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A Rock Paper Scissors Brainteaser
New submitter arsheive (609065) writes with a link to this interesting RPS brainteaser: "How do you play against an opponent who _must_ throw Rock 50% of the time, and how much would you be willing to pay to play against them?"
The odds are even odds. If he must throw rock 50% of the time, and he knows that you know that data, the other 50% of the time he will throw scissors to beat your paper....
We were all warned a long time ago that MS products sucked, remember the Magic 8 Ball said, "Outlook not so good"
You tie 50% of the time, win 25% of the time, and lose 25% of the time. Pay? Nothing.
100% paper strategy will win 50% of the time. Of the remaining 50% of games played, (assuming even distribution of the remaining picks) 25% will be losses and 25% will be tied. Thus, you'd be assured a win-loss-tie ratio of 2-1-1, which is quite good. If their remaining options are not distributed evenly, this changes things. You'd want to look at their play to see whether there are any discernable patterns, such that you know that Rock will be played for certain every other move, for example. Then you just sync Paper moves to their Rock moves, and play Scissors or Rock randomly for the other half.
You see? You see? Your stupid minds! Stupid! Stupid!
With no such restriction, random choices on both sides lead to 33% win, 33% draw, 33% loss, right? With the opponent throwing Rock 50% of the time, assuming the other 50% is evenly divided between Paper and Scissors, if I always throw paper I'll win 50% of the time, lose 25% of the time, and draw 25% of the time.
So depending how the betting works, I'd be pretty willing.
Expect them to play scissors a lot to beat your paper. Play rock as often as they play scissors.
Rock, paper, scissors, lizard, Spock!
Scissors cut paper
Paper covers rock
Rock crushes lizard
Lizard poisons Spock
Spock smashes scissors
Scissors decapitate lizard
Lizard eats paper
Paper disproves Spock
Spock vaporizes rock
Rock crushes scissors
Political debates have me rolling my eyes so much I think I got optical whiplash. I should sue. - Foamy The Squirrel
I think the winning strategy is to randomly throw 50% paper to cover his rock. I'm just guessing though. No idea how much to pay.
For all intensive purposes, "whom" is no longer a word. That begs the question, "who cares"?
If you actually enjoy this question, watch the Kaiji Anime or read the manga, you won't be disappointed.
If you don't, run far away from it.
How would you measure 50%? When does the game end? If my opponent strictly goes by 50 25 25 then Yes, I'd be happy yo play for an infinite amount of money, however given the play I am going to be playing, my opponent may opt/adapt for 50 0 50 instead and we are stalemate again .. So the game is rigged, conditions flawed, how would you measure those 50%?
Oooh, not rock again!!!!
Assuming we randomly use 50% paper and 50% rock, we get:
- on rounds he is forced to play rock, we get half victory and half tie, so 25% win, 25% tie so far
- on other rounds he can:
- always use scissors, which will turn into our 25% win, 25% lose - we win overall, 50% win, 25% tie, 25% lose
- always use rock, which will turn into our 25% tie, 25%win - we win overall, 50% win, 50% tie
- always use paper, which will turn into 25% win, 25% lose - we win overall, 50% win, 25% tie, 25% lose
- any kind of random mix of above, which can only move between results above, as our picks are completely random as well
So, he best strategy is to always play scissors or rock, for 50:25:25 result. If ties are repeated, we win 2 out of 3 times.
Any better strategy?
I've been having students in my introductory programming courses work on this class of problem for a few years.. They all seem to really enjoy it. I code up bots to play RPS with certain biases just like the OP and they have to program a single player that identifies the bias in an opponent and adjusts its play to give it an advantage. They all routinely can generate solutions that perform far better than random against predictable, dumb bots, but things get very interesting when I throw the students' bots against each other in a throwdown tournament. :)
If i know the number of rounds in advance (atleast your oponent has to right if he needs to calculate the 50%). I would recalculate the odds after every round taking into account the number of rock move remaining for him...
Like drawing to an inside flush, an "optimized" strategy is not necessarily what the opponent plays. There is no reason inherent in the description to make assumptions about the opponent's other play. They may also be constrained to play paper the other fifty percent of the time, and to play paper , then rock, then paper, then rock. In the real world, don't assume that the minimal description of the problem gives all the important data.
there is a factual problem with the summary.TFA says it all so better read it. If not read this (I hope not to mess up it too much).
