Runs

[ddubois]

Given a deck of size X, with Y red cards, and X-Y blue cards, if we were to pull cards off the top of the deck until we hit a blue card, what's the average number of cards we will pull? What's the probability that we will reveal Z red cards before we hit our first blue card?

I think the answer to the second question can be computed like:
Y/X * (Y-1)/(X-1) ... * (Y-Z)/(X-Z)
I assume this shortens to some simple factorial, maybe:
Y!/X! * (X-Z-1)!/(Y-Z-1)!
But the answer to the first question I don't have a clue how to do.

By the way, the typical scenario in practise for the game I'm playing would usually be around 30-45 cards in the deck, and somewhere between 4 and 10 blue cards.

PS: If you guessed this was Goblin Charbelcher math, you win the prize.

[MikeL05]

Haven't heard that name in at least a year. How about playing a real deck that doesn't rely on stupid stuff like this? (i.e. why play poker when you can play chess)

Want an answer? Here's an equation for you:

-(Time) + -(Money) + (Girls*0) = a really bad game

That said, it is my favorite game of all time and I miss it terribly.

[BruceZ]

[ QUOTE ]
Given a deck of size X, with Y red cards, and X-Y blue cards, if we were to pull cards off the top of the deck until we hit a blue card, what's the average number of cards we will pull?

[/ QUOTE ]

Each red card has a probability of 1/(X-Y+1) of coming before all of the X-Y blue cards. The average number of red cards that come before the first blue card is Y/(X-Y+1). The average number of cards that we pull, including the first blue card, is Y/(X-Y+1) + 1 = (X+1)/(X-Y+1).


[ QUOTE ]
What's the probability that we will reveal Z red cards before we hit our first blue card?

I think the answer to the second question can be computed like:
Y/X * (Y-1)/(X-1) ... * (Y-Z)/(X-Z)
I assume this shortens to some simple factorial, maybe:
Y!/X! * (X-Z-1)!/(Y-Z-1)!

[/ QUOTE ]

Close, you just went 1 too far:

Y/X * (Y-1)/(X-1) ... * (Y-Z+1)/(X-Z+1)

= Y!/X! * (X-Z)!/(Y-Z)!

= P(Y,Z)/P(X,Z) = C(Y,Z)/C(X,Z)