Question 1178071: Roll a die once. Then roll it as many times as the outcome from the first roll. Getting the special number "3" on any roll means a win. What is the expected number of wins from this experiment?
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem step by step.
**Understanding the Experiment**
1. **First Roll:** Roll a six-sided die once. Let the outcome be X. X can be 1, 2, 3, 4, 5, or 6.
2. **Subsequent Rolls:** Roll the die X more times.
3. **Wins:** A win occurs if a "3" is rolled in any of the subsequent rolls.
4. **Goal:** Find the expected number of wins.
**Calculating Probabilities and Expected Wins**
1. **Probability of First Roll Outcomes:**
* P(X = 1) = P(X = 2) = P(X = 3) = P(X = 4) = P(X = 5) = P(X = 6) = 1/6
2. **Probability of a Win in a Single Roll:**
* P(Win) = 1/6
* P(No Win) = 5/6
3. **Expected Wins Given the First Roll (X):**
* Let W be the number of wins.
* If X = 1, W ~ Bernoulli(1/6), E[W|X=1] = 1/6
* If X = 2, W ~ Binomial(2, 1/6), E[W|X=2] = 2 * (1/6) = 2/6
* If X = 3, W ~ Binomial(3, 1/6), E[W|X=3] = 3 * (1/6) = 3/6
* If X = 4, W ~ Binomial(4, 1/6), E[W|X=4] = 4 * (1/6) = 4/6
* If X = 5, W ~ Binomial(5, 1/6), E[W|X=5] = 5 * (1/6) = 5/6
* If X = 6, W ~ Binomial(6, 1/6), E[W|X=6] = 6 * (1/6) = 6/6 = 1
4. **Expected Number of Wins (E[W]):**
* E[W] = Σ [E[W|X=x] * P(X=x)]
* E[W] = (1/6 * 1/6) + (2/6 * 1/6) + (3/6 * 1/6) + (4/6 * 1/6) + (5/6 * 1/6) + (6/6 * 1/6)
* E[W] = (1/36) + (2/36) + (3/36) + (4/36) + (5/36) + (6/36)
* E[W] = (1 + 2 + 3 + 4 + 5 + 6) / 36
* E[W] = 21 / 36
* E[W] = 7 / 12
**Therefore, the expected number of wins from this experiment is 7/12.**
Answer by ikleyn(52943)  (Show Source):
You can put this solution on YOUR website! .
Roll a die once. Then roll it as many times as the outcome from the first roll.
Getting the special number "3" on any roll means a win.
What is the expected number of wins from this experiment?
~~~~~~~~~~~~~~~~~~~~~~~~~~~
Regarding this problem, I have two notices.
First notice is that the problem's formulation is mathematically incomplete.
To be complete, it should say
"Getting the special number "3" on any roll means a win and stopping further rolling".
This correction is almost obvious, but it is important in the analysis.
Second notice is that the solution in the post by @CPhill is INCORRECT.
It is INCORRECT, since there are errors in his analysis.
Below I will show these errors, but I will not provide a whole solution with complete corrections.
Why ? - - - - Because @CPhill is a pseudonym for the artificial intelligence,
and this solution belongs not to @CPhill, but the artificial intelligence.
So, for AI it will be just a great benefit to get my pointing to their error.
I myself have no any desire to work here for or instead of AI.
So, I will copy-paste here the part of the @CPhill's text with pointing the errors.
3. **Expected Wins Given the First Roll (X):**
* Let W be the number of wins.
* If X = 1, W ~ Bernoulli(1/6), E[W|X=1] = 1/6
* If X = 2, W ~ Binomial(2, 1/6), E[W|X=2] = 2 * (1/6) = 2/6 <<<---=== error. The winning "3" can be obtained at
the first of the two rolls (and then the game stops),
or at the second roll. It should be correctly counted.
* If X = 3, W ~ Binomial(3, 1/6), E[W|X=3] = 3 * (1/6) = 3/6 <<<---=== error. The case of getting X = 3 was just analyzed above,
and it was just led to a stop/break earlier.
* If X = 4, W ~ Binomial(4, 1/6), E[W|X=4] = 4 * (1/6) = 4/6 <<<---=== similar error: outcome "3" can be obtained in any of 4 rolls,
leading to stop. It should be accounted.
* If X = 5, W ~ Binomial(5, 1/6), E[W|X=5] = 5 * (1/6) = 5/6 <<<---=== similar error
* If X = 6, W ~ Binomial(6, 1/6), E[W|X=6] = 6 * (1/6) = 6/6 = 1 <<<---=== similar error.
So, this artificial intelligence should fix his brain accordingly and consistently.
////////////////////////////////////
Regarding the post by @CPhill . . .
Keep in mind that @CPhill is a pseudonym for the Google artificial intelligence.
The artificial intelligence is like a baby now. It is in the experimental stage
of development and can make mistakes and produce nonsense without any embarrassment.
It has no feeling of shame - it is shameless.
This time, again, it made an error.
Although the @CPhill' solution are copy-paste Google AI solutions, there is one essential difference.
Every time, Google AI makes a note at the end of its solutions that Google AI is experimental
and can make errors/mistakes.
All @CPhill' solutions are copy-paste of Google AI solutions, with one difference:
@PChill never makes this notice and never says that his solutions are copy-past that of Google.
So, he NEVER SAYS TRUTH.
Every time, @CPhill embarrassed to tell the truth.
But I am not embarrassing to tell the truth, as it is my duty at this forum.
And the last my comment.
When you obtain such posts from @CPhill, remember, that NOBODY is responsible for their correctness,
until the specialists and experts will check and confirm their correctness.
Without it, their reliability is ZERO and their creadability is ZERO, too.
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
Hello, @CPhill, don't you think, that it is just a time came to express your thanks to me
for my tireless work on finding, identifying, pointing and correcting your errors ?
Otherwise, I feel myself uncomfortably about your bad manners.
|
|
|
