In a game you have 1/36 probability of winning $85 and a 35/36 probability of losing $4. What is the expected value?

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Answer 1 · 2016-11-01T15:05:59

#EV~~-$1.53#

We can find this by adding together the probabilities times the values of each probability, so it's:

#P_1V_1+P_2V_2=EV#

Substituting in we get:

#1/36(85)+(35/36)(-4)=EV#

#85/36-140/36=EV#

#-55/36=EV#

#EV~~-$1.53#