The Second Hearts Problem

Question

According to the last part of these lecture notes, if we have a standard deck of playing cards and turn cards until the first heart appears, the probability that the next card is a heart is $\color{red}{1/4}$.

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Without using probability theory, this looks to me like we are computing:

$$E[ \ \mathbb P(Y_{N+1} = 1 | Y_N = 1, \{Y_i= 0\}_{\{i < N\}}) \ ]$$

where:

1. $N$ is a discrete random variable whose range is $\{1,2,...,40\}$, indicating the first time a heart is turned over (I guess $N$ is a stopping time, but I don't believe I'm using probability theory)

2. $\{Y_i\}_{i \in \{1,2,...,52\}}$ are binary random variables indicating whether or not the $i$th card is a heart,

as follows:

$$E\left[ \ \mathbb P(Y_{N+1} = 1 | Y_N = 1, \{Y_i= 0\}_{\{i < N\}}) \ \right]$$

$$= E\left[ \ \sum_{n=1}^{40} \mathbb P(Y_{n+1} = 1 | Y_n = 1, \{Y_i= 0\}_{\{i < n\}} ) 1_{\{N=n\}} \ \right]$$

$$= E\left[ \ \sum_{n=1}^{40} \frac{12}{52-n} 1_{\{N=n\}} \ \right]$$

$$= \sum_{n=1}^{40} \frac{12}{52-n} E\left[ \ 1_{\{N=n\}} \ \right]$$

$$= \sum_{n=1}^{40} \frac{12}{52-n} \mathbb P(N=n)$$

$$= \sum_{n=1}^{40} \frac{12}{52-n} \frac{13 \cdot P(39,n-1)}{P(52,n)} = \color{red}{1/4}$$

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Questions:

So what then is the following quantity (in plain English):

$$\mathbb P(Y_{N+1} = 1 | Y_N = 1, \{Y_i= 0\}_{\{i < N\}}) = \sum_{n=1}^{40} \frac{12}{52-n} 1_{\{N=n\}}$$

?

That to me looks like

$$P((n+1) \text{th card is hearts} \ | \ n\text{th card is first heart})$$

So how is that different from what we are computing which again is

$$E[ \ \mathbb P(Y_{N+1} = 1 | Y_N = 1, \{Y_i= 0\}_{\{i < N\}}) \ ]$$

?

The latter to me seems something like expected probability while the former is probability.

Answer 1

We should compute for

$$E[ \ \mathbb P(Y_{N+1} = 1 | Y_N = 1, \{Y_i= 0\}_{\{i < N\}}) \ ]$$

rather than

$$\mathbb P(Y_{N+1} = 1 | Y_N = 1, \{Y_i= 0\}_{\{i < N\}}) = \sum_{n=1}^{40} \frac{12}{52-n} 1_{\{N=n\}}$$

because the latter computes

$$P((n+1) \text{th card is hearts} \ | \ n\text{th card is first heart})$$

meaning a certain card is already given to be the first heart while the question does not assume a certain card is given to be the first heart.

Answer 2

$\mathbb P(Y_{N+1} = 1 \mid Y_N = 1, \{Y_i= 0\}_{\{i < N\}})$ is a random variable (I hope that term exists in plain English?) that's a function of the random variable $N$.

You're comparing it to $P((n+1) \text{-th card is hearts} \mid n\text{th card is hearts})$, by which I suspect you mean more precisely $P((n+1) \text{th card is hearts} \mid n\text{-th card is the first hearts})$. This is similar, but with a small $n$: $\mathbb P(Y_{n+1} = 1 \mid Y_n = 1, \{Y_i= 0\}_{\{i < n\}})$. This is a (conditional) probability, not a random variable; it has a particular value for every particular value of $n$.

You ask how this differs from the quantity $E\left[\mathbb P(Y_{N+1} = 1 \mid Y_N = 1, \{Y_i= 0\}_{\{i < N\}})\right]$ being computed. This is the expected value of the random variable $\mathbb P(Y_{N+1} = 1 \mid Y_N = 1, \{Y_i= 0\}_{\{i < N\}})$, where $N$ takes all values of $n$, not a particular value.

In concrete terms, we have

\begin{align} \mathbb P(Y_{n+1} = 1 \mid Y_n = 1, \{Y_i= 0\}_{\{i < n\}})&=\frac{12}{52-n}\;,\\ \mathbb P(Y_{N+1} = 1\mid Y_N = 1, \{Y_i= 0\}_{\{i < N\}}) &= \sum_{n=1}^{40} \frac{12}{52-n} 1_{\{N=n\}}\;,\\ E\left[\mathbb P(Y_{N+1} = 1 \mid Y_N = 1, \{Y_i= 0\}_{\{i < N\}})\right]&=\sum_{n=1}^{40} \frac{12}{52-n} E\left[1_{\{N=n\}}\right] \\&=\sum_{n=1}^{40} \frac{12}{52-n} \mathbb P(N=n)\;. \end{align}

I don't feel sure that I've addressed your question, but if not, hopefully I've at least provided a basis on which we can address it further.