Return to Answer

Note. At the time of writing, the accepted answer to this question was that of Douglas Zare.

The accepted answer to this question is incorrect, albeit for what appears to be a relatively minor reason. I discovered this while answering a special case of the same question over at [math.SE][1], where it was observed that in the special case $b=3$ and $n=10$, the formula from this post gave an absurdly large value.

After consulting the literature myself, I found the correct formula in Theorem 2 of [Stadje [1990]][1] (specifically equation (2.15) therein) with $p=1$ and $l=s=n$ and $m=b$. The desired expectation equals $$ \binom{n}{b}\sum_{j=0}^{n-1}\frac{(-1)^{n-j+1}\binom{n}{j}}{\binom{n}{b}-\binom{j}{b}}. $$

Comparing this formula with the (incorrect) one from the accepted answer, we change variables in the sum to $s=n-j$, yielding

$$ \sum_{s=1}^{n}\frac{(-1)^{s+1}\binom{n}{s}}{1-\binom{n-s}{b}/\binom{n}{b}}, $$ whereas the incorrect accepted answer states $$ \sum_{s=1}^{n-b}\frac{(-1)^{s+1}\binom{n}{s}}{1-\binom{n-s}{b}/\binom{n}{b}}. $$ The only difference is in the range of the summation, and it arises due to a mistake made by the accepted answer in the $i=0$ case of the following manipulation: $$ \sum_{S\not=\varnothing}(-1)^{|S|+1}\left(\frac{\binom{n-|S|}{b}}{\binom{n}{b}}\right)^i\overset{!}{=}\sum_{s=1}^{n-b}(-1)^{s+1}\binom{n}{s}\left(\frac{\binom{n-s}{b}}{\binom{n}{b}}\right)^i. $$ The equality holds when $i\not=0$, but when $i=0$ the terms with $s>n-b$ do in fact contribute. The corrected formula reads as follows: $$ \sum_{S\not=\varnothing}(-1)^{|S|+1}\left(\frac{\binom{n-|S|}{b}}{\binom{n}{b}}\right)^i=\sum_{s=1}^{n}(-1)^{s+1}\binom{n}{s}\left(\frac{\binom{n-s}{b}}{\binom{n}{b}}\right)^i. $$ The rest of the reasoning in the accepted answer is correct, and with this fix yields the correct answer. [1]: https://math.stackexchange.com/questions/3278200/box-of-10-chocolates/3278285#3278230

Note. At the time of writing, the accepted answer to this question was that of Douglas Zare.

The accepted answer to this question is incorrect, albeit for what appears to be a relatively minor reason. I discovered this while answering a special case of the same question over at math.SE, where it was observed that in the special case $b=3$ and $n=10$, the formula from this post gave an absurdly large value.

After consulting the literature myself, I found the correct formula in Theorem 2 of Stadje [1990] (specifically equation (2.15) therein) with $p=1$ and $l=s=n$ and $m=b$. The desired expectation equals $$ \binom{n}{b}\sum_{j=0}^{n-1}\frac{(-1)^{n-j+1}\binom{n}{j}}{\binom{n}{b}-\binom{j}{b}}. $$

Comparing this formula with the (incorrect) one from the accepted answer, we change variables in the sum to $s=n-j$, yielding

$$ \sum_{s=1}^{n}\frac{(-1)^{s+1}\binom{n}{s}}{1-\binom{n-s}{b}/\binom{n}{b}}, $$ whereas the incorrect accepted answer states $$ \sum_{s=1}^{n-b}\frac{(-1)^{s+1}\binom{n}{s}}{1-\binom{n-s}{b}/\binom{n}{b}}. $$ The only difference is in the range of the summation, and it arises due to a mistake made by the accepted answer in the $i=0$ case of the following manipulation: $$ \sum_{S\not=\varnothing}(-1)^{|S|+1}\left(\frac{\binom{n-|S|}{b}}{\binom{n}{b}}\right)^i\overset{!}{=}\sum_{s=1}^{n-b}(-1)^{s+1}\binom{n}{s}\left(\frac{\binom{n-s}{b}}{\binom{n}{b}}\right)^i. $$ The equality holds when $i\not=0$, but when $i=0$ the terms with $s>n-b$ do in fact contribute. The corrected formula reads as follows: $$ \sum_{S\not=\varnothing}(-1)^{|S|+1}\left(\frac{\binom{n-|S|}{b}}{\binom{n}{b}}\right)^i=\sum_{s=1}^{n}(-1)^{s+1}\binom{n}{s}\left(\frac{\binom{n-s}{b}}{\binom{n}{b}}\right)^i. $$ The rest of the reasoning in the accepted answer is correct, and with this fix yields the correct answer.

