Ro-Sham-Bo-Fu Game Theory

[PairTheBoard]

At the "Moola" website where they give you a penny and let you play games with a chance to double it up 30 times, one of the games is a variation of Rock-Paper-Scissors called Ro-Sham-Bo-Fu.

It's the same Rock-Paper-Scissors except you play for a number of points, N, and you are also given 2 bonus points if you don't use the specified tainted object. For example, you might be playing for N=6 points and, win or lose, you get 2 bonus points if you avoid use of the Rock. Depending on N, what's the optimum frequencies with which you should be choosing your objects?

For example, if the Rock is tainted and you are playing for N=1 point then it's clear you would always choose scissors. Let's call Scissors the Superior non-tainted object and Paper the Inferior non-tainted object in this case. So you always have a Tainted Object, Superior Object, and Inferior Object.

I think as N gets larger the optimum frequency for the Tainted Object goes up along with the optimum frequency for the Inferior Object. Letting "Op" denote optimum frequency I'm guessing it's always true that,

Op(Superior) > Op(Tainted} > Op(Inferior)

and that these approach equality as N gets large.

What does Game Theory have to say about this game?


You know, if this Moola thing got popular enough and you could achieve a consistent edge, playing a Kelly Criteria could turn into some real money.

The Hi-Lo game there could use some analysis as well.

PairTheBoard

[AaronBrown]

I assume that you are just trying to maximize your own score, not beat the other player. In that case, this is a complex game for P > 2. For P < 2, both players should always pick the superior object. For P > 2, the equilibrium is for each player to pick the tainted object with probability 1/3, the superior object (the one that beats the tainted object) with probability 1/3 + 2/(3P), and the inferior object with probability 1/3 - 2/(3P). For large P, this approaches the usual 1/3, 1/3, 1/3 strategy.

This gives a long-term value of 4/3 to both players, when they can get a sure 2 every time by agreeing to always pick either the superior or inferior object. So you get into Prisoner's Dilemma considerations and the answer depends on how you resolve that issue. For example, you could always play the superior object, and count on the other player reciprocating, and do better than the equilibrium. But if the other player knows you will always pick the superior object, she can do better by picking the inferior one. Similarly, you could always pick the inferior one, but if she knows you're doing that and P > 2, she can do better picking the tainted one.

In general, if the other player picks the tainted object with probability less than 1/3, you choose only between the tainted object and the inferior ones. These are the interesting strategies because although none are equilibrium, both players are better off the less each one picks the tainted object. If the other player picks the tainted object with probability greater than 1/3, you choose only between the tainted object and the superior one. These are destructive strategies. One is an equilibrium, both players pick the tainted strategy half the time and the superior strategy half the time. Each player makes 1 on average, versus the 4/3 of the other equilibrium and the 2 of the cooperative strategy.

[PairTheBoard]

Actually, at the Moola site the object of the game is to get more points than the other player in 6 throws. The number of points you are playing for varies randomly from I believe 2 to 6. In case of ties I believe the points accumulate and you play for them on the next throw.

[ QUOTE ]
I assume that you are just trying to maximize your own score, not beat the other player. In that case, this is a complex game for P > 2. For P < 2, both players should always pick the superior object. For P > 2, the equilibrium is for each player to pick the tainted object with probability 1/3, the superior object ?(the one that beats the tainted object)? with probability 1/3 + 2/(3P), and the inferior object with probability 1/3 - 2/(3P). For large P, this approaches the usual 1/3, 1/3, 1/3 strategy.



[/ QUOTE ]

In the questioned highlighted phrase, I'm not sure if you mispoke or if you are changing the terminology at that point. With the Rock Tainted I identifed the Scissors as the Superior Object because it beats the other bonus, non-tainted object, Paper. With the number of points P=1 you seem to be keeping with my terminology when you point out that "For P < 2, both players should always pick the superior object." Clearly in this case you would never pick Rock because you could get at most 1 point beating Scissors while Scissors and Paper get at least 2 bonus points. Between Scissors and Paper, Scissors is the "superior" choice because it beats Paper. So you would always pick Scissors. Since you say you would always pick the "Superior" in this case I conclude you mean to keep with my terminology for Scissors being the "Superior" object.

