#1
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Runs
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. |
#2
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Re: Runs
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. |
#3
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Re: Runs
[ 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) |
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