Discuss 'Two-Finger' Dakota here

[Lepton]

Puzzle, allez-y!

[Fuldu]

I must be missing something. To me this looks like a symmetric game where, if Skinny had a winning strategy, his opponent could employ the same strategy and expect the same earnings. Since it appears to be a zero-sum game, under best play assumptions you'd have an expected winnings of zero for both players. Since that's clearly not what is being presented, I return to that I must be misunderstanding the nature of this game.

If you expect the other player to play randomly (equally like to show 1 or 2 fingers, equally likely to guess 1 or 2 fingers, and the two decisions completely independent of one another), then I would play show 1 finger, guess 2 fingers, 100% of the time. Half the time, nothing happens, a quarter of the time you lose two chips, and the last quarter of the time you gain three chips. That makes an average gain of 1/4 chip per round. {2,2} would have the same expected outcome, but with greater volatility, so I prefer {1,2}.

[Lucky Wizard]

I would randomly pick between {1,2} and {2,2}, because it would make the other guy less likely to catch on.

[kevinatilusa]

If Skinny had a "winning" strategy, it would be to con you into believing that there is an effective random strategy, then pick the counter strategy which would be effective against the strategy that he told you.

[wordcross]

taking Lucky Wizard's strategy a bit further I would choose any of the four options besides [2,1] at random, but perhaps choose [1,2] more often. This would keep the other person from figuring out that A, you put down one finger all the time, and/or B, that you choose 2 fingers all the time. Otherwise, it would be too easy for him to stymie you one way or another.

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GLmathgrant: "We had a vote, and I unanimously agreed that I'm not a schizo."

[Persona]

He holds out a finger and a thumb, claiming that the thumb is/is not a finger depending on what would make the most profit for him.

[xyno512]

i'll have to agree with the finger thumb.
but what if he does count the thumb against a player?
then he cannot use the same strategy again until he finds and un-tainted bunch of suckers.

[Peekay]

See, what I'd do is hold out two fingers on one hand, one finger on the other. Which ever one matches the kid's number I say is my guess, and the other is my number. I never lose!

[Griffin]

Game theory says that the optimal strategy (assuming no tricks like Peekay and others have suggested) is to play (guess 1, show 2) 5/12 of the time, and to play (guess 2, show 1) 7/12 of the time. Using this strategy you would be expected to win a chip for every twelve rounds that your opponent is stupid enough to play either (guess 1, show 1) or (guess 2, show 2). However, like Fuldu said, your opponent could use a symmetric strategy and not only would you not win anything, the game would get boring very quickly.

[Tec_Picasso]

To put the most chips into the game each time Skinny would have to always put in two fingers and guess the three or four, but it would only work for a period of time until the other person caught on. His odds of taking the most chips would be slightly better, as he could have the chance at the three and four, while the other player might pick a two, where Skinny already has thrown out two and the other must still put in at least one finger. I am guessing that it is something along these lines. There's a similar game called "Odds and Evens" but it is a 50-50 chance to win or lose.

[lurker]

A small correction, Tec_Picasso: the players don't have to guess what the total number of fingers shown is, but what the opponent's number is. That is, one would always guess either 1 or 2, and if Skinny's opponent always 'guesses' 2, he's dramatically improved his chances against your proposed strategy.

[Mr Stoofer]

Fuldu's math is right in principle when he says that {2,2} and {1,2} have the same average outcome - i.e. you win 1/4 chip per game.

Assuming that your opponent has worked this out and is adopting this strategy, you should: show 1 finger and you will either beat him or draw. But if he has worked this out, you should ... etc etc ad infinitum. You end up going on forever.



[This message has been edited by Mr Stoofer (edited 01-15-2004 04:18 AM).]

[Scotty Rudell]

Skinny doesn't seem particularly prone to pick on kids (in his other puzzle appearances). Yet this time he does. The implication to the puzzle seems to be that the "kids" are not going to immediately catch on to the strategy, if ever. That's the only way Skinny could gain an edge. What gets me is this: if the "kids" are not assumed to be smart enough to stump Skinny's plan, then why is Skinny wasting his time playing such a small-margin game? Skinny always stacks the odds in his favor; it's out of character for him to be nickle-and-dimeing them, when most games have a better return. Anyone up for some Roulette?

[hively]

I think you are overlooking something in the original text of the puzzle. It states:

"If only one person choses right, ** THEY ** win a number of chips equal to the total number of fingers..."

This seems to suggest that the players are rewarded for cooperation, doesn't it?

[rydiafan]

I assume the author was mistakenly using "they" as a gender neutral pronoun.

[hively]

While I agree with that assumption about "neutral pronouns" in most prose, I find that with puzzles it is extraordinarily hazardous to casually make assumptions that in effect "fill in" your own meaning into the puzzle's text. On the contrary, I would presume that the puzzle's author labored over the selection of the words very carefully before publishing it, taking an inordinate amount of time to review every turn of a phrase and to make certain that it is precise and accurate, and most importantly solvable.

