24 Oct 2023

Posted on:

24 Oct 2023

0

Hello, I have a question, in the experiment of throwing 2 dice simultaneously (that is, you cannot distinguish between dice 1 and dice 2) the sample space is:

(1, 1)(1, 2)(1, 3)(1, 4)(1, 5)(1, 6)(2, 2)(2, 3)(2, 4)(2, 5)(2, 6)
(3, 3)(3, 4)(3, 5)(3, 6)(4, 4)(4, 5)(4, 6)(5, 5)(5, 6)(6, 6)

My question is whether we are faced with equiprobable events.
I think so, but I would like you to give me feedback.
Regards and thank you very much in advance

Instructor
Posted on:

24 Oct 2023

1

Hi Enrique,

When you roll two dice simultaneously and you cannot distinguish between dice 1 and dice 2, the events in your provided sample space are indeed equiprobable. However, your sample space does not account for all possible outcomes.

The sample space you've provided is a condensed version that represents only distinct sums and combinations. It is missing combinations like (2, 1), (3, 1), (3, 2), etc., which are essentially the mirror images of some of the pairs you've already listed. The reason you've left these out is that, since you can't distinguish between the dice, (1, 2) is the same event as (2, 1) for your purposes. In other words, both represent the event "one die shows a 1 and the other shows a 2."

Given that you've treated (1, 2) as the same event as (2, 1) and so on, you have 21 distinct outcomes in your sample space. Each of these outcomes has the same probability:

P(any specific outcome) = 1 / 21

However, keep in mind that if you were distinguishing between the dice, there would be 36 possible outcomes, and each outcome would have a probability of 1/36. In that case, the probability of rolling a sum of 3 would be the sum of the probabilities of (1, 2) and (2, 1), or 2/36 = 1/18. In your non-distinguishing case, the probability of "rolling a 1 and a 2 in some order" is 1/21.

So yes, in your provided sample space where you do not distinguish between dice, the events are equiprobable. Each of the 21 outcomes has an equal chance of happening.

Best,

Ned