Welcome to the Slashdot Beta site -- learn more here. Use the link in the footer or click here to return to the Classic version of Slashdot.

Thank you!

Before you choose to head back to the Classic look of the site, we'd appreciate it if you share your thoughts on the Beta; your feedback is what drives our ongoing development.

Beta is different and we value you taking the time to try it out. Please take a look at the changes we've made in Beta and learn more about it. Thanks for reading, and for making the site better!

Actually, you will generally know an upperbound on the length of whatever is encypted given the encrypted text. That is something, so it is not quite perfect (In theory anyway).

I think your point count against you./. is fairly large. It would be reasonable to apply the same standards as for a large business. Hence, you should standardize your communication. In this case that means English and metric. It is ok that you use it amongst yourself but it would be nice of you to try to keep it only amongst yourself:) The standard custom in Europe is more or less that if someone wouldnt understand otherwise, you speak English (at least for the larger places I have worked). Similarly, it would be nice if you wrote in metric if people wouldn't understand otherwise (which I think it can be assumed that some wont on a page as big as/.).

I personally do not mind yards and feet too much but I dislike miles since it depends on the country.

The question isnt how much you bet against his bet but how much you are willing to pay him to play a given game with a fixed value for the outcomes (i.e winning losing and tieing). Note this is different from a bet in exactly the way you mention with the 3rd player, except that you pay your oppeont and not someone else - i.e you lost your payment even if you tie.

If you win 75% and him the rest and you get 1 for win 0 for tie and 1 for lose, you get 1 with 75% chance and lose 1 with 25% chance (0.5 on average) and should then be willing to give him upto 0.5 before playing the game because you then come up even (equvialently the game is fair/even if you get 0.5 for winning (1 minus the 0.5 you paid upfront), -0.5 for tieing and -1.5 for losing). Similarly if you pay him 0.5 and you win 1 with 50% chance and tie and get 0 with the remaning you are still even.

Game theory theories are math theories and not physics theories. Building on some given assumptions (i.e. what we measure is what you are willing to pay upfront) the theories are correct (if we look away from posible errors in the proofs - that said this is von Neumanns normal form games we are looking at right now and the proofs are correct under the assumptions used).

Not really:) The idea is that you subtract his winnings from yours (in the game you get the same for winning as he get) and not divide them. And both 1/2-1/3 and !/3-1/6 is equal to 1/6. Note that you do not replay if you both get the same but the game is instead a draw (this is not explicitly mentioned as far as I can see, but it is what is meant). Replaying the game makes it somewhat more complicated (which is why he must mean draw in case of ties - he seems to want an analytical solution you can do by hand). An exact solution in case of replaying on ties can be found by modeling the game as a stochastic game ala. Everett (you need to define what happens if you keep on replaying infinitely many times though) and then solve it fast using some algorithm in Kristoffer Arnsfelt Hansen, Michal Koucký, Niels Lauritzen, Peter Bro Miltersen, Elias P. Tsigaridas: Exact Algorithms for Solving Stochastic Games. I am currently writing a paper on an alternate, simpler way of solving such games, but it is not done yet:(

Say you increase to paper 2/3 - x and rock 1/3 + x. Then against paper 50% you win 1/3 - x/2 of the time and he wins 1/6+ x/2 of the time. Hence you get 1/6 - x on avg. On the other hand against scissors 50% you win 1/2 of the time (after having picked yours he got one option which is losing and he play each with 50%) and he wins 1/3 -x/2 of the time for 1/6-x/2. Finding the x that maximizes the smallest is easy. It is x=0. Note that you get 1/6 against the strategy I mentioned for the other guy because you do not do anything really stupid (= scissors). Also, note that maximizing the smallest is the most important thing if he is smart enough to figure out your strategy (since he will answer with the strategy leading to the smallest nr.)