With that in mind, I feel that the existence of "they" needs to be evaluated and addressed.

[Birdman]

What if Skinny always had one finger out, and always guessed two? When he wins, he would get three chips (2+1). The only way he would lose would be if the kid had one finger out and guessed one finger, which means the kid could only win a total of two.

[StormWatch]

I think Birdman is on to something...

You only have to guess the opponents # of fingers: 1 or 2. That makes essentially 4 possibly outcomes (both right, both wrong, you or opponent correct). Since you can only win on one of these, I think a possible strategy is attempting to maximize your winnings while limiting your opponents. Assuming it's a clever kid:

By initially showing 1 finger, and guessing 2, you get:
Your Correct - Opponent is wrong = 3 Chips for you
Your Correct - Opponent is correct( or wrong/wrong) = Nada for both
Your Wrong - Opponent is correct = 2 chips for him.

The moment your opponent guesses twice on a 1/1 combo, assume he's caught on to your strategy and switch your fingers to 2 guessing 1 - again it's 3 chips for you. - Return back to the other strategy or block him by guessing ones until he tries to alternate again

Not sure if this would hold over the long haul, but I haven't figured out how he's cheating yet.

[This message has been edited by StormWatch (edited 01-16-2004 10:55 AM).]

[kevinatilusa]

What if you exactly follow Birdman's strategy yourself? By symmetry, Dakota's back to breaking even.

[dnwq]

Well, if he plays a game with everyone - kids do get bored quickly, after all - and he uses that strategy for each, he'd gain slowly, wouldn't he?

Even if the other player catches on and starts mirroring, he'd not only have to catch up, chances are he'd give up and surrender his money. After all, it's more likely to be his parent's money...

And how many kids think through the strategies for a game before playing?

[Anony Moose]

In response to Hively, I would suggest that the author of this particular puzzle did not laboriously pore over the wording of the puzzle. If he had, he might have caught the grammatical error in, "If only one person choses right," or the fragmented, "whereby he is pulling a very nice."

To more directly answer the puzzle, <INV> it would seem that if the kids are smart (as just about everyone mentioned in these puzzles seem to be), Skinny will have met his match on the playground. This is a zero-sum game, with both players on equal footing, and Skinny will, on average, win no chips per hand. </INV>

[gravedigger0586]

I'm new here, although i have bee watching the site for quite a while. Where do we go to discuss the answer to the MAS Minihunt Metapuzzle #9?

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"Give me leave. Here lies the water-good. Here stands the man-good. If the man go to this water and drown himself, it is, will he, nill he, he goes. Mark you that. But if the water comes to him and drown him, he drowns not himself, argal, he that is not guilty of his own death shortens not his own life."
Shakespeare's HAMLET, first gravedigger(clown)

[sadisticgnome]

FYI: A quick Google search showed Morra to be a game played in Spain, but using all 5 fingers, and guessing the sum of the fingers shown. I'm sure it doesn't help, but good to know.

[sadisticgnome]

There seems to be an alternate strategy here, Skinny could watch the mouth of the person opposing him, and assuming Skinny has lightning reflexes he could change the number of fingers he is about to show, or change his guess so that no one wins. The shape of the mouth is very different between pronouncing "one" and "two" and anyone not on to Skinny wouldn't know to disguise it. however this assumes that skinny has superhuman reflexes, plus it's a little bit of a cheat, even for Skinny, but I thought I'd throw it out there.

[Suspence]

Supporting the point made by Anony Moose, "saunders" should be "saunters".

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I hate people who try to write interesting things in their signature

[James not to sure]

There are four diffrent possible ways each can play giving 16 possible outcomes, each giving there own return or loss, if you always play either guess two put two fingers up, or guess two put one up, then either of these will only allow a draw a draw a loss or a win 1 greater than the posibble loss, if each outcome averages out it will result in an overall average profit of one chip per every four games so .25 chips per game. It would be better to always guess two put one up as although it would yeild the same average prize the win to loss ratio would be better

[Repairman]

Fixed your light.

[Swordsman Kirby]

I think I got something.

-Always guess two. Randomize whatever you have. He gets one coin per four games on average.

[Wire Jockey]

Using 4 total fingers. Maximum pay off. Both always flash two fingers and alternate guessing number opposite each other, one is designated to start with guess of one and the other two. Each turn they alternate 1,2,1,2.

Using 3 total fingers, alternate between 1 and 2 fingers and always use the same number guess which is predetermined, each uses the opposite number guess. Player A always guesses "One" and Player B always guesses "Two"

It would help to understand this problem if the directions were edited. One sentence makes no sense. "Skinny goes on to explain that he's found a strategy whereby he is pulling a very nice. "

I interpret the sentence as; "Skinny goes on to explain that he's found a strategy whereby the players can reap the rewards."

