Rock Paper Scissors

Thursday 15th July 2021

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In the May 2021 North East regional newsletterOpens a new window, we asked 'What is the probability of winning at Rock, Paper, Scissors?' and we asked this because we believe it forms the basis for an interesting discussion…

Some may automatically give the answer of \(\frac{1}{2}\) which is correct (sort of – depending on how things are defined), whilst others may intuitively say \(\frac{1}{3}\) which is also sort of correct (depending on how you interpret the situation). So this question regarding the probability of winning forms the basis of an interesting discussion, because you can challenge anyone’s answer and drill down into assumptions, interpretations and definitions.

Why \(\frac{1}{3}\)? Well...

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With 9 possible outcomes, 3 of which are winning ones, we may say that we get \(\frac{1}{3}\). However, in the rules laid out last time, we specifically said that if there was a draw then the round just replays until someone wins… so how might this look? Well, we can rearrange the space so that we still have 9 cells but so that we’ve grouped the winning ones (green) together, the losing ones (blue), and the draws (red) down the central diagonal. And if we ‘draw’ then, within each of those draw cells, we simply reiterate the same procedure again… and again… and again etc… and you can see perhaps that overall the probability of winning will eventually reach \(\frac{1}{2}\)…

etc…

But can we actually say the probability of winning is \(\frac{1}{2}\)? Sure – if we continue on to the infinite stage, constantly drawing as if playing in front of a mirror… but the probability of this happening is zero! So, can the probability of winning be \(\frac{1}{2}\)? What do you think? More next time!

By Jeremy Dawson and Michael Gibson

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