This is right yes:) Well, assuming that you know that he is playing 50% rock (and not more - that is btw. right - he would worse off if he played rock with even higher probability). Also, to be truly formal about it you should argue that it is a maximum and not a minimum you found:)

His nash strategy (minimax strategy) is to play rock 1/2, scissors 1/3 and paper 1/6 (it is true that rock 1/2 and scissors 1/2 is a best response to the nash strategy, but that does not mean it is optimal against abitrary strategies). I have a long earlier post showing the strategies and so on:)

Ok lets see: you play 2/3 paper (I shorten the fraction I hope that is ok:) ) and 1/6 scissors and 1/6 rock. You play against the strategy 1/2 rock, 1/3 scissors, 1/6 paper. Fast version: lets look at a random round in which you play rock: In those you win 1/3 against his scissors and lose 1/6 against paper, thus you get 1/6 on avg. Next, random round in which you play paper: In those you get 1/2 against his rock, and lose 1/3 against his scissors, i.e. again you gain 1/6. Next, random round in which you play scissors: In those you get 1/6 against his paper and lose 1/2 against his rock, i.e. you LOSE 1/3. On avg you play rock 2/3 of the time and get 1/6 in those rounds, scissors in 1/6 of the time and LOSE 1/3 and paper 1/6 of the time and get 1/6. Thus, on avg. 2/3*1/6+1/6*(-1/3)+1/6*1/6=1/12. This is below your lower bound so there is something wrong with it. (the reason is that you lose whenever a bit on avg. whenever you do not play paper).

My strategies, played against each other gives 1/6. Thus, you can not say that yours is better always. I can argue that against ANY strategy mine gets 1/6. You can not get better than 1/12 (because you get that against mine strategy for the other player). Thus, yours can not be optimal sorry:(

"you" are not the guy playing rock 50% of the time. "you" are the guy beating on the poor guy playing rock 50% of the time. The optimal choice is to play rock 1/3 and paper 2/3 (his is to play rock 1/2 and paper 1/6 and scissors 1/3).

Since I already explained the optimal solution to the basic question mentioned in the summery lets solve the bonus question too (my solution also matches the solution given in comments on the article side so it should be good (and said to be correct by the author) - note currently no answer with a high score is correct - mine has 1).

The bonus question is that you play two rounds, and your oppoent must play atleast rock once. So, if he plays something not rock in the first round he must play rock in the second and loss (you just play paper). If he plays rock in the first he can play 1/3 all in the second (which leads to a draw like normal). Thus, if he plays rock first it is like normal RPS (because he get 0 in the next). Otherwise you get one free win (for the second round).

Thus, we can model the first game of the bonus question as (where the numbers is the number of rounds he wins on avg given the choice in round 1):
R P S R 0 -1 1 P 0 -1 -2 S -2 0 -1

Where you pick columns and him rows. We see that rock dominates paper for the row player. We get

R P S R 0 -1 1 S -2 0 -1

For the column player, the choice of rock now dominates scissors. We get

R P R 0 -1 S -2 0

Playing rock 1/3 and paper 2/3 for the collumn player gives -2/3 wins on avg. Similarly, the row player can get -2/3 wins on avg by playing rock 2/3 and scissors 1/3.

There is a flaw in your reasoning. You do not know that your oppoent flipped so you can not condition on it like you do here (you can not play paper all the time if he "flips" rock because you do not know his coin flip). If you think about it you should NEVER play scissors. In the best case for you he plays rock 50% and paper 50% and you get 0 in expectation and clearly you got an advantage so 0 is not good.

The optimal strategy is to play 1/3 rock, 2/3 paper. It gives at least 1/6 against anything he could play. He can similarly ensure that you can not get more than 1/6 a game by playing rock with probability 1/2, paper with probability 1/6 and scissors with probability 1/3. Your strategy would get less than 1/6 against that (more precisely, you get 1/6 if you play either rock or paper and you lose 1/3 if you play scissors. Therefore you get 1/6*5/6-1/3*1/6=1/12 which is less than the 1/6 you get for playing 1/3 rock and 2/3 paper).

You can get an advantage. The important point is to notice that you should not play scissors ever. You can only get 0 in expectation IF he plays paper 50% and rock 50% and he gets an advantage otherwise and 0 is not good for you:/ See my above post for further details (spoiler: The optimal choice for you is 1/3 rock, 2/3 paper).

Sorry:/ There are some mistakes in the last part. The strategy for player 1 wins 1/2-1/6=1/3 and not 1/6 as claimed. Also, the strategy for player 2 wins 2/3 against pure rock and not 1/3 as claimed. Still, it just makes it even clearer that you should not play rock with probability more than 50% as player 1 and not play scissors at all.

