Can you help me get a way to explain all of this to someone, especially for step 2 how do they get the “327=19” and ect Question-when 3 people play rock paper scissors -what is the probability of 1 loser and why is it like that for the outcome. Step 1 The sample space contains 3*3*3 = 27 3×3×3=outcom
Can you help me get a way to explain all of this to someone, especially for step 2 how do they get the “327=19” and ect
Question-when 3 people play rock paper scissors
-what is the
Step 1
The sample space contains 3*3*3 = 27 3×3×3=outcomes, as you say.
There are different situations. If all players pick the same symbol, there is not a player that stands out (3 outcomes). If the three players pick three different symbols, there is also not a player that stands out (6 outcomes).
You call these 3+6 = 9 outcomes draws.
he remaining situations corresponds to two of the players having the same symbol, and the last player having a different symbol. If the symbol held by the two players beats the symbol of the remaining player, you use the terminology that the two players win together (9 outcomes). However, if the symbol shared by the two players is weaker than the symbol chosen by the last player, then the last player wins alone (9 outcomes).
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Step 2
You count the number of outcomes where player A wins alone as 3, corresponding to probability that A wins alone on 327=19.
You also correctly count the number of outcomes where player A wins together with another player as 6, corresponding to probability 627 = 29 . But in 3 of these 6 outcomes, player B also wins together with another player, and in the other 3 it is player C which also wins. So this overlaps.
This why you cannot say that the (correct!) probability 13 that A wins either alone or together with one other player does not leave room for the situation where we have a draw.
If we denote a draw with three different symbols ABC, the situation where A and B win together as AB, and the situation where A wins alone with A, and finally a draw with a single symbol repeated threes times O, then we have:
P(ABC) = 627 P(AB) = P(BC) = P(CA) = 327 P(A)=P(B)=P(C)= 327 P(O) = 327
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