[dARKhORN]

Every one keeps trying to find a way that Skinny D can find an advantage to win more often; this is not going to happen. No matter what strategy you employ your chance of winning more often do not increase. I believe the strategy lies in trying to maximize your winnings while, at the same time, minimizing the winnings of your opponent. This is done by call in 2 fingers.
When you call 2 fingers of three thins are going to happen; win, lose, or draw. In the event of a draw nothing is lost. If you win and you call 2 fingers you will win 3 or 4 chips depending on how many fingers you throw. If you lose and you call 2 fingers you opponent can only win 2 or 3 chips because he would have to win by throwing one finger plus the 1 or 2 fingers you throw.
Using this strategy you will win as often as you lose but you will win more chips when you win and lose less chips when you lose.
There is a flaw in this strategy. If your opponent figures out way you are doing, they can always throw 1 finder completely derailing your chances of winning. For this reason I would recommend calling every 3rd or 4th run, just to keep things fresh.
If Skinny D uses this strategy statistically he should come out ahead.
On average skinny will come out ahead 1 chip every 4 runs.

[dARKhORN]

Also, Skinny will win an average of 3 chips each win

[kevinatilusa]

Ah! I misread the original problem and assumed that the game is zero-sum (which I guess it isn't). I also misread it a bit in my analysis below, as I have each player guessing the total number of fingers rather than the number the other player is showing (it's equivalent though, my notation will just seem strange).

WARNING: Lots of (possibly erroneous) math and stuff below

Assumptions made: Each player cares only about his own winnings...it does not do him better or worse if his opponent gets fewer or more chips. He plays each game independently, and does not base his strategy on his opponent's prior moves. Chips won are won from some third party, and not from the other player (else by symmetry Skinny could never do better than break even). THERE IS NO COLLUSION BETWEEN THE PLAYERS.

Suppose the optimal strategy for player 1 is to pick 1 with probability p, guess 3 fingers with probability q if he picks 1, r if he picks 2.

Suppose the optimal strategy for player 2 is to pick 1 finger with probability a, guess 3 fingers with probability b if he picks 1, c if he picks 2. The goal is to figure out which strategies of (a,b,c,p,q,r) lead to Nash equilibria.

In the table below I have listed all 16 possibilities. The first two columns give the number of fingers 1 and 2 hold up, respectively. The next two give the guesses of 1 and 2 respectively. The next two give the winnings of 1 and 2, respectively, and the last gives the odds of this occuring.

1 1 2 2 0 0 p(1-q)a(1-b)
1 1 2 3 2 0 p(1-q)ab
1 1 3 2 0 2 pqa(1-b)
1 1 3 3 0 0 pqab
1 2 2 4 0 0 p(1-q)(1-a)(1-c)
1 2 2 3 0 3 p(1-q)(1-a)c
1 2 3 4 3 0 pq(1-a)(1-c)
1 2 2 4 0 0 pq(1-a)c
2 1 4 2 0 0 (1-p)(1-r)a(1-b)
2 1 4 3 0 3 (1-p)(1-r)ab
2 1 3 2 3 0 (1-p)ra(1-b)
2 1 3 3 0 0 (1-p)rab
2 2 4 4 0 0 (1-p)(1-r)(1-a)(1-c)
2 2 4 3 4 0 (1-p)(1-r)(1-a)c
2 2 3 4 0 4 (1-p)r(1-a)(1-c)
2 2 3 3 0 0 (1-p)r(1-a)c

Note that if p or a is 1 or 0, the other player has a cast-iron strategy to always guess right, so player the player always picking the same thing can never win. But for player 1 to want to pick randomly, he must be indifferent between picking 1 and picking 2 fingers. His expected winnings from picking 1 finger is 2(1-q)ab+3q(1-a)(1-c). His expected winnings from picking 2 fingers is 3ra(1-b)+4(1-r)(1-a)c. Thus we have

2(1-q)ab+3q(1-a)(1-c)=3ra(1-b)+4(1-r)(1-a)c, and (by the same argument for 2)
2pq(1-b)+3(1-p)(1-r)b=3p(1-q)c+4(1-p)r(1-c).

Assuming b,c,q,r lie strictly between 0 and 1 is a little bit trickier, since the other player can't necessarily take advantage of your consistency. (I still believe it to be the case though).

Assuming this is true, we have 4 other equations relating indifference between calling 2/3 for each player (after they show 1 finger) and 3/4 for each player (after they show 2 fingers)

2ab=3(1-a)(1-c)
3a(1-b)=4(1-a)c
2pq=3(1-p)(1-r)
3p(1-q)=4(1-p)r

6 equations, 6 unknowns. Unfortunately non-linear. As it turns out, there is only one solution to them, and it has a=p=10/17, b=q=3/5, c=r=3/7 (if I had known beforehand that these equalities would hold it would have been easier, but I couldn't see why there couldn't be two different equilibria with players following different strategies).