It is not required of the opponent to play rock 50% of the time. The referee is using a fair coin to determine if your opponent is to play rock or not. If s/he is not forced to play rock s/he is free to chose allowing him also to chose rock 100% of time too if s/he so wishes.
"Never go in against a Sicilian when death is on the line!"
If the opponent plays P or S on the first move, his second move is determined, so there's a greater than random chance he will want to play R first. So the safest thing is to play paper twice. Or: paper first, and if the opponent played rock, then play anything at random.
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).
Over the long term, the strategy must converge to stable, therefore true random can be the only optimized strategy.
50% of time opponent must play R. The remaining 50% of the time they can equally choose R,P,S.
R -> P = 1/2 + 1/3 * 1/2
S -> R = 1/3 * 1/2
P -> S = 1/3 * 1/2
For a guaranteed win, roll a die: 1-4 => P, 5 => R, 6 => S.
By how much? Consider you are random as above and opponent is fixed wlog at 100% R. You win 2/3 of the time and lose 1/6 of the time.
The expected payoff to play is 2/3, the expected cost is 1/6. Payoff - cost = 1/2, so the most you should be willing to pay to play a $100 game is $150.
You should play paper 4/6 of the time, rock 1/6, and scissors 1/6 of the time.
The key (if you RFTA) is that whether or not your opponent plays rock is determined by a coin toss. So really you are playing a compound game. You are playing a coin toss and rock paper scissors (RPS). Since the coin toss determines your opponents move, you can think of it as playing 50% coin toss and 50% RPS. The RPS is a subgame of the coin toss.
Since the coin toss is the dominate game, you play with win that first. But instead of heads/tails, it is paper/other. The answer to the coin toss is a 50/50 guess of heads/tails, so the answer to the paper/other is 50% paper, 50% other.
The "other" is the RPS game. And since the answer to the RPS game is 1/3 rock, 1/3 paper, 1/3 scissors, we know what the solution to the other 50% of the game is.
So the equations are:choice = (Coin Toss) + (RPS) so: paper = 1/2 + 1/3, rock = 0 + 1/3, scissors = 0 + 1/3. Or paper = 4/6, rock = 1/6, scissors = 1/6.
The payoff is 16 2/3.
The defender should do 1/3 paper 2/3 scissors. (When not forced to play rock.)
This is just a standard mini max problem. The payoff matrix looks like:
0 -50 50
100 50 0
-100 0 -50
(With row and columns being in RPS order.)
Nobody seems to be asking what the definition of "50% of the time" is ?
Over what interval/time frame?
If the opponent throws scissors or paper on his first move is he then required to throw rock on his second move?
First, make sure you read TFA, since it explains what the summary doesn't: how the 50% is determined and how the opponent can play in the non-forced turns.
If you play using a deterministic algorithm, for example always play paper, the opponent can figure it out and beat you on all the non-forced turns. At best you'll get an even result.
If you play using a random algorithm, the opponent can figure out the frequencies you're using and compensate for that. For example, if you decide to play paper 50% of the time and rock and scissors 25% of the time, you'd win against an opponent playing rock 50% of the time and paper and scissors 25% of the time. However, if the opponent decides to play rock 50% of the time and scissors the other 50%, the result is even again. If the opponent would be forced to play rock more than 50% of the time, there is no room to compensate and you would win consistently with 100% paper. I think that with 50% rock, there is enough room to respond to any frequency distribution you can come up with, although I have no proof for that.
You could change your algorithms during play, but if there isn't any algorithm that results in an advantage when playing it consistently, gaining an advantage from changing your algorithm would depend on how well your opponent responds to your changes. In other words, you're playing mind games. I don't think the 50% rock restriction is going to be of any help here.
at his head
If I play rock all the time, I tie 50% of the time, lose 25% and win 25%. Perfect tie overall.
If I play scissors all the time, I lose 50% of the time, win 25% and tie 25%. Lose two out of three.
If I play paper all the time, I win 50% of the time, tie 25% and lose 25%. Win twice as much as lose.
Go for paper.
If the opponent is human and both players can see each other, the opponent won't be able to completely hide his foreknowledge of when he must throw rock. Learn his body language and tells, then throw paper whenever he inevitably gives away his move. If you're good enough to successfully predict, better than chance, when he must throw rock, then over a large enough number of games you can throw randomly the rest of the time and still make a profit.