For P=3 and Frequenies of 1/3 for Tainted-Rock, 1/3+2/9 for Paper - "the superior object ?(the one that beats the tainted object)? - and 1/3-2/9 for Scissors; Compare to the strategy that always picks Scissors. It's expected number of points won is:

(1/3)*2 vs Rock (1/3)*3
(1/3+2/9)*5 vs Paper (1/3+2/9)*2
(1/3-2/9)*2 vs Scissors (1/3-2/9)*2

Or 11/3 for Always Scissors vs 7/3 for the other strategy.


So I have to conclude you mispoke when you said, "superior object (the one that beats the tainted object) with probability 1/3 + 2/(3P)" and meant to say, "superior object (the untainted one that beats the other untainted object) with probability 1/3 + 2/(3P)"

I was puzzled by your comment, "I assume that you are just trying to maximize your own score, not beat the other player. In that case, this is a complex game for P > 2."

For the Moola game it seems you are trying to beat the other player by maximizing your score.

Now as I think about it some more it doesn't make sense that your optimum frequecy for the Tainted Object jumps from zero to 1/3 as P crosses 2. It also doesn't make sense when you say, "This gives a long-term value of 4/3 to both players, when they can get a sure 2 every time by agreeing to always pick either the superior or inferior object." What are your units here?

So, how complex a problem is it for maximizing your points won? That's what we need to solve to get rich and make all the moola at Moola.

Thanks,

PairTheBoard

[PairTheBoard]

Aha! So for Rock-Tainted, 2 bonus points, and playing for P points with P<2, the frequencies for Rock-Paper-Scissors are 0-0-1. Aaron says that for P>2 the frequency for Rock suddenly jumps to 1/3, with Rock-Paper-Scissors frequencies 1/3 - (1/3 - 2/3P) - (1/3 + 2/3P)

Why is this jump reasonable? Well compare the two at P=2. Aaron's frequencies are 1/3 - 0 - 2/3. When played against Always Scissors the points go:

(1/3)*2 vs (1/3)*2
(0)*4 vs (0)*2
(2/3)*2 vs (2/3)*2

A tie. And as P grows larger than 2 it becomes more profitable to take a chance with Paper beating Rock even though Paper gets beat so bad by Scissors.

If Aaron's frequencies are correct it is suprising to me. I suspect most players give Rock too small a frequency for P=3 and too large a frequency for say P=9.

Thanks Aaron. Are you sure about your results? I had 16 cents riding on this game a little while ago.

PairTheBoard

[AaronBrown]

The game is different if you are trying to outscore the other person. Now there's no question of cooperating to get a better joint score. Zero sum games are easier.

Assuming you want to maximize your expected score, which is a good assumption early in the game (later in the game you may want to get or avoid a specific score, or want to increase or decrease variance), the strategy is similar to the equilibrium one above, except you only deviate half as much from 1/3, 1/3, 1/3.

I apologize for switching the terminology on you. I used "superior" to mean superior to the tainted object. I'll switch back to yours, so "superior" means superior to the other non-tainted object. So if Rock is tainted, Paper is inferior and Scissors is superior.

You should pick the tainted object 1/3 of the time, the inferior object 1/3 - 1/(3P) and the superior object 1/3 + 1/(3P). So if P = 1 and Rock is tainted, you pick Rock 1/3 and Scissors 2/3. You then don't care what he does. If he picks Rock, you come out even whether you picked Rock or Scissors. If he picks Paper, you lose 4 the 1/3 of the time you pick Rock, and win 2 the 2/3 of the time you pick Scissors. If he picks Scissors, you come out even whether you pick Rock or Scissors. Since you come out even whatever he does, you obviously don't care how he randomizes his choices. This is not the best way to win, obviously, but it is a strategy that cannot be beaten in expected value.

If P = 2, you get Rock 1/3, Paper 1/6 and Scissors 1/2. Once again, as P gets large, you approach 1/3, 1/3, 1/3.

In general, if he picks the tainted object, you get 0 the 1/3 of the time you pick the tainted object, 2*(1 + P) the 1/3 - 1/(3P) of the time you pick the inferior object and 2*(1 - P) the 1/3 + 1/(3P) of the time you pick the inferior object. That comes out to an expected value of zero. You get the same thing if he picks the superior or inferior object.

With any other strategy, there is an assignment of probabilities he can use to get a positive expected value.