Therefore Skinny's "optimal" strategy is to hold out 1 finger 10/17 of the time, 2 fingers the remaining 7/17. When he holds out 1 finger, he should guess that his opponent is holding out 2 fingers 3/5 of the time. When he holds out 2 fingers, he should guess that his opponent is holding out 2 fingers 3/7 of the time.

By following this strategy, he will make (on average) 2*(1-q)*a*b+3*q*(1-a)*(1-c)=2*(2/5)*(2/17)*(3/5)+3*(3/5)*(7/17)*(4/7)=12/17 of a chip regardless of his opponent's strategy. Conversely, if his opponent follows this strategy Skinny will be held to 12/17 of a chip expected value regardless of what strategy Skinny follows. Thus this is an optimal strategy (minus the caveat way up there about b, c, q, and r possibly being 1).

Anyone have a pleasanter proof for this interpretation of the rules?

[Griffin]

Kevin, I'm not sure where you got the impression that it's not a zero sum game (Based on the puzzle description, I would think that it is, and this page seems to agree). However, your interpretation does seem to make the puzzle a little more interesting.

[asmussen]

I don't agree with the people who think that the money is coming from an unmentioned 3rd party who is just giving away money for no reason with nothing to gain. Although you can interpret one sentence of the puzzle that way because of the use of the word 'they', there are also multiple indications that this is not the case if you read the entire puzzle. For example, it says that Skinny 'walks off to con more kids out of their lunch money'. If both players win money in this game, he's not conning the other players out of anything. Even if they win less than him, they're getting a good deal. Also, it says "Wisely, you decline his offer for a low-stakes game." If you can only win money at this game, exactly how wise is it to decline to play?

[Jeff]

He could just always pick 3. If the other person picks 2, he'd know they will show 1 finger, (the only way they could win if they picked 2 is if they showed 1 and he showed 1, so they'd have to show 1 finger) so you would show 2 fingers and win (2+1=3). If they chooses 4, he'd know they'll show 2 fingers (they could only win if both showed 2 fingers, so they'd have to show 2 fingers) so he'd know to show 1 finger and 1+2=3, so he'd win. odds of losing 0%

He=Skinny
They=The kids

[Speeder]

Okay, there are sixteen combinations of showing/guessing, and assuming there are no strategies being employed this leads to each player winning 12 chips each for every 16 turns equalling 0.75 chips each per turn, thus giving an equal chance of winning for each player.

If Skinny Dakota guesses 2 each time and shows 1, and assuming the other player retains no strategy, then this leads to 4 combinations where Skinny will win an average of 3 chips and his opponent will win only 2. Thus making Skinny's win per turn equal to 0.75 only this time his opponent's chips per turn is reduced to 0.5 (a nice ratio for Skinny.)

However, his opponent would soon cotton on to this strategy and need only retaliate by showing 1 finger and guessing 1 finger each time which would render the stratgey useless. So, I would suggest that Skinny shows 1 or 2 fingers at random, but always guesses 2. This would mean out of the (now) 8 combinations Skinny would expect to win 7 chips making his chips per turn 0.875 (his opponent would only win 0.625 per turn.)

His opponent has only one option to level the odds and that is to also guess 2 each time while varying the number of fingers shown. This would put the odds back to evens but, by the time the opponent realises, Skinny would hope to have already cleared up.

So the answer to the puzzle I (optimistically) believe is; on average, Skinny would win 0.875 chips per hand using his strategy.

Speeder.

[Speeder]

P.S. frequently choosing 1 as a guess is a bad idea as the odds of guessing correctly remain the same, but the amount of chips won per turn will reduce as a consequence.

P.P.S. I forgot to mention that Skinny guessing 2 each time and showing 2 each time leads to winning 1 chip per hand on average (as opposed to the opponent winning an average of 0.75 per hand if he retains a random strategy) but this would soon be spotted by the opponent and then foiled by their guessing 2 and showing 1 each time.

I stick with my answer posted previously.

Speeder.



[This message has been edited by Speeder (edited 03-23-2004 06:16 AM).]

[Jeff]

never mind about what i said. i misread it. i thought they had to guess the TOTAL number of fingers shown, not what the other person would show. duh.

[Electrician]

*fixes lightbulb*

[burntarticwolf]

wellone the rules were a little confusing i understand it as you show the fingers and prdict what your opponint will do at the same time so he could quickly chang his fingers to make it wrong if he was fast enough right when he heard his opponint te hopyfly guess theres correct making it so he never loses any chips

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live to learn and let mothernatur be your teacher