# Extended "stone paper scissors" game

In the game "stone paper scissors" we have 3 options. Now extend the game. There are 5 opportunities: $$1,2,3,4,5$$. The rules: $$1\rightarrow 1,2; 2\rightarrow 3,4; 3\rightarrow 4; 4\rightarrow 1,5; 5\rightarrow 1,2,3$$.($$a\rightarrow b$$ means that a hit b). The chance to win with $$1,2,3,4,5$$ are $$a,b,c,d,e$$ in this order. What are a,b,c,d,e?

P.S. I think we choose an option with a chance equal to the chance of winning with it. For example we choose the first option with $a$ chance.

Note by Páll Márton
2 weeks, 3 days ago

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\left\{\begin{matrix}\begin{aligned} a&=\left(b+c\right)+a\left(b+c\right)+a^2\left(b+c\right)+\cdots =\left(b+c\right)\left(a^0+a^1+a^2+\cdots \right)=\dfrac{b+c}{1-a}\\ b&=\left(c+d\right)+b\left(c+d\right)+b^2\left(c+d\right)+\cdots =\left(c+d\right)\left(b^0+b^1+b^2+\cdots \right)=\dfrac{c+d}{1-b}\\ c&=(d)+c(d)+c^2(d)+\cdots =d(c^0+c^1+c^2+\cdots )=\dfrac{d}{1-c}\\ d&=(a+e)+d(a+e)+d^2(a+e)+\cdots =(a+e)(d^0+d^1+d^2+\cdots )=\dfrac{a+e}{1-d}\\ e&=(a+b+c)+e(a+b+c)+e^2(a+b+c)+\cdots =(a+b+c)(e^0+e^1+e^2+\cdots )=\dfrac{a+b+c}{1-e} \end{aligned}\end{matrix}\right.

But there aren't solutions.

- 2 weeks, 3 days ago

Hey @Páll Márton, hope you aren't ignoring me... you do understand that it wasn't me but my cousin right? Yajat and Hamza are too angry i guess, but are you open?? Please say yes, i really need to talk with a friend.

- 1 week, 4 days ago

Yes? But I should learn... Have you friends? I think I'm only a mate for you.

- 1 week, 4 days ago

And this isn't a friend chanel, this is a math not :)

- 1 week, 4 days ago

yes, of course, and you are a friend, but let's stay with math here, I think extending a stone paper scissor to 5 is awesome idea.

- 1 week, 4 days ago

This isn't my idea. working?

- 1 week, 4 days ago

This isn't an uploaded image :)

- 1 week, 4 days ago

Nice, Star Trek spock

- 1 week, 4 days ago

The bottom comment. I don't have idea too

- 2 weeks, 2 days ago

Hm.

- 2 weeks, 2 days ago