Two standard dice are rolled. What is the expected number of 1's obtained? Express your answer as a common fraction.
I don't know the difference between "the expected value of 1's obtained" and the probability of obtaining at least one 1 on a 2 dice roll. However, I shall attempt to answer the latter:
The probability of NOT getting a 1 on the first die is: 5/6, and:
The probability of NOT getting a 1 on the second die is also 5/6. Then:
The probability of getting at least one 1 would be:
1 - [5/6 x 5/6] =1 - [25/36] = 11/36 =~30.56%
+2440 The expected value is the probability-weighted average.
There are two dice, each with a probability of (1/6) of rolling a one, so multiply the probability by two (2). (1/6) *(2) = (1/3)
The expected number of 1's is (1/3).
GA
+118609 Ginger, you have counted the probability of getting 2 ones twice, that is why your answer is too high :)
+2440 Hi Melody,
I calculated “the expected number of 1's obtained” on a single roll of two dice.
If this were a roll of one die, then the expectation would be (1/6). However there are two dice, so the expectation is weighted by a factor of two (2), giving 2*(1/6)=(1/3).
Though mathematical expectation and probability are closely related, they are not the same thing. One notable difference is a probability is always a value between zero (0) and one (1), and an expectation may be equal to any (real) number. This value depends on the quantity of interest and on the measured probability. Some probabilities do not have a mathematical expectation, but all mathematical expectations have at least one associated probability.
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My troll post, below, contrasts the mathematical expectation to that of a Binomial CDF.
GA
+118609 Ok Ginger.
I do not know enough about this to argue with you.
I suppose I should do some homework :)
+33615 Here’s another way of looking at it:
Suppose you roll the two dice, A and B, N times.
The probability of getting a 1 on A and not on B is (1/6)(5/6) → 5/36
The probability of getting a 1 on B and not on A is (5/6)(1/6) → 5/36
So the number of single 1s you expect to get is N*10/36
The probability of getting a 1 on both A and B is (1/6)(1/6) → 1/36
So the number of pairs of 1s you expect to get is just N*1/36. But for each pair you see two 1s.
This means that the number of 1s you would expect to see in N throws of two dice is:
(10/36 + 2*1/36)*N→ 12N/36 or N/3
The expectation on a single throw of the pair is therefore just 1/3.
If someone asked me to turn my back while they rolled two dice onto a table top, and then asked me how many 1's I would expect to see when I turned round to look at them, I would say that I expected to see none.
The likelyhood of there being either one or two 1's is roughly 30.56% so it's more than twice as likely that there will be no 1's as there being either one or two.
It will not matter how many times we repeat this, my answer will always be the same, none.
If though I was told that we were going to do this say 100 times, and on how many times would I expect my expectation to be wrong, well that's a different question.
+2440 If someone asked me to turn my back while they rolled two dice onto a table top, and then asked me how many 1's I would expect to see when I turned round to look at them, I would say that I expected to see none.
Really? That’s blòódy amazing! So then, if you watched the dice roll then you would expect to see ones every time. You must have telekinetic powers.
Perhaps, when you are not watching, you are unable to control your telekinetic powers to produce a one (1), and in fact cause the dice to roll to every number except a one. That is still impressive!
Of course, you might just have very low expectations. Are you pessimistic by nature?
If you are not using telekinetic powers and this still happens then maybe you’ve entered the quantum realm, where the very act of measuring something changes the result. This relates to the Heisenberg Uncertainty Principle. (I’d explain this, but I’m not sure how to dumb-it-down to your education and intelligence level.)
However, if you resist using your telekinetic powers, and leave the dice to roll naturally –without influence, then the mathematical expectation for rolling two dice and having at least one (1) “single pip” show on one die is (1/3). (Note: in case you didn’t know, a “single pip” means a “one” on a die.)
Anyway, this question is about mathematical expectation, not your personal expectations.
What you posted is a Binomial CDF. This is the Cumulative Distribution Function for a binomial probability, not mathematical expectation. The mathematical expectation does not change –no matter how many times you do this experiment. If it did change, then the Binomial CDF would also change.
It’s great to have (another) “erudite” idiot to troll! We genetically enhanced chimps truly enjoy trolling “erudite” idiots. If you stay around long enough, I’ll give you a name.
The expectation is 92%, that you will continue posting for at least six weeks. What’s the Binomial CDF for this?
My dog, Mr. Peabody, and cat, DC Copper, have reviewed my post, now they want to shoot craps. DC says he wants to test his telekinetic powers for rolling snake eyes. This should be entertaining!
GA
