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Ask The Wizard #418


Image source: YouTube

 

Season 2 of The Squid Game features a game called Rock Paper Scissors Minus One, explained in this video. The rules are based on the classic version of Rock Paper Scissors, with these modifications.

  • At the same time, both players play two symbols, one with each hand.
  • After examining what both player examine what was thrown, at the count of three, each takes back one symbol with on hand, leaving the other hand.
  • The game is scored according to the standard rules: rock beats scissors, paper beats rock, scissors beats paper.

The recruiter in the YouTube video uses the following situation as an example:

  • Triangle plays scissors and paper
  • Circle plays rock and paper

The recruiter says that triangle should then play paper because he can't lose. Is this strategy correct? I tend to think that Circle would probably predict this strategy and go for the tie by playing paper as well. Thus, maybe Triangle should at least consider taking a chance and playing scissors. What is the optimal strategy for both players?

anonymous

After doing some calculus, I find in the optimal strategy for both sides is as follows:

  • Triangle: Paper with probability 2/3, scissors with 1/3.
  • Circle: Paper with probability 2/3, rock with 1/3.

If at least one player follows this strategy, the following will be the probability of each outcome for Triangle:

  • Win = 4/9
  • Tie = 4/9
  • Loss = 1/9

For Circle, it is the opposite, as follows:

  • Win = 1/9
  • Tie = 4/9
  • Loss = 4/9

The bottom line is the Recruiter is absolutely wrong in his advice. Here is my general advice for any situation:

  • Always play two different symbols to begin with.
  • If you play the same two symbols as your opponent, play the symbol that would beat the other symbol you played. For example, if both play rock and paper, then play paper.
  • If both players play one symbol in common, then both sides should play that common symbol with probability 2/3 and the other one 1/3.

Links

What is the expected number of rolls needed to roll the same side twice on a six-sided die? What about other Platonic Solids? Is there a formula for any n-sided die?

anonymous

The answer for a six-sided die is 1223/324 =~ 3.774691 rolls.

The following table shows the answer for various numbers of sides.

Sides Expected
Rolls
4 3.218750
6 3.774691
8 4.245018
12 5.036074
20 6.293585
50 9.543127
100 13.209961
200 18.398444

 

Here is a formula for a six-sided die. For any other sided die, change the 6 to the desired number of sides.

To solve such difficult integrals I recommend integral-calculator.com.

My thanks to Wizard of Vegas member Ace2 for his help with this question.

Suppose you are given the opportunity to roll a die until it lands on a 6. You will win the square, in dollars, of the number of rolls needed. For example, if it took 6 rolls you would win $25. What is the expected win of this game?

anonymous

$66

 

Here is my solution (PDF).