Now since you got into probabilities, I read that problem a few months ago. A game is set between player A and B in the following fashion:
1) The first player writes the numbers from 1 to 18 on three 6-sided dices (empty) without repetition.
2) The second player investigates the dices and chooses one.
3) The first player takes one of the two dices left and discards the last
4) Both players throw their dices. The one that shows higher number wins.
Question: Which of the players has better chance of winning. If you were player A would you go first or second and why?
Originally posted by ilywrinAt a glance.
Now since you got into probabilities, I read that problem a few months ago. A game is set between player A and B in the following fashion:
1) The first player writes the numbers from 1 to 18 on three 6-sided dices (empty) without repetition.
2) The second player investigates the dices and chooses one.
3) The first player takes one of the two dices left a ...[text shortened]... players has better chance of winning. If you were player A would you go first or second and why?
Player A when writing on the dice must weight the averages to be equal otherwise Player B will pick the highest average leaving Player A at a dis-advantage, this means player A weights the averages on all the dice as 19, i.e.
18, 1
17, 2
16, 3
Then it doesn't matter what is picked.
Player 1 might as well go 2nd in the hopes player 2 is a donkey and miss weights the averages and gives him an advantage. Or in poker terms, 'gives him a chance to make a mistake' while if he went first 'he'd be playing optimal making his opponent have 0 Expected Value'.
I think? But it seems a bit short.
Originally posted by ark13Read this. Hope this helps
Sorry, I know we already passed this, but can someone explain to me why you're twice as likely to get the car if you switch doors in the gameshow puzzle. I know there was a puzzle like this before but I'm too slow to understand what they were talking about.
http://exploringdata.cqu.edu.au/montyexp.htm
Originally posted by ilywrinSomething tells me that player 1's chance of winning is 2/3, but I'll have to work out a way of realising that.
But the whole point is that there exists a strategy that may give you more than 50% chance of wining. And [b]no hoping that the other player will make a mistake does not count as optimal startegy 🙂[/b]
Edit: Hmm, maybe not. But I can manage odds of 7/12:
Die A has 14 13 12 11 10 1
Die B has 17 16 15 4 3 2
Die C has 18 9 8 7 6 5
Die B beats die A with prob. 7/12
Die C beats die B with prob. 7/12
Die A beats die C with prob. 25/36
Originally posted by sarathianBoth are exapmples of elliptic integrals.
OK, TRY these two mathematical problems : -
(a) Evaluate the definite integral of SQRT( a + cos x ) .dx for x going from 0 to pi , where a > 1.
(b) Evaluate the definite integral of SQRT{ a^2 + (cos x) ...[text shortened]... meter a.
This is quirky and difficult and solvable.
You have just found a long-sought pair of ancient treasure chests. One chest is plated with silver, and the other is plated with gold. According to legend, one of the two chests is filled with great treasure, whereas the other chest houses a man-eating anaconda that can rip your head off. Faced with a dilemma, you then notice that there are inscriptions on the chests:
Silver Chest: This chest contains the anaconda.
Gold Chest: One of these two inscriptions is true.
Based on these inscriptions, which chest should you open?
Originally posted by THUDandBLUNDERSilver
You have just found a long-sought pair of ancient treasure chests. One chest is plated with silver, and the other is plated with gold. According to legend, one of the two chests is filled with great treasure, whereas the other chest houses a man-eating anaconda that can rip your head off. Faced with a dilemma, you then notice that there are inscriptio ...[text shortened]... hese two inscriptions is true.
Based on these inscriptions, which chest should you open?
Silver
(Since: If the silver inscription is true, then so is the gold - cannot be
If the gold is true, then it is the only true one, therefore the silver is false and does not contain the snake)
EDIT: Of course, the gold inscription may be true as well as the silver - depends on the interpretation. The gold inscription is NOT necessarily saying "ONLY one of these...." It could be that it means "At least one of these..." etc
Originally posted by AlcraSilver, eh? So you presumably believe that the anaconda is in the Gold chest.
Silver
(Since: If the silver inscription is true, then so is the gold - cannot be
If the gold is true, then it is the only true one, therefore the silver is false and does not contain the snake)
EDIT: Of course, the gold inscription ma ...[text shortened]... " It could be that it means "At least one of these..." etc
Firstly, assume that the Gold inscription means 'one, and only one, inscription is true':
If the anaconda is in the Gold chest, then the Silver inscription is false, and the Gold inscription is indeterminate (if it is true, then it is true; and if it is false, then it is false). Just as a contradiction is not the same as being false, this indeterminate situation is not the same as being true.
Secondly, assume that the Gold inscription means 'at least one of these statements is true':
If the anaconda is in the Gold chest, then the Silver inscription is false, and the Gold inscription is again indeterminate, not true or false.
.
1) Take two regular tetrahedra of unit side length and glue them together so that 2 faces exactly coincide. How many faces does the resulting solid have?
2) Take a regular tetrahedron and a square-based pyramid, with all sides unit length and glue 2 triangular faces together so that they exactly coincide. How many faces does the resulting solid have?
Originally posted by The PlumberAnd you're assuming that what's in the chests makes no difference.
You're assuming that what's written on the chests makes any difference............and open the silver chest.😲
Seems like you are too eager to get to grips with an anaconda's plumbing.
😛