If your opponent must throw rock 50% of the time, then you throw paper 100% of the time.
You will win AT LEAST 51% of the time, because you get the 50% gifted to you, and the other non-0% of the times that your opponent throws paper will cause a rematch.
For the bonus question you can set up a similar payoff matrix. For the case where the restricted player picks rock on the first round the value of the second round will be zero. When rock is not picked in the first round by the resticted player the value of the second game will be 50 to the unrestricted player.
So the payoff matrix for the first round that includes the value of the second round looks like:
0 0 100
50 50 0
-50 100 50
(Again rows and columns are in RPS order.)
The payoff to the unrestricted player is 33 1/3 for both rounds combined.
The unrestricted player should use 2/3 paper 1/3 rock in the first around. In the second round either just paper or 1/3 RPS depending on whether or not the restricted player must play rock.
The resticted player should player should play 2/3 rock 1/3 scissors in the first round.
There is a program to calculate these results at:
http://wolff.to/bruno/strategy...
If there were no betting caps, I would progressive bet on paper. Bet 1 dollar paper. If lose, bet 2 dollar. If lose, bet 4 dollar. If lose, bet 8 dollar, et al. Once you win you reset back to zero. This naturally has drawbacks and potential for huge failure, but I've had good luck with it in blackjack and roulette if betting caps are high and I have about $10000 to play around with.
'We are trying to prove ourselves wrong as quickly as possible, because only in that way can we find progress.' RPF
There are a fair number good ideas here that are getting close to the correct solution here, but there is also a lot of stuff that is pretty far off so I'll be posting a video solution to my youtube channel for this Monday. If your interested in the solution you can subscribe to my youtube channel and you'll see it when I post it on Monday :)
https://www.youtube.com/channe...
Personally, I always throw 'gun' if there is sufficient cash involved.
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.
To figure out what you should do, first assume your opponent is rational, and will make good choices whenever he is able. Since he knows that you will play a paper-heavy strategy to counter his rock-heavy strategy, it would not be rational to voluntarily choose more rocks. That could only make things worse.
But if he tried to exploit your paper-heavy strategy by throwing scissors on turns when he gets a choice, you'd have a perfect strategy against this: All rock. On forced rock, you get a redo, and on non-forced rock, he does scissors and you win. So on first pass, I think that your opponent should favor paper when he can choose. It's not like you'll ever do scissors. That's auto-lose half the time - basically a complete surrender of your advantage. So paper is a safe move for your opponent.
The problem is that if he did 50% rock and 50% paper, then all-paper will be your perfect counter. He won't let you do that, so he'll have to throw in some scissors. Just how many? It looks now like you will both simultaneously have to determine optimal strategies in order to answer that question, and this will require derivatives of two sets of optimal-choice equations - so that you can solve for the two maxima. Sounds like a fun problem!
The best strategy is to randomly choose paper 2/3 of the time and randomly choose rock 1/3 of the time. You will be expected to win at least 2/3*1/2 = 1/3 of the time and tie 1/3*1/2 = 1/6 of the time when the opponent randomly chooses rock. When free to choose, the best option for the opponent is then to always choose scissors which results in you winning 1/3*1/2 = 1/6 of the time and losing 2/3*1/2 = 1/3 of the time. So, overall you will be expected to win 1/2 the time, tie 1/6 of the time, and lose 1/3 of the time.
This is a typical game theory type problem. One simply needs to optimize a max-min problem. If the probability of the player choosing rock, paper, scissors is x,y,z, respectively, and the probability of the opponent choosing is a,b,c, then one optimizes (Wins - Losses):
max_{x,y,z}min_{a,b,c} (b-c)*x + (c-a)*y + (a-b)*z subject to x + y + z = 1, x,y,z >= 0, a + b + c = 1, a >= 1/2, b,c >= 0.
There are no solutions in the interior since those points boil down to the original rock paper scissors game. So, one only needs to check the boundary, when b = 1/2 and, thus, c = 1/2 - a. If you fix x,y,z then the extreme points occur at a = 0 and a = 1/2 giving a Win-Loss of x/2 - y/2 when y >= (x+z)/2 and y/2 - z/2 otherwise . Again the interesting behavior only happens at the boundary and, so, setting y = (x+z)/2 gives a Win-Loss of x/4 - z/4 which is maximized when z = 0, and, therefore 1 = x + y = x + x/2 =(3/2)*x, yielding x = 2/3, y = 1/3, and a Win-Loss of 1/6.
