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I had posted an urn probability problem that didn't have good motivation. I'd like to try to explain the motivation here, and reintroduce the problem.

Consider binary sequences of length $2n$. Let's say we put a marker in such a sequence as soon as we see a total of $n$ 0's or $n$ 1's, reading left to right. For example, if $n=4$, then the sequence 00101011 would receive a marker thus: 001010|11. Now write down the bits to the right of the marker. In the case of our example, this would be 11. Do this for every binary sequence of length $2n$. We observe that we have written down $2n\binom{2n}{n}$ bits, half 0's and half 1's. It is possible to prove this observation using binomial coefficient identities, but I wonder whether there is a simple bijective proof.

The previous urn problem was an equivalent probabilistic formulation of the case $n=5$.

I had posted an urn probability problem that didn't have good motivation. I'd like to try to explain the motivation here, and reintroduce the problem.

Consider binary sequences of length $2n$. Let's say we put a marker in such a sequence as soon as we see a total of $n$ 0's or $n$ 1's, reading left to right. For example, if $n=4$, then the sequence 00101011 would receive a marker thus: 001010|11. Now write down the bits to the right of the marker. In the case of our example, this would be 11. Do this for every binary sequence of length $2n$. We observe that we have written down $2n\binom{2n}{n}$ bits, half 0's and half 1's. It is possible to prove this observation using binomial coefficient identities, but I wonder whether there is a simple bijective proof.

The previous urn problem was an equivalent probabilistic formulation of the case $n=5$.

I had posted an urn probability problem that didn't have good motivation. I'd like to try to explain the motivation here, and reintroduce the problem.

Consider binary sequences of length $2n$. Let's say we put a marker in such a sequence as soon as we see a total of $n$ 0's or $n$ 1's, reading left to right. For example, if $n=4$, then the sequence 00101011 would receive a marker thus: 001010|11. Now write down the bits to the right of the marker. In the case of our example, this would be 11. Do this for every binary sequence of legnth $2n$. We observe that we have written down $2n\binom{2n}{n}$ bits, half 0's and half 1's. It is possible to prove this observation using binomial coefficient identities, but I wonder whether there is a simple bijective proof.

The previous urn problem was an equivalent probabilistic formulation of the case $n=5$.

I had posted an urn probability problem that didn't have good motivation. I'd like to try to explain the motivation here, and reintroduce the problem.

Consider binary sequences of length $2n$. Let's say we put a marker in such a sequence as soon as we see a total of $n$ 0's or $n$ 1's, reading left to right. For example, if $n=4$, then the sequence 00101011 would receive a marker thus: 001010|11. Now write down the bits to the right of the marker. In the case of our example, this would be 11. Do this for every binary sequence of length $2n$. We observe that we have written down $2n\binom{2n}{n}$ bits, half 0's and half 1's. It is possible to prove this observation using binomial coefficient identities, but I wonder whether there is a simple bijective proof.

The previous urn problem was an equivalent probabilistic formulation of the case $n=5$.

I had posted an urn probability problem that didn't have good motivation. I'd like to try to explain the motivation here, and reintroduce the problem.

Consider binary sequences of length 2n. Let's say we put a marker in such a sequence as soon as we see a total of n 0's or n 1's, reading left to right. For example, if n=4, then the sequence 00101011 would receive a marker thus: 001010|11. Now write down the bits to the right of the marker. In the case of our example, this would be 11. Do this for every binary sequence of legnth 2n. We observe that we have written down 2n*Binomial(2n,n) bits, half 0's and half 1's. It is possible to prove this observation using binomial coefficient identities, but I wonder whether there is a simple bijective proof.

The previous urn problem was an equivalent probabilistic formulation of the case n=5.

Thank you.

I had posted an urn probability problem that didn't have good motivation. I'd like to try to explain the motivation here, and reintroduce the problem.

Consider binary sequences of length $2n$. Let's say we put a marker in such a sequence as soon as we see a total of $n$ 0's or $n$ 1's, reading left to right. For example, if $n=4$, then the sequence 00101011 would receive a marker thus: 001010|11. Now write down the bits to the right of the marker. In the case of our example, this would be 11. Do this for every binary sequence of legnth $2n$. We observe that we have written down $2n\binom{2n}{n}$ bits, half 0's and half 1's. It is possible to prove this observation using binomial coefficient identities, but I wonder whether there is a simple bijective proof.

The previous urn problem was an equivalent probabilistic formulation of the case $n=5$.