Also, to be more precise, the strategy for player 1 is to play rock with probability 1/2, paper with probability 1/6 and scissors with probability 1/3.

Lets call the guy with the restriction player 1 and the other player 2. If you think about it player 1 got 3 "pure" strategies (as in: each other strategy he can play can be seen as a mixture of these 3): (1) rock 100%, (2) paper 50%/rock 50% and (3) scissors 50%/rock 50%. Against (1) rock gives 0, paper -1 and scissors 1. Against (2) rock gives 1/2, paper -1/2 and scissors 0. Against (3) rock gives -1/2, paper 0 and scissors 1/2. In each case, the number is the probability of player 1 winning minus player 2 winning.

We see that player 2 should not play scissors, because he will never gain anything from it and clearly the optimal strategy should be able to gain something (we will see that it is 1/6). Then, knowing that, paper 50%/rock 50% is better than rock 100% for player 1: If player 2 plays rock, player 1 gets 1/2 (instead of 0) and if player 2 plays scissors player 1 gains -1/2 (instead of -1).

Hence, we are down to: (2) paper 50%/rock 50% and (3) scissors 50%/ rock 50% vs. rock and paper. If player 1 plays (2) with pr. 1/3 and (3) with probability 2/3, he loses 1/6 against both rock and paper. If player 2 plays rock with probability 1/3 and paper with probability 2/3 he gets 1/6 against (1) and 1/6 against (2). This is optimal since each player have a way to guarantee that player 1 loses 1/6 and player 2 wins 1/6. If either had a better strategy it would break the other players guarantee (note that the given strategy for player 1 wins 1/6 against scissors, again showing that it is a bad strategy for player 2 and player 2's strategy wins 1/3 against rock 100% showing that it is a bad strategy).

They actually do not need to do any framing or anything. They simply give up all the files Snowden got at one time, everybody gets angry at them and in a month the furor over all this will have died down. Currently people are getting annoyed at them whenever Snowden releases a new file and that can continue for a long time yet, which seemes way worse for them. Sort of the real version of the boiling frog. In reality it jumps out when it gets too hot, but it might not if it gets warm only for a short time.

Lyx gives you both options. E.g. I want to use the fancy macro I made previously, so I write \macroname like I would in your latex by hand. I need to take the union of two sets and can't remember the syntax (it is \cup or \bigcup but this is a example - I could write \bigcup, but this is a example) so I find it in the menu...

The real strength, to me, is that per default you see the equations as they would look in print (and not like \frac{stuff}{other stuff}). It is (to me atleast) alot easier to read 10 lines of inequations using symbols with 1 super script and 2 subscribts if I can see them as the way they are in the output instead of unstructed text. It is not really a solution to have output at the side, since you have to look back and forth between them.

LyX bad sides are that it cant convert that well from pure latex and that the algorithms (I am a ph.d. student in comp. sci.) are not the best looking.

He might be. The question does not say one way or the other.

The idea is that there are 198 posibillities for two children and the day they are born (as in: the first is a boy on a monday the second is a girl on sunday - there are 198 different statements like that). If you go through it you will see that excatly 27 of them contains a boy on a tuesday. Of those 27, 13 had two boys. Each of the events are equally likely - well not in the real world, but it would require a bit more info otherwise. So you end up with 13/27.

A non-numerical version goes as follows. To easen the reading I will use unordered pairs:

There are three sets. A set where the boy on a tuesday is born first, F, one there he is born second, S, and one where there are two boys on a tuesday, B. The S and F are equally large and the number of girl boy pairs are equally large in both (similary for boy boy pairs), but they contain a slightly larger probabillity for boy, girl pairs over boy, boy pairs(because neither S nor F contains two boys on a tuesday - they are in B - but they DO contain the posbillity that it is a boy on a tuesday and a girl on a tuesday). The difference is excatly such that if you add B to e.g. F the numbers of boy, girl pairs are equal to the number of boy, boy pairs. Therefore you will have a slightly larger probabillity for girl, boy pairs (because the number of boy, boy in F+B is equal to the number of boy, girl in F+B, but the number of girl, boy pairs are slightly larger than the number of boy, boy pairs in S).