There are four playing cards face down on a table, two are red two are black. You must guess which two are red by pointing to them, and implicitly which two are black. Assuming all configurations are equally likely, and that you do not have psychic abilities, find the probability that exactly j of your guesses are correct for j = 0, 1, 2, 3, 4.
Blitzstein, Introduction to Probability, exercise 1-30.
My thoughts:
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First, I do not fully understand the j guesses part. I assume it means that you play n rounds of the game, and must find the probability of
0, 1, 2, 3, 4of the n games being won…? -
As all configurations are equally likely, this reads as an ordered problem without replacement when picking cards in each round. Therefore probability of selecting two red cards is
1/6 -
I think the general answer is
P(j wins of n Games) = (1/6)**j * (5/6)**(n-j)but I have invented the variable n because the initial question appears vague.
Best Answer
This is a very vague and confusing question. However the # of way to choose 2 out of 4 is clear $$\binom{4}{2}=6$$
Let r be red card and b be black card. Assume the cards are in rrbb order.
You guess the red cards, and the rest of 2 cards assigned to black implicitly. All possible outcomes:
rrbb
rbbr
rbrb
brrb
brbr
bbrr
j is equal to 4, 2, 2, 2, 2, 0 respectively. Therefore, $$P(j=4)=1/6$$ $$P(j=2)=4/6$$ $$P(j=1)=P(j=3)=0$$ $$P(j=0)=1/6$$