Probability of not losing if picking 2 choices in paper rock scissors game? In a game of paper, rock, scissors, the probability of winning or tying (not losing) with any one option is 2/3. This is because scissors beats paper, and scissors is tied with scissors but loses to the remaining option of scissors vs rock. So I added each instance of winning then put that sum over the total choices given. This should also be the same probability of choosing only one option, to play. For example, if choosing only scissors against a random opponent, then the chance of winning or tying is also 2/3.
However, when it comes to selecting only two choices to play in any given match, I am getting a little lost. I am thinking that if the chance of winning or tying is 2/3 for any choice, then the chance of picking any two choices is [P(choice 1)] + [P(choice 2)] = 2/3 + 2/3, which gives me 4/3. But I believe probability should never be greater than 1 so I must be calculating this wrong. I also think picking solely two choices would give a greater chance of winning or tying compared to choosing 1 choice.
What formulas would you suggest for this kind of problem? I would also greatly appreciate if you'd care to show how you would calculate this probability. I am thinking of using "n pick k" formula. I would like to continue to use this scenario to build on calculating different amounts of choices.
I attempted to include a table of the possible outcomes but I am not sure if I can format it to include the y-axis headers. Here is an image that may give a better look at a paper rock scissors chart https://slideplayer.com/slide/8708708/26/images/4/Rock%2C+paper%2C+scissors+What+is+the+probability+that+both+players+will+show+the+same+hands+in+a+game+of+rock%2C+paper%2C+scissors.jpg




Scissors
Paper
Rock




SS
SP
SR


PS
PP
PR


RS
RP
RR



 A: I regard the answers linked to in the comments as unnecessarily complicated.  For this particular problem, I advise elbow grease.  That is, because of the dependence of events, based on the two choices selected, I think this is not a problem that should be attacked with elegance.
Instead, individual cases should be manually examined, to see what's going on.  This is what I mean by elbow grease.
Suppose, for example that you choose rock + paper. 
If your opponent chooses paper, you tie.  Otherwise you win.
Further, similar results will pertain if you choose [rock + scissors] or [paper + scissors].
Also, because of the inherent symmetry of the game, if your intuition is sophisticated enough, then without manual examination of the other choices, you could conclude that the [rock + scissors] choice and the [paper + scissors] choice will give similar results.
So, regardless of which two (different) choices are made, the probabilities are:

*

*$(2/3)$ that you win.

*$(1/3)$ that you tie.


For what it's worth, once you accept the above results, you can expand your intuition by regarding the game as an example of (mod $3$) clock arithmetic.
That is, consider the inequalities:

*

*0 < 1, 1 < 2, 2 < 0.

You choose any two numbers.  
For example, you choose 0 and 1.  
If your opponent chooses 1, you tie.  If he chooses anything else, you win.
Then, consider choosing instead, [1 and 2] or [2 and 0].  It becomes easier to grasp that overall results are the same in both cases as with the choice of [0 and 1].
