Probability of drawing the nearest room card in Cluedo In the game Cluedo you start from a specific position on the map which you chose before looking at the cards you have. What is the probability that you have the card of the a)nearest room and b)both of the two nearest rooms to your starting character. (Since most of the time people would not go to a room which they have the card of, the strategy my calculations are based on)
additional info

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*2 player game, i.e. each player gets 9 cards

*I have calculated the probability of getting every hand of cards in cluedo, (i.e. what is the probability of having 8room 1weapon 0person, 7room 2weapon 0person etc. however the specific room or person or weapon is not defined). I am then calculating the probability of having the nearest room for every set of cards.

*I think that for the nearest room it should be 1/9 chance that you have the nearest room? if you have 1 room card in your hand. then 2/9 chance that you have the nearest room when you have 2 room cards in your hand, and so on and so forth

*Previously I was using the calculations of (1c1 x 7c7 )/ 9c8 to find the probability of having the nearest room when you have 8 room cards, then (1c1 x7c6)/9c7 when you have 7 room cards, etc. But the values did not make sense as it cannot be that there is a 1/9 chance of having the nearest room card if you have 8 room cards....

*Another calculation I used was (1c1 x 7c7)/ 8c8, (1c1 x 7x6)/(8c7)... Which made the probability of having the nearest room 1 when you had the 8 room cards. This is also wrong because the nearest room card could be the murder card which you do not own.

*For the calculation of having both the nearest rooms, I was previously using (1c1 x 1c1) / 9c2, (1c1 x 1c1 x 6c1) / 9c3 etc...But since using this method for one room proves to be wrong, I think that this is wrong too.

Another problem I am facing with my calculations is that I had drawn a tree diagram and then it was such that having 2 nearest rooms was an event after having the nearest room (i.e. I multiplied both together). However, I suspect that since they are not independent events, since having the two nearest rooms involve having the nearest room. I think that they might need to be in the same column of the tree diagram? But I am not sure.

 A: Let's use some conditional probability.
Let $A$ be the event the closest room is chosen as the secret room (it is not in the deck anymore). Let $B$ be the event that the closest room is in your hand. Obviously:
$$P(A\cap B)=0$$
Thus, we have the formula:
$$\begin{align*}P(B) & =P(B|A)P(A)+P(B|\neg A)P(\neg A)\\ & =P(B|\neg A)P(\neg A)\end{align*}$$
And this is much easier to calculate. $$P(\neg A)=\dfrac{8}{9}$$
And $P(B|\neg A)$ is found by assuming a room other than the closest room is removed and two other cards are also removed. This leaves $18$ distinct cards ($8$ rooms, $5$ people, $5$ weapons).
$$P(B|\neg A)=\dfrac{\dbinom{1}{1}\dbinom{17}{8}}{\dbinom{18}{9}}=\dfrac{1}{2}$$
So, by the formula above, we have:
$$P(B)=\dfrac{1}{2}\cdot \dfrac{8}{9}=\dfrac{4}{9}$$
If $C$ is the event that one of the two nearest rooms is the secret room, and $D$ is the event that both the nearest two rooms are in your hand, we similarly have probabilities:
$$P(\neg C)=\dfrac{7}{9}$$
$$P(D|\neg C)=\dfrac{\dbinom{2}{2}\dbinom{16}{7}}{\dbinom{18}{9}}=\dfrac{4}{17}$$
So, the probability both of the nearest rooms are in your starting hand is:
$$P(D)=\dfrac{4}{17}\cdot \dfrac{7}{9}=\dfrac{28}{153}$$
Note: while your method is possible, it is very easy to make arithmetic errors. This method simplifies the problem significantly so that the probabilities are much easier to calculate.