[PairTheBoard]

[ QUOTE ]
The game is different if you are trying to outscore the other person. Now there's no question of cooperating to get a better joint score. Zero sum games are easier.


[/ QUOTE ]

You are trying to outscore your opponent but it's not a zero sum game. The points are awarded by the House to the winning player. In case of a Tie the points are contested on the next throw. Most points scored in 6 throws wins. There are 2 Bonus points for using a Non-Tainted Object. It looks like you're now thinking it's 1 bonus point. If you use a Non-Tainted Object you receive the 2 bonus points win or lose. The number of points put up by the House each throw which you play for varies from I believe 2 to 6 - although you may play for more if the last throw was a tie.

Even though it's not zero-sum I think you still want to maximize your Expected Points, at least on the first throw. However, I'm not sure now which of your solutions apply.

Thanks,

PairTheBoard

[gull]

The reward is zero-sum (i.e., if something is good for you, it's bad for your opponent). The specific scoring method is irrelevant.

[AaronBrown]

No, I'm always using 2 bonus points.

My question is what happens after the six rounds. Do you just win (+1) or lose (-1)? Or does the number of points you scored matter? For example, suppose you could make a deal with the other player to always pick the superior non-tainted object, so you each got the bonus points each round. You would each get +12, which is a good result. Does that make sense, or would it just result in both of you tying and getting 0 outcome for the game.

[PairTheBoard]

[ QUOTE ]
No, I'm always using 2 bonus points.

My question is what happens after the six rounds. Do you just win (+1) or lose (-1)? Or does the number of points you scored matter? For example, suppose you could make a deal with the other player to always pick the superior non-tainted object, so you each got the bonus points each round. You would each get +12, which is a good result. Does that make sense, or would it just result in both of you tying and getting 0 outcome for the game.

[/ QUOTE ]

Essentially it's +1 or -1. Actually you are betting a predetermined amount with your opponent. The one with the most points after 6 rounds wins the bet. There are always 3 points put up for grabs each round, along with the 2 free bonus points for picking a NonTainted Object. If the round ties, the 3 points carries over to the next round and you play for 6. If tied the next round is for 9. So there are extra considerations later in the match depending on how much you are ahead or behind.

For example, if you are tied going into the last round it really doesn't matter that there are bonus points. Whoever wins the throw wins. Another example would be if you are 4 points behind with two rounds to go. With Rock Tainted and knowing your opponent will pick scissors you are better off picking scissors than rock, thereby letting the 3 points ride to the last round. That way your opponent can't hide behind scissors in the last round.

So you are really trying to maximize the difference between the points you score and the points your opponent scores. If Rock is tainted, picking Paper vs your opponent's Rock pick gives you a 5 point lead. That's the best result. So, what's the optimum first round strategy?

To avoid confusion I suggest we consistently make Rock the Tainted Object and refer to frequencies for Rock, Paper, and Scissors under this assumption.

PairTheBoard

[mykey1961]

This game does present a rather interesting situation.

Lets say you are behind by 1 point going into the last round, and the previous round was not a tie.

We'll assume Rock is the tainted object.

The payout table for the final round would be
<font class="small">Code:</font><hr /><pre>
R P S
R +0 -5 +1
P +5 +0 -3
S -1 +3 +0
</pre><hr />

Assume we are playing Rows

The final score in points would be:

<font class="small">Code:</font><hr /><pre>
R P S
R -1 -6 +0
P +4 -1 -4
S -2 +2 -1
</pre><hr />

In terms of Wins, Loses, and Ties:

<font class="small">Code:</font><hr /><pre>
R P S
R L L T
P W L L
S L W L
</pre><hr />

In terms of units of money won and lost:

<font class="small">Code:</font><hr /><pre>
R P S
R -1 -1 +0
P +1 -1 -1
S -1 +1 -1
</pre><hr />

The optimal Row strategy is: Rock 50%, Paper 25%, and Scissors 25%
The optimal Column strategy is: Rock 25%, Paper 25%, and Scissors 50%
Giving an average result of 1/2 unit lost for the Row strategy.

p(W) = 1/8
p(L) = 5/8
p(T) = 1/4
EV = -1/2

My question is: Is this really optimal for the Row strategy?
And is it stable?