Because in classical probability, the results of simultaneous throws and sequential throws are different. I believe that although these English texts can be understood, we need to consider more factors if we want to take a more rational view. I’m just making a suggestion.
1 Like
Hi @Chris.X
In classical probability, the result of throwing two dice whether you throw them simultaneously or one after the other is treated the same way mathematically. That’s because what matters is the set of all possible outcomes, not the physical order of throwing.
Feel free to correct me if I misunderstood your point.
yes, you misunderstood
1 Like
Could you please explain further?
Are these not independent events?
Could anyone explain how simultaneou and sequential throws are different in classical probability?
1 Like
I just got an explanation from AI:
https://chat.inceptionlabs.ai/s/c066e1d4-e4f7-439f-ad0e-f373497a3fd7
It was in Chinese, don’t worry about translating those.
Hi @Chris.X,
I see your point. Let’s consider two cases.
- In the first case dice are rolled sequentially or simultaneously, but dice are labeled (assuming that the dice are fair), like {\rm dice}_1, {\rm dice}_2. Here each outcome of the experiment is an ordered pair (i, j), where i, j \in \{1, 2, 3, 4, 5, 6\}, and the sample space is \Omega = \{(i,j)∣i,j∈ \{1,2,3,4,5,6\} \}, |\Omega| = 36. Each elementary outcome is equally likely P(\omega) = {1 \over 36}, for all \omega \in \Omega. So for any event A \subseteq \Omega we have P(A) = {|A| \over 36}. The event “sum is 10” contains all ordered pair with sum 10: A_{10} = \{(4, 6), (5, 5), (6, 4)\}. Therefore, P(A_{10}) = {3 \over 36} = {1 \over 12}.
- In the second case If you do not care about the order or the dice are not labeled, so the pair (4,6) is the same as (6,4), i.e. unordered pair \{4, 6\}. The sample space \Omega' = \{\{i,j\}∣ 1 \le i \le j \le 6\}, and |\Omega'| = 21. In this case probability measure is not uniform since some outcomes like \{4, 6\} come from two permutations (4,6) and (6,4), but some outcomes like \{ 5, 5 \} come from one permutation (5, 5). Therefore, P'(\{i, j\}) = \cases{{1 \over 18}, {\rm if}\ i \ne j \cr {1 \over 36}, {\rm if}\ i = j}. This defines P' correctly since \sum_{\omega' \in \Omega'} = {6 \over 36} + {15 \over 18} = 1. The event “sum is 10” contains all unordered pairs with sum 10: A'_{10} = \{ \{4,6\}, \{5,5\} \}. Therefore, P'(A'_{10}) = {1 \over 18} + {1 \over 36} = {1 \over 12}.
1 Like
Your raise query was actually raised by other learners too and it was advised to add an example framing for better understanding of the question.
@lucas.coutinho I think for another post you have replied to the learner about the same for this issue, I hope this question has been updated in the course.
Regards
DP
Hey all!
In the case of independent dice throws, there is no difference between sequential and simultaneous throws. As mentioned by @Alireza_Saei, the key factor is the set of possible outcomes, which remains the same since the dice throws are independent events.
@Chris.X I couldn’t access the explanation you sent. If there’s a specific aspect you are referring to that might cause a difference between sequential and simultaneous rolls within the context of how they influence probability, please feel free to elaborate so we can discuss it here.
@Deepti_Prasad this other post is referring to another question, which was replaced after further discussion with the team.
Cheers,
Lucas
2 Likes
Hi @conscell,
Thanks for such a good explanation with example, I understand the difference now. 
Okay Lucas I just wanted to make sure if this wasn’t persistent feedback from learners, so tagged your comment, so you are notified for clarification.