This problem from ISI MStat 2018 PSB (Problem 6) is called the Elchanan Mossel's Dice Paradox. The problem has a paradoxical nature, but there is always a way out.
A fair 6 -sided die is rolled repeatedly until a 6 is obtained. Find the expected number of rolls conditioned on the event that none of the rolls yielded an odd number.
Let \(X_{1}, X_{2}, \cdots\) be the throws of a die. Let
$$
{T}=\min\{{n: X_{n}=6}\}
$$
Then \(T\) ~ Geo(\(p =\frac{1}{6}\))
But, here it is given that none of the rolls are odd numbers. So,
$$
{T}=\min\{{n: X_{n}=6} | X_n = \text{even}\} = \min\{{n: X_{n}=6} | X_n = \{2, 4, 6\}\}
$$
Then \(T\) ~ Geo(\(p =\frac{1}{3}\)) Since, there are three posibilities in the reduced (conditional) sample space.
So, \(E(T) =3\).
Obviously, this is false. But you are not getting why it is false right? Scroll Down!
It went wrong in observing the given condition of the problem. Observe that it is given that none of the rolls are odd till the roll you got success, not for all the rolls beyond that also.
So, $$
{T}=\min\{{n: X_{n}=6} | X_n = \text{even}, n \leq T\} = \min\{{n: X_{n}=6} | X_n = \{2, 4, 6\}, n \leq T\}
$$
So, we are essentially seeing that the sample space didn't get reduced all along, it got reduced till that point of the roll. This is where the paradox marches in.
We are thinking of the experiment as we are picking up only \( \{ 2, 4, 6\} \) in the experiment and rolling. No!
The idea is to think from a different perspective as with the case of every elegant solution. Let's reconstruct the experiment in a different way. It is like the following. Remember, we need to exclude the odd numbers, so just throw them away and start anew.
Idea
If you get \(\{1, 3, 5\}\), start counting the number of rolls again from the beginning. Stop when you get 6. This is the exact formulation of the waiting time to get a 6 without getting any odd numbers till that toss. We will show that our success is when we get \(\{1, 3, 5, 6\}\) in this experiment.
Mathematical Form
Let \(\tau\) be the time required to get an outcome different from \(\{2,4\}\) Then \(E(\tau | X_{\tau}=j)\) is independent of \(j\) for \(j \in \{1,3,5,6\}\) because it is same for all \( j\). Thus, by the smooting property of \(E\left(\tau | X_{\tau}=j\right)=E(\tau)\).
Observe, \(\tau\) ~ Geo( \( p = \frac{4}{6}\)). Hence, \( E(\tau) = \frac{3}{2}\).
\(T=\min \{n: X_{n}=6\}\)
We need to calculate \( \mathbb{E}(T | X_{1}, \cdots, X_T \in \{2,4,6\})\).
For that we need to find out the conditional probabilities \(\mathrm{P}\left(\mathrm{T}=\mathrm{k} | \mathrm{X}{1}, \cdots, \mathrm{X}{\mathrm{T}} \in{2,4,6}\right)\) and that is given by
$$
\frac{\mathrm{P}\left(\mathrm{T}=\mathrm{k} \cap\left(\mathrm{X}_{1}, \cdots, \mathrm{X}_{\mathrm{T}} \in \{2,4,6\}\right)\right)}{\mathrm{P}\left(\mathrm{X}_{1}, \cdots, \mathrm{X}_{\mathrm{T}} \in \{2,4,6\} \right)}=\frac{\mathrm{P}\left(X_{\mathrm{k}}=6, \mathrm{X}_{1}, \cdots, \mathrm{X}_{\mathrm{k}-1} \in \{2,4\} \right)}{\mathrm{P}\left(\mathrm{X}_{1}, \cdots, \mathrm{X}_{\mathrm{T}} \in \{2,4,6\} \right)}=\frac{1}{6}\left(\frac{1}{3}\right)^{\mathrm{k}-1} \frac{1}{\alpha}
$$
where \(\alpha=\mathrm{P}\left(\mathrm{X}_{1}, \cdots, \mathrm{X}_{\mathrm{T}} \in \{2,4,6\} \right)\) . Thus \(\mathrm{T} |\left(\mathrm{X}_{1}, \cdots, \mathrm{X}_{\mathrm{T}} \in \{2,4,6\} \right)\) follows a geometric distribution with parameter \(\frac{2}{3}\) and consequently its expectation is \(\frac{3}{2}\).
Stay Tuned! Stay Blessed!
import random
times = 0 #number of times a successful (all-even) sequence was rolled
rolls = 0 #total of all number of rolls it took to get a 6, on successful sequences
curr = 0
alleven = True
for x in range(0, 100000):
num = random.randint(1,6)
if num % 2 != 0:
alleven = False
else:
if num == 6:
if alleven:
times += 1
rolls += curr + 1
curr = 0
alleven = True
else:
curr += 1
print(rolls * 1.0 / times)
#1.51506456241
Source: mathstackexachangeStay Tuned! Stay Blessed!
