Odds of Winning

[blimbert]

Hello - can you please help me calculate the odds of being selected for an upgraded hotel room? The rooms are randomly assigned at check in.

There are 5 rooms of the same price. 2 of these are deemed "prestige" (which have higher perceived value (rooftop, view, etc)), the other 3 are of lesser perceived value.

Given that the room assignment at check in is random:
1). What are the odds of receiving on of the "prestige" rooms if I make 1 reservation?

2). What are the odds of receiving on of the "prestige" rooms if I make 2 reservations?

Thank you!

 

[pka]

(H)elp me calculate the odds of being selected for an upgraded hotel room? The rooms are randomly assigned at check in.
There are 5 rooms of the same price. 2 of these are deemed "prestige" (which have higher perceived value (rooftop, view, etc)), the other 3 are of lesser perceived value.
Given that the room assignment at check in is random:
1). What are the odds of receiving on of the "prestige" rooms if I make 1 reservation?
2). What are the odds of receiving on of the "prestige" rooms if I make 2 reservations?

1) \(\mathcal{P}(prestige)=\dfrac{2}{5}\) WHY?
For #2) do you mean two prestige or do you mean at least one prestige?

 

[blimbert]

Excellent point. It is meant to get just 1 of 2 prestige.

 

[Deleted member 4993]

Excellent point. It is meant to get just 1 of 2 prestige.

Do you want to exclude the chance of getting both prestige rooms? The choices are:

• both are regular rooms
• at least one is prestige room where
  • exactly one is prestige room
  • both are prestige rooms

 

[pka]

Excellent point. It is meant to get just 1 of 2 prestige.

\(\mathcal{P}(prestige_1, regular_1 )+\mathcal{P}(regular_2, prestige_2)=\dfrac{3}{36}+\dfrac{3}{36}=\dfrac{1}{6}\)

 

[blimbert]

Do you want to exclude the chance of getting both prestige rooms? The choices are:

• both are regular rooms
• at least one is prestige room where
  • exactly one is prestige room
  • both are prestige rooms

Yes, I do not need to know the odds of getting both prestige rooms, only need one of them. Thank you!

 

[blimbert]

\(\mathcal{P}(prestige_1, regular_1 )+\mathcal{P}(regular_2, prestige_2)=\dfrac{3}{36}+\dfrac{3}{36}=\dfrac{1}{6}\)

Thank you, however I do not understand the result? If the answer to the first question was 2/5, why would the odds decrease if I made two reservations and only want one prestige?

 

[pka]

Thank you, however I do not understand the result? If the answer to the first question was 2/5, why would the odds decrease if I made two reservations and only want one prestige?

Sorry that was a awful job of counting. It should be \(\dfrac{6}{25}+\dfrac{6}{25}=\dfrac{12}{25}\)

 

[blimbert]

Sorry that was a awful job of counting. It should be \(\dfrac{6}{25}+\dfrac{6}{25}=\dfrac{12}{25}\)

Thank you! So that I may understand and learn, can you please explain how you arrived at the numerator and denominator?

 

[pka]

Thank you! So that I may understand and learn, can you please explain how you arrived at the numerator and denominator?

I considered the ordered pairs formed by \(\{P.P.R,R,R\}\times\{P.P.R,R,R\}\) there are 25 pairs.
You wanted exactly one P from two reservations; there are three \((P,R)'s\) three \((R,P)'s\).

 

[blimbert]

I considered the ordered pairs formed by \(\{P.P.R,R,R\}\times\{P.P.R,R,R\}\) there are 25 pairs.
You wanted exactly one P from two reservations; there are three \((P,R)'s\) three \((R,P)'s\).

THANK YOU!