The accepted answer to this question is incorrect, albeit for what appears to be a relatively minor reason. I discovered this while answering a special case of the same question over at [math.SE][1], where it was observed that in the special case $b=3$ and $n=10$, the formula from this post gave an absurdly large value.

After consulting the literature myself, I found the correct formula in Theorem 2 of [Stadje [1990]][1] (specifically equation (2.15) therein) with $p=1$ and $l=s=n$ and $m=b$. The desired expectation equals $$ \binom{n}{b}\sum_{j=0}^{n-1}\frac{(-1)^{n-j+1}\binom{n}{j}}{\binom{n}{b}-\binom{j}{b}}. $$

Comparing this formula with the (incorrect) one from the accepted answer, we change variables in the sum to $s=n-j$, yielding

$$ \sum_{s=1}^{n}\frac{(-1)^{s+1}\binom{n}{s}}{1-\binom{n-s}{b}/\binom{n}{b}}, $$ whereas the incorrect accepted answer states $$ \sum_{s=1}^{n-b}\frac{(-1)^{s+1}\binom{n}{s}}{1-\binom{n-s}{b}/\binom{n}{b}}. $$ The only difference is in the range of the summation, and it arises due to a mistake made by the accepted answer in the $i=0$ case of the following manipulation: $$ \sum_{S\not=\varnothing}(-1)^{|S|+1}\left(\frac{\binom{n-|S|}{b}}{\binom{n}{b}}\right)^i\overset{!}{=}\sum_{s=1}^{n-b}(-1)^{s+1}\binom{n}{s}\left(\frac{\binom{n-|S|}{b}}{\binom{n}{b}}\right)^i. $$ The equality holds when $i\not=0$, but when $i=0$ the terms with $s>n-b$ do in fact contribute. The corrected formula reads as follows: $$ \sum_{S\not=\varnothing}(-1)^{|S|+1}\left(\frac{\binom{n-|S|}{b}}{\binom{n}{b}}\right)^i=\sum_{s=1}^{n}(-1)^{s+1}\binom{n}{s}\left(\frac{\binom{n-|S|}{b}}{\binom{n}{b}}\right)^i. $$ The rest of the reasoning in the accepted answer is correct, and with this fix yields the correct answer. [1]: https://math.stackexchange.com/questions/3278200/box-of-10-chocolates/3278285#3278230

Note. At the time of writing, the accepted answer to this question was that of Douglas Zare.

The accepted answer to this question is incorrect, albeit for what appears to be a relatively minor reason. I discovered this while answering a special case of the same question over at [math.SE][1], where it was observed that in the special case $b=3$ and $n=10$, the formula from this post gave an absurdly large value.

After consulting the literature myself, I found the correct formula in Theorem 2 of [Stadje [1990]][1] (specifically equation (2.15) therein) with $p=1$ and $l=s=n$ and $m=b$. The desired expectation equals $$ \binom{n}{b}\sum_{j=0}^{n-1}\frac{(-1)^{n-j+1}\binom{n}{j}}{\binom{n}{b}-\binom{j}{b}}. $$

Comparing this formula with the (incorrect) one from the accepted answer, we change variables in the sum to $s=n-j$, yielding

$$ \sum_{s=1}^{n}\frac{(-1)^{s+1}\binom{n}{s}}{1-\binom{n-s}{b}/\binom{n}{b}}, $$ whereas the incorrect accepted answer states $$ \sum_{s=1}^{n-b}\frac{(-1)^{s+1}\binom{n}{s}}{1-\binom{n-s}{b}/\binom{n}{b}}. $$ The only difference is in the range of the summation, and it arises due to a mistake made by the accepted answer in the $i=0$ case of the following manipulation: $$ \sum_{S\not=\varnothing}(-1)^{|S|+1}\left(\frac{\binom{n-|S|}{b}}{\binom{n}{b}}\right)^i\overset{!}{=}\sum_{s=1}^{n-b}(-1)^{s+1}\binom{n}{s}\left(\frac{\binom{n-s}{b}}{\binom{n}{b}}\right)^i. $$ The equality holds when $i\not=0$, but when $i=0$ the terms with $s>n-b$ do in fact contribute. The corrected formula reads as follows: $$ \sum_{S\not=\varnothing}(-1)^{|S|+1}\left(\frac{\binom{n-|S|}{b}}{\binom{n}{b}}\right)^i=\sum_{s=1}^{n}(-1)^{s+1}\binom{n}{s}\left(\frac{\binom{n-s}{b}}{\binom{n}{b}}\right)^i. $$ The rest of the reasoning in the accepted answer is correct, and with this fix yields the correct answer. [1]: https://math.stackexchange.com/questions/3278200/box-of-10-chocolates/3278285#3278230