This problem from ISI MStat 2018 PSB (Problem 6) is called the Elchanan Mossel's Dice Paradox. The problem has a paradoxical nature, but there is always a way out.
A fair 6 -sided die is rolled repeatedly until a 6 is obtained. Find the expected number of rolls conditioned on the event that none of the rolls yielded an odd number.
Let \(X_{1}, X_{2}, \cdots\) be the throws of a die. Let
$$
{T}=\min\{{n: X_{n}=6}\}
$$
Then \(T\) ~ Geo(\(p =\frac{1}{6}\))
But, here it is given that none of the rolls are odd numbers. So,
$$
{T}=\min\{{n: X_{n}=6} | X_n = \text{even}\} = \min\{{n: X_{n}=6} | X_n = \{2, 4, 6\}\}
$$
Then \(T\) ~ Geo(\(p =\frac{1}{3}\)) Since, there are three posibilities in the reduced (conditional) sample space.
So, \(E(T) =3\).
Obviously, this is false. But you are not getting why it is false right? Scroll Down!
It went wrong in observing the given condition of the problem. Observe that it is given that none of the rolls are odd till the roll you got success, not for all the rolls beyond that also.
So, $$
{T}=\min\{{n: X_{n}=6} | X_n = \text{even}, n \leq T\} = \min\{{n: X_{n}=6} | X_n = \{2, 4, 6\}, n \leq T\}
$$
So, we are essentially seeing that the sample space didn't get reduced all along, it got reduced till that point of the roll. This is where the paradox marches in.
We are thinking of the experiment as we are picking up only \( \{ 2, 4, 6\} \) in the experiment and rolling. No!
The idea is to think from a different perspective as with the case of every elegant solution. Let's reconstruct the experiment in a different way. It is like the following. Remember, we need to exclude the odd numbers, so just throw them away and start anew.
Idea
If you get \(\{1, 3, 5\}\), start counting the number of rolls again from the beginning. Stop when you get 6. This is the exact formulation of the waiting time to get a 6 without getting any odd numbers till that toss. We will show that our success is when we get \(\{1, 3, 5, 6\}\) in this experiment.
Mathematical Form
Let \(\tau\) be the time required to get an outcome different from \(\{2,4\}\) Then \(E(\tau | X_{\tau}=j)\) is independent of \(j\) for \(j \in \{1,3,5,6\}\) because it is same for all \( j\). Thus, by the smooting property of \(E\left(\tau | X_{\tau}=j\right)=E(\tau)\).
Observe, \(\tau\) ~ Geo( \( p = \frac{4}{6}\)). Hence, \( E(\tau) = \frac{3}{2}\).
\(T=\min \{n: X_{n}=6\}\)
We need to calculate \( \mathbb{E}(T | X_{1}, \cdots, X_T \in \{2,4,6\})\).
For that we need to find out the conditional probabilities \(\mathrm{P}\left(\mathrm{T}=\mathrm{k} | \mathrm{X}{1}, \cdots, \mathrm{X}{\mathrm{T}} \in{2,4,6}\right)\) and that is given by
$$
\frac{\mathrm{P}\left(\mathrm{T}=\mathrm{k} \cap\left(\mathrm{X}_{1}, \cdots, \mathrm{X}_{\mathrm{T}} \in \{2,4,6\}\right)\right)}{\mathrm{P}\left(\mathrm{X}_{1}, \cdots, \mathrm{X}_{\mathrm{T}} \in \{2,4,6\} \right)}=\frac{\mathrm{P}\left(X_{\mathrm{k}}=6, \mathrm{X}_{1}, \cdots, \mathrm{X}_{\mathrm{k}-1} \in \{2,4\} \right)}{\mathrm{P}\left(\mathrm{X}_{1}, \cdots, \mathrm{X}_{\mathrm{T}} \in \{2,4,6\} \right)}=\frac{1}{6}\left(\frac{1}{3}\right)^{\mathrm{k}-1} \frac{1}{\alpha}
$$
where \(\alpha=\mathrm{P}\left(\mathrm{X}_{1}, \cdots, \mathrm{X}_{\mathrm{T}} \in \{2,4,6\} \right)\) . Thus \(\mathrm{T} |\left(\mathrm{X}_{1}, \cdots, \mathrm{X}_{\mathrm{T}} \in \{2,4,6\} \right)\) follows a geometric distribution with parameter \(\frac{2}{3}\) and consequently its expectation is \(\frac{3}{2}\).
Stay Tuned! Stay Blessed!
import random
times = 0 #number of times a successful (all-even) sequence was rolled
rolls = 0 #total of all number of rolls it took to get a 6, on successful sequences
curr = 0
alleven = True
for x in range(0, 100000):
num = random.randint(1,6)
if num % 2 != 0:
alleven = False
else:
if num == 6:
if alleven:
times += 1
rolls += curr + 1
curr = 0
alleven = True
else:
curr += 1
print(rolls * 1.0 / times)
#1.51506456241
Source: mathstackexachangeStay Tuned! Stay Blessed!
