In the card game bridge, the 52 cards are dealt out equally to 4 players – called East, West, North, and South. Bridge is played between 2 teams of 2 people each. North-South is one team of partners & West-East is the other team.
North, the declarer, is tackling the heart suit. North knows his hand & his partner South's hand (In bridge, the one of the 4 people's hand is opened after the 1st trick & then everyone can see the hand).
In hearts, South has AKJxx & North has xxx (x denotes a low card). The remaining 5 heart cards are split between East & West. A 3-2 split has the highest probability.
Assuming a 3-2 split, what is the probability of the Queen of Hearts being with the opponent who has 2 heart cards as compared to the one who has 3 heart cards.
Many Bridge Books say that there is a higher probability of the Queen being in the hand which has 3 heart cards, but I don't understand how/why this is true. Shouldn't each of the remaining 5 cards have an equal probability of being in either hand with a 3-2 split?
I think you are mixing up two different statments.
The $HQ$ is equally likely to be $1$ of the $5$ outstanding hearts, so it has probability of $3/5$ to be in the hand with $3$ hearts for the given $3-2$ break.
Given both $E$ and $W$ has equal numbers of vacant spaces. Then they are equally likely to have the $3$ hearts hand, and thus this implies the $HQ$ is equally likely to be held by $E$ or $W$.
This statement does not contradict to the statement above. The latter statement states the marginally probability of $HQ$ to be held by $E$ or $W$, but the first statement states the probability of $HQ$ to be held by the $3$-hearts hand - it does not specify whether $E$ or $W$ to hold the $3$-hearts hand.