The accepted answer to this question is incorrect, albeit for what appears to be a relatively minor reason. I discovered this while answering a special case of the same question over at [math.SE][1], where it was observed that in the special case $b=3$ and $n=10$, the formula from this post gave an absurdly large value.

After consulting the literature myself, I found the correct formula in Theorem 2 of [Stadje [1990]][1] (specifically equation (2.15) therein) with $p=1$ and $l=s=n$ and $m=b$. The desired expectation equals $$ \binom{n}{b}\sum_{j=0}^{n-1}\frac{(-1)^{n-j+1}\binom{n}{j}}{\binom{n}{b}-\binom{j}{b}}. $$

Comparing this formula with the (incorrect) one from the accepted answer, we change variables in the sum to $s=n-j$, yielding

$$ \sum_{s=1}^{n}\frac{(-1)^{s+1}\binom{n}{s}}{1-\binom{n-s}{b}/\binom{n}{b}}, $$ whereas the incorrect accepted answer states $$ \sum_{s=1}^{n-b}\frac{(-1)^{s+1}\binom{n}{s}}{1-\binom{n-s}{b}/\binom{n}{b}}. $$ The only difference is in the range of the summation. [1]: https://math.stackexchange.com/questions/3278200/box-of-10-chocolates/3278285#3278230

The accepted answer to this question is incorrect, albeit for what appears to be a relatively minor reason. I discovered this while answering a special case of the same question over at [math.SE][1], where it was observed that in the special case $b=3$ and $n=10$, the formula from this post gave an absurdly large value.

After consulting the literature myself, I found the correct formula in Theorem 2 of [Stadje [1990]][1] (specifically equation (2.15) therein) with $p=1$ and $l=s=n$ and $m=b$. The desired expectation equals $$ \binom{n}{b}\sum_{j=0}^{n-1}\frac{(-1)^{n-j+1}\binom{n}{j}}{\binom{n}{b}-\binom{j}{b}}. $$

Comparing this formula with the (incorrect) one from the accepted answer, we change variables in the sum to $s=n-j$, yielding

$$ \sum_{s=1}^{n}\frac{(-1)^{s+1}\binom{n}{s}}{1-\binom{n-s}{b}/\binom{n}{b}}, $$ whereas the incorrect accepted answer states $$ \sum_{s=1}^{n-b}\frac{(-1)^{s+1}\binom{n}{s}}{1-\binom{n-s}{b}/\binom{n}{b}}. $$ The only difference is in the range of the summation, and it arises due to a mistake made by the accepted answer in the $i=0$ case of the following manipulation: $$ \sum_{S\not=\varnothing}(-1)^{|S|+1}\left(\frac{\binom{n-|S|}{b}}{\binom{n}{b}}\right)^i\overset{!}{=}\sum_{s=1}^{n-b}(-1)^{s+1}\binom{n}{s}\left(\frac{\binom{n-|S|}{b}}{\binom{n}{b}}\right)^i. $$ The equality holds when $i\not=0$, but when $i=0$ the terms with $s>n-b$ do in fact contribute. The corrected formula reads as follows: $$ \sum_{S\not=\varnothing}(-1)^{|S|+1}\left(\frac{\binom{n-|S|}{b}}{\binom{n}{b}}\right)^i=\sum_{s=1}^{n}(-1)^{s+1}\binom{n}{s}\left(\frac{\binom{n-|S|}{b}}{\binom{n}{b}}\right)^i. $$ The rest of the reasoning in the accepted answer is correct, and with this fix yields the correct answer. [1]: https://math.stackexchange.com/questions/3278200/box-of-10-chocolates/3278285#3278230