for a bonus :)
$endgroup$
– Peter
Jan 16 at 23:11
$begingroup$
Did you try to calculate the probability that the first one, the second one and so one gets the bonus ?
$endgroup$
– Peter
Jan 16 at 23:13
$begingroup$
This question is very poorly specified. What does "first customer" mean? What "line" are you talking about? Please clarify,
$endgroup$
– Rob Arthan
Jan 16 at 23:51
$begingroup$
Also if you're in place number 367 or more, it's impossible to win the bonus (pigeonhole principle).
$endgroup$
– Daniel Schepler
Jan 17 at 0:07
$begingroup$
Intuitively, you don't want to be too near the front (else the chance of getting a duplicate will be too small) nor do you want to be too far back (else somebody will beat you to a match). The expressions for the probabilities are likely to be a bit unwieldy, so I'd just compute them numerically and search for the max.
$endgroup$
– lulu
Jan 17 at 0:17