@boboquack X2 is the event "the dice takes 2 rolls to get the first 6, given that the rolls are even". You then claim P(X2) = P(A2 | B2), which is not true since in the case where the first roll is a 6 the second roll is not guaranteed to be even
To make more explicit what was wrong with your earlier answer @bobo: assume that we continue rolling the dice after we get a 6, but now we don't care whether we get odds or evens. Clearly nothing changes about the expected value the problem asks about, that is, we don't actually care what happens after we roll the 6. But by saying that "we rolled two evens" was a given, you were caring about what happened after the 6, since a sequence such as 6 1 would have been ruled out.
Here is an amazing probability puzzle that is currently going viral in the discrete mathematics community:
You throw a (single) die until you get $6$. What is the expected number of throws (including the throw giving $6$) conditioned on the event that all throws gave even numbers?
well the easy explanation is: chance is geometric with r = 2/3 s_n = (2/3) / (1 - 2/3) = (2/3) / (1/3) = 2 chance with starting value of 1/3 = 2 * 1/3 = 2/3 rolls = 1/(2/3) = 3/2
@ASCII-only , that is wonderfully intuitive. My take on what you said: If it is geometric, the factor must be 2/3 because that is the ongoing probability of continuing the streak.
Someone (willingly) takes part in an experiment. They are told the rules of the experiment:
They will be put to sleep. A coin will be flipped. If heads, they will be woken up on Monday and Tuesday, for a total of two wake-ups. If tails, they will be woken up on Monday only.
Now, each time they get woken up, they will be asked if they believe the coin was heads/tails and then they will be given something to make them forget that they had woken up. Then they will be put to sleep again.
The question is what probability the subject of the experiment should assign to the proposition "The coin came up heads". [EDITED to fix a tense error]
Note that in many probability questions you can translate that question into something like "What odds should you be prepared to bet at, so as to end up not losing money whichever side of the bet you take?" but it's not so clear that that translation is the right one here.
If you do the experiment many many times, then the coin comes up heads on half the coin-flips, but two thirds of the times you get asked the question (because you get asked it twice when it's heads)
It may be worth saying that unlike the Monty Hall problem, where there is a single right answer that everyone competent ultimately agrees on but it's easy to get it wrong at first, here you can find highly competent people arguing for (1) p=1/2, (2) p=2/3, and (3) there is no fact of the matter as to what p is. It's more a philosophy-of-probability question than a probability question.
@ASCII-only let's consider another experiment: if the tails land, you get woken up on Monday. If the heads land, you don't get woken up at all. Now, clearly both outcomes are equally likely. But, what should you bet on when you get woken up?
So, if your strategy is to choose one option with certainty, it doesn't matter how many times you wake up. If you chose the wrong answer the first time, you would choose the wrong answer again the second time.
However, if when waking up in the morning, you flip your own coin and claimed its result, you would have a 5 in 8 chance of being correct.
CCCC: Member with unworldly enthusiasm earlier harbored primarily certain feeling (10)
(I wanted to say "enthusiasm for the unworldly", but then "for the" would have been unused. But I think this still works, @Sp3000 can opine after it's solved)
The only scenario I can think of in which this comment makes any sense is where "unworldly enthusiasm" and "enthusiasm for the unworldly" are not equivalent - that is, not used for their mutual meaning, but rather where (at least) one of the two words is an indicator. The likely indicator is then 'unworldly', with a subtraction sense, as noted by others. But if that's what it is, "world" cannot be subtracted literally from surrounding tex, making it indirect either in that something synonymous with "world" is subtracted instead, or in that "world" is subtracted from something synonymous wi…
(Literal meaning of "enthusiasm", by the way, is "having a god in you". Unworldly indeed. I make no claim that this has anything to do with the current clue.
@Rubio Youu're reading too much. You weren't happy about an extraneous "are"; I didn't think you would be happy with an extraneous "for" and an extraneous "the"
Heres a sequence of numbers:
133, 173, 30, 133, 176, 160, 171, 63, 137, 176, 160, 60, 177
I'm sure they are an important part of a puzzle i'm trying to solve which i found hidden in a image - but i cant identify the cipher used.
Does anyone here have any ideas?
Here is an amazing probability puzzle that is currently going viral in the discrete mathematics community:
You throw a (single) die until you get $6$. What is the expected number of throws (including the throw giving $6$) conditioned on the event that all throws gave even numbers?
Heres a sequence of numbers:
133, 173, 30, 133, 176, 160, 171, 63, 137, 176, 160, 60, 177
I'm sure they are an important part of a puzzle i'm trying to solve which i found hidden in a image - but i cant identify the cipher used.
Does anyone here have any ideas?