... is coming monday on the author's youtube channel https://www.youtube.com/channe... -- he commented with this but is evidently bad at slashdot so I'm hopping this comment gets read even if his doesn't ...
@AlexSheive
It says that rock *must* be thrown 50% of the time. Not that there is a 50/50 chance of the choice being rock for each game. So the number of games *has* to be set out before gaming commences. As you draw closer to the last game, the odds will change to how likely rock will be thrown. This is very similar to counting cards in a blackjack.
"When life gives you lemons, don't make lemonade. Make life take the lemons back!" -- Cave Johnson
You play all paper, all the way. You beat him in all the forced turns, giving you an even score. However it will cost him a few lost points on the non-forced turns to figure out your all paper strategy and switch to an all scissor strategy. At that point, haven't you already won? Opponent will only win 50% of the remainder of the games but is already behind, so will never win. For extra points, you can switch to rock at random intervals for 1 turn only, after opponent picks an all-scissor strategy on the non-forced turns. Those will give you a 50/50 chance of a draw (on the forced turns) or a win against scissors (on the non forced turns).
51 to 49 is a win as good as 99 to 1 in RPS.
Always play paper. Worst case scenario is a perfect stalemate. Over the long term it's not possible for you to lose, only tie.
Do your own statistics homework.
Ok, my current guess would be around 1/3 rock and 2/3 paper. Opponent cannot replicate this strategy, because of requirement of having at least 50% rock. If he goes 50% scissors, he will have 1/6 tie, 2/6 lose, 1/6 lose and 2/6 win, so 3:2 in my favor. With 50% paper, he gets 1/6 tie, 2/6 lose, 1/6 win, 2/6 tie, so 2:1 in my favor. 100% rock is obvious lose. I think that mixing scissors and paper will be clearly worse than pure scissors on his part (because with scissors, he has at least 2/3 chances of winning after he got a choice).
From random testing, it seems that optimal ratio of rock is around 38% (not 1/3). Around that point, pure paper and pure scissor strategies for opponent seems to get equal and I'm winning around 62% of cases.
There is probably some interesting math behind that...
doesn't include a consideration of my lifespan. If I only live 100 years my opponent can play scissors every single game and play rock continuously after I am dead and the restriction would still be satisfied. In other words, "having to play rock 50 percent of the time" doesn't give you any relevant information about your opponent's behaviour.
Opponent throws rock more than he wants to => You shouldn't throw scissors at all (assumption)
Let your optimal throw be according to probability { R: 1-x, P: x, S: 0 }
Opponent doesn't want to throw R at all in this case; he should respond by matching optimal probabilities on what he can choose: { R: 1/2, P: (1-x)/2, S: x/2 }
You continue to assume that you shouldn't throw any scissors; you respond by matching optimal probabilities on R and P: { R: x/(x+1), P: 1/(x+1), S: 0 }
Since your throw probabilities are optimal 1-x = x/(x+1) and x = 1/(x+1)
Guess what? x = 1/(x+1) is the definition of the golden ratio conjugate! (x = 0.618034) And it works in the first equation as well .5, P: 0.190983, S: 0.309017 }
Therefore your optimal throw is according to probabilities { R: 0.381966, P: 0.618034, S: 0 }
Therefore your opponent throws according to probabilities { R:
You will win (2x-x^2)/2 = 0.427051 of the time, lose (1-2x+2x^2)/2 = 0.263932 of the time, and draw (1-x^2)/2 = 0.309017 of the time. When draws are dropped you win 0.618034 (x) of the time.
I believe this is the optimal solution (but I still have to confirm the above assumption)
You kick him in the testicles and let him keep the chicken.
So assuming you get 1 point for winning -1 for losing and 0 for tying you can get an expected outcome of 1/6. Here's how:
Play Paper with 2/3rds probability and Rock with 1/3rd probability. Your opponent must play an affine linear combination of the following three strategies:
SP = 1/2( Rock + Paper )
SS = 1/2( Rock + Scissors )
SR = Rock
(alternatively, they pick what to play, and with 50% probability their play in changed to Rock)
Against SP, my strategy wins 1/3 of the time and loses 1/6 of the time for a total of 1/6.
