Impact of standard vs random blind choice of card I have no idea if there is a better way to word this problem, so please feel free to suggest improvements.
Here is the scenario:
Jack shows Jill ten playing cards, face down. One of those cards is a Joker, the other nine are not. The probability of Jill drawing the Joker is 1/10.
Simple enough.
Now suppose we repeat the experiment ten times, each time Jack shuffles the cards to get a perfectly random order.
The question: Would the probability of selecting the Joker remain the same if Jill always chooses a random card, as opposed to always picking the same card (say, third from the left)?
Intuitively it feels to me like the probability should remain the same, because the probability for any card she chooses is always 1/10, per draw. But I have this nagging feeling that I should somehow account for the probability of a given card being the winner several times.
How should I calculate this?
 A: Let the card be placed on spots $1,2,\dots,10$ and let $J$ denote the spot for the Joker.
Let $X$ denote the spot chosen by Jill.
Then:
$$P(J=X)=\sum_{k=1}^{10}P(J=X\mid X=k)P(X=k)=\sum_{k=1}^{10}P(J=k)P(X=k)=$$$$\sum_{k=1}^{10}\frac1{10}P(X=k)=\frac1{10}\sum_{k=1}^{10}P(X=k)=\frac1{10}\cdot1=\frac1{10}$$
The second equality is based on independence.
This shows the the probability distribution of $X$ does not affect the outcome.
A: This is a nice question and the answer is Yes! To see why this happens if the shuffling is ideal (i.e. the ambiguity is maximum) the opponent is indifferent between choosing card from one constant position and choosing the random position card, in both cases he/she doesn't know the Joker. In a mathematical way let $X$ be the position the Joker possesses after the shuffle (which is unknown to us) and $Y$ be the random position chosen by the opponent. If the opponent always chooses ,say position $1\le i\le 10$ the probability of choosing the Joker would be $$\Pr(X=i)=\dfrac{1}{10}$$and if he/she chooses random position the probability of choosing the Joker again would become$$\Pr(Y=X)=\sum_{\text{position }x}\Pr(X=x)\Pr(Y=X|X=x)=\dfrac{1}{10}\sum\Pr(Y=X|X=x)=\dfrac{1}{10}$$which is $\dfrac{1}{10}$ again.
