I have two probability-related problems I can't quite formulate. I would appreciate help from any Tempers with related knowledge. There's a Temper with a user title "mad statistician" or something along those lines, maybe he could help.
Anywhoo, the problems:
1
EDIT: Solved, thanks to Ericthegreat
2
A card game is played using a deck of 52 cards (poker deck). There are four players, each player is dealt 10 cards, and they can only look at the top 4 of those cards. Out of the remaining cards, one card is flipped over and shown. Players can then place one of their face-up cards on the revealed card, but only if the number on the card is one less or one more than the number on the revealed card (cards having values from 1 to 13, in 4 colours). What are the odds that out of the players' 16 face-up cards, not a single card can be placed on the revealed card?
So it's 40 cards out of 52, that's the binomial coefficient, then there's 16 out of 40 and 1 out of the remaining 12, so the question is, what are the odds that all the 16 revealed cards will be numbers 2 or more below or above, or the same as, the number of the revealed card? The revealed card is 1 out of 52; if it's a 1 or a 13 there are 4 cards that can be placed on it (2s in any color and 12s in any color, respectively), for the rest there are 8 that can be placed on it, What are the odds that out of the 16 face-up cards, none can be placed on the revealed card?
Can I ignore the order in which I draw the cards, and say I reveal the card first and then deal the rest (13 possible strengths)? In that case I have 16 out of 51 cards (binomial) face up, 4 or 8 of which can be played (depending on the first one). Do i separate the possibilities? I have 4 times 2 cases where only 4 cards can be played (if available), and 11 times 4 cases where 8 cards can be played, the odds of the first case are 2 against 13, the odds of the second case are 11 against 13, so i calculate conditional probability for the face-up cards for the two separate sets of cases and then add them, or do I multiply them by 2/13 and 11/13 respectively? I'm lost.
And an additional question, what about a deck of 32 cards (8 strengths, 4 colours), 6 cards dealt to each player (3 face up, 3 face down)?
Halp pls, thank you
Anywhoo, the problems:
1
EDIT: Solved, thanks to Ericthegreat
There's a boarding queue at an airport, 100 passengers are waiting to fill 100 seats. The first passenger is a loud obnoxious asshole that doesn't care about reservations, and he will take a random seat. The passenger whose seat that was will have to take a random empty seat. The passenger whose seat that one was will have to take a random empty seat too, and so on, until the last passenger is seated. What are the odds that the last passenger ends up in his allocated seat (the seat that he reserved)?
Passengers in between the one who took a random seat and the one whose seat it was take their allocated seats. This is important.
I have no idea where to start. I was doing fine until I realized that last point
The asshole has a 1/100 chance of sitting in his own seat. If he doesn't, the guy whose seat was taken (1/99 chance of it being the second passenger, third, fourth etc.) has (101 - his place in line) places to choose from, and that's a 1/(101-his place in line) chance of taking the asshole's place, and they cancel each other out. If not, the next person has (101-his place in line) places to choose from, and so on, until one of them sits in the asshole's seat and the mixup is neutralized and the rest of the line is seated as scheduled.
Passengers in between the one who took a random seat and the one whose seat it was take their allocated seats. This is important.
I have no idea where to start. I was doing fine until I realized that last point
The asshole has a 1/100 chance of sitting in his own seat. If he doesn't, the guy whose seat was taken (1/99 chance of it being the second passenger, third, fourth etc.) has (101 - his place in line) places to choose from, and that's a 1/(101-his place in line) chance of taking the asshole's place, and they cancel each other out. If not, the next person has (101-his place in line) places to choose from, and so on, until one of them sits in the asshole's seat and the mixup is neutralized and the rest of the line is seated as scheduled.
2
A card game is played using a deck of 52 cards (poker deck). There are four players, each player is dealt 10 cards, and they can only look at the top 4 of those cards. Out of the remaining cards, one card is flipped over and shown. Players can then place one of their face-up cards on the revealed card