Against SS, it wins with 1/2 probability and loses with with 1/3 probability for a total of 1/6.
Against SR, it wins with 1/3 probability and loses with 0 probability for a total of 1/3.
On the other hand, the opponent can hold you to an expected score of 1/6 by playing 1/3 SP + 2/3 SS, or in other words playing Rock 1/2 of the time, Paper 1/6 of the time and Scissors 1/3 of the time. Against this Rock and Paper each get an outcome of 1/6, and Scissors gets and outcome of -1/3.
Thus, the best you can do is 1/6.
For those interested in how I got this, note that we just need to compute a maximin strategy for the zero-sum game where your options are R,P,S and the opponent's options and SP,SS,SR. Just write out a payoff matrix and compute it.
As some have said, the optimum is a mixed strategy of playing paper 2/3 of the time and rock the rest of the time.
Let's say we're player 1 and they're player 2. The way to calculate this exactly is to first observe that the general form for each players mixed strategy is:
P(r1) + P(p1) + P(s1) = 1
1/2 + P(p2) + P(s2) = 1
where, for example, P(r1) is the probability of a rock being played by player 1.
This means that for a given P(r1), P(p1) and P(p2),
P(s1) = 1 - P(r1) - P(p1) [1]
P(s2) = 1/2 - P(p2) [2]
Now observe that the probability of player 1 winning, P(W), is the sum of the probabilities of each of the winning configurations occurring,
P(w) = P((r1 and s2) or (p1 and r2) or (s1 and p2))
= P(r1 and s2) + P(p1 and r2) + P(s1 and p2)
= P(r1)*P(s2) + P(p1)*P(r2) + P(s1)*P(p2)
After substituting from [1] and [2] we get the following,
P(W) = P(r1)*(1/2 - P(p2)) + P(p1)*(1/2) + (1 - P(r1) - P(p1))*P(p2)
Let's make this easier to read:
x = P(r1)
y = P(p1)
z = P(p2)
So then we have:
P(W) = x*(1/2 - z) + y/2 + (1 - x - y)*z
Similarly, we find that the probability of losing, P(L), is:
P(L) = x*P(p2) + y*(1/2 - z)+ (1 - x - y)/2
Now if we gain by 1 for winning, gain by -1 for losing (aka make a loss by 1) and nothing happens when we draw, then the first players expected gains, F, is given by:
F(x, y, z) = (1)*P(W) + (0)*P(D) + (-1)*P(L) = x + y/2 - 1/2 + z*(1 - 3*x)
[Note that we didn't bother finding the probability of drawing because it has no effect on the value of the expected gains]
Now, recall that player 1 controls x and y (the probability of player 1 playing rock and paper respectively) and player 2 controls z (the probability of player 2 playing paper). Be aware that x and y can at least be zero and at most be 1 minus the other. Also note that z can only range from 0 to 1/2 (because the other half of the time player 2 must throw a rock). Finally, you should understand that player 1 wants to maximize F and player 2 wants to minimize F.
From this equation it is clear that when x is less than or equal to a 1/3, player 2 should minimize z in order to minimize F. So in this case,
F(x, y, z) = F(x, y, 0) = x + y/2 - 1/2
and player 1 would maximize this by setting x = 1/3 (as was assumed for this case) and y = 0 which gives F(1/3, 0, 0) = 1/3 - 1/2 = 1/6.
If on the other hand x were set to a probability greater than or equal to 1/3, player 2 would want to maximize z. This would give,
F(x, y, z) = F(x, y, 1/2) = x + y/2 - 1/2 + (1/2)*(1 - 3*x) = (y - x)/2
Player 1 would maximize this by minimizing x (x = 1/3) and maximizing y (y = 2/3 as x + y = 1),
F(x, y, z) = F(1/3, 2/3, 1/2) = ((2/3) - (1/3))/2 = 1/6.
Finally it's nice to observe that when player 1 sets x = 0, the second player's choice of z has no effect on the gains function whatsoever.
So, in conclusion, the optimal strategy for player 1 is x = 1/3 and y = 2/3 where x is the probability of player 1 playing rock and y is the probability of player 1 playing paper.
Since he can expect you to tthrow paper pretty often, he'd want to make scissors take up at least some of the non-rock ones he can throw, so if you throw rock, he would lose and if you were countering one of his rocks, you'd negate.
