[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 136
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Toss two coins. If two eagles come out, throw another third.
How many tails on average will be on the table after each round with this strategy?
// No catch, two or three coins on the table after each round. We count, write down the result, then another round. And so on.
The second one is more difficult.
Roll two dice. If you get a double (two equal numbers of dice), roll another one.
How many points are on the table on average after the round?
Second question to this problem: if we add the number of dice on the table to the number of points, we get X. What is the average value of X ?
1/4*0 + 1/4*1 + 1/4*1 + 1/4*0.5 = 5/8. That's tails.
Откуда, откуда... из Перельмана, не знаете, что ли...
Просто это, похоже, чуть ли не первая задачка здесь от Richie, которая так мощно всколыхнула общественность. Ну, конечно, не считая задачи о какашках.
Ну понятно, что плотность в центре Земли побольше, и это будет влиять на движуху. Остальное типа температуры, давления на кирпич не действует.
So you think the timing will be "slightly different" ?1/4*0 + 1/4*1 + 1/4*1 + 1/4*0.5 = 5/8. Это решек.
Erm... that's not how it works for me. We have five outcomes:
1. Tails = 2.
2. = 1
3. = 1
4. = 0
5. = 1
The average p = 1.
Who is right?
Я кажись догадался про кирпич, часовые пояса вот!
!!!!!!!! -))))))
Кто прав?
Hee hee. They're both wrong. I totally lied, of course, but Lech was lying too.
The correct formula is: 1/4*2 + 1/4*1 + 1/4*1 + 1/4*0.5 = 1/2+5/8 = 9/8.
// I'll count the cubes tomorrow. That's it. Thanks.....
Yeah, 9/8. My wife threw me out in the cold for a soda, but she got her way, didn't let me think it through. During the hike, I realized what a gibberish I'd written.
The funny thing is, there's something that confuses me, too. Why do heads-up (OR) and tails-up (RO) count as different outcomes? Well... it just seems that way to me; I don't know any other arguments yet :)
I mean, why not: 1/3*2 (two tails) + 1/3*1 (heads and tails) + 1/3*0.5 (two heads and half of tails) = 7/6?
That is, it appears as if the trials are not modelled as simultaneous tossing of two dimes, but independent tossing of one each "with a slight delay" to somehow distinguish between OR and RO sequences.
The reality proves this "virtual delay": if 10 tiles are tossed simultaneously, from the thrower's point of view there are only 11 outcomes (from 0 to 10 tails), which seem to be equally probable because for him they are atomic, indivisible. However, a sophisticated terver will immediately tell us about Bernoulli's scheme, according to which the probability of 0 tails is much less than 1/11, the probability of 5 tails is the highest, and 10 tails is again very low.
There is another point: the sequence RRRRRRRR is equal to RO, although the former has 10 tails and the latter has 5. The unequal probability of 10 and 5 tails arises only because the sequence with 10 tails is unique, i.e. it is rare, while the sequences with 5 tails are numerous, 10*9*8*7*6/5! = 252, i.e. they are not rare at all.
How about heat?
Here's a simple problem:
Two identical vessels are put on two identical burners at the same time. One has 1l, (N1) the other has 0.5l (N2) of water of the same temperature. At the moment the half-litre boils, another half-litre of water of the original (preheating) temperature is poured in there. The burners are put out. In which vessel will the temperature be higher?