I don't know the meaning of the word 'don't' - J
Thanks for that. If Slashdot used an RTF SYSTEM LIKE THE REST OF THE CIVILISED WORLD, your angle brackets would have been no problem....
Shoes for Industry. Shoes for the Dead.
As others have pointed out, the game has positive expectation for the player facing the one who is forced to play rock 50% of the time.
So the answer to the question "how much would you be willing to pay to play against them?" is simple - you play as many games as you possibly can, for as much money as you possibly can. Every bet has positive expectation so every bet makes you money.
The only constraint is that you size each bet according to the size of your bankroll with respect to Gambler's Ruin. In other words, you don't put your whole (or even a significant percentage of) bankroll on the line in one game because you can lose it, and then you won't have any money left to make more positive expectation bets.
In each game, you bet an amount where the probability of you going on a losing streak from your current bankroll value which ends up costing you your entire bankroll is sufficiently tiny, given the statistical variance of results.
Based on @ShanghaiBill's solution, I wrote a solver and simulation in Python (that also fixes ShanghaiBill's buggy pinning of "him.rock" to 0.5 - the player could in theory, choose to play rock at more than 50% probability). Use Pypy for speedy execution - I uploaded the code to GitHub: https://github.com/ttsiodras/R...
Let r, p and s be the probabilities we play rock, paper and scissors on a given game respectively. It follows that,
r + p + s = 1
where r, p and s are elements of the interval [0,1]
For our opponent we will use a similar notation, only with uppercase letters,
R + P + S = 1
where R is an element of [1/2, 1] and p and s are elements of [0, 1/2]
Rearranging, we find that,
s = 1 - r - p
S = 1 - R - P
Now the expected gains, G, we get from any one game is given by,
G = AW + 0D + (-A)L = A(W - L)
where A is the amount exchanged between players and W, D and L is the probability of winning, drawing and losing respectively.
The probability of winning on single game, W, is given by the sum of the probabilities of each of the winning configurations occurring is
W = rS + pR + sP
Similarly the probability of losing on a game is
L = Rs + Pr + Sp
and so the expected gains is given by
G = A( r - R + P - p + 3(pR - Pr) )
Now note that our opponent must play rock at least half the time, so whenever we play scissors, at minimum we will lose half of the time and so it will never improve our gains. Therefore we should never play scissors. In the worst case scenario, our opponent is aware that we will never play scissors and that playing rock more than he has to will never improve his gains. This means
R = 1/2
G = A( r + (p-1)/2 + P(1-3r) )
Assuming the opponent is aware of this, he will do what he can to minimize G. This would mean that if (1-3r) is negative then he would maximize P (the only variable he controls). Similarly if (1-3r) is positive, he will minimize P. Finally if (1-3r) is 0, he cannot affect G.
In all of these scenarios, if we set r = 1/3 (or at least as close as possible to it) we will maximize G. This means that we should play rock a third of the time and paper the rest of the time. Our expected gains per game is G = A/6 so if A = $100 we should only be willing to pay $16.66 or less if we wish to make any gains per game.
Use the Kobayashi Maru solution.
Star Trek transporters are just 3d printers.
He could throw whatever he likes in the short term, and claim he was going to make up the 50% rock in the long term. How would you disprove this?
Solution with some explanation of game theory concepts, by the article author: https://www.youtube.com/watch?...
then on the 1st round rock is 50:50; but rock is thrown then on the 2n round they can't throw rock again (that would be 100% rock when they _MUST_ throw rock 50% of the time); or if they didn't throw rock on the 1st round then they'd have to throw rock on the 2nd round, etc. ;-)
Funny Tom stfu disappearing after that post (not). Tom's busy "eating his words". Tom's polite (now that apk humbled him http://slashdot.org/comments.p... after that libel of Tom's for Tom's numerous mistakes). Tom doesn't talk with his mouth full (of his own words he had to eat).
And libeler: How'd "eating your words" taste? See here http://slashdot.org/comments.p... were they flavorful (lol) seasoned with "the bitter taste of SELF-defeat" + YOUR FOOT IN YOUR MOUTH you bigmouth libelous Open SORES bullshitter?
As to the rest of my subject, let's let TOM speak shall we:
"I'm having great conversations on this site with one of my alias accounts" - by Tom (822) on Monday April 07, 2014 @02:29PM (#46686259) Homepage
FROM -> http://slashdot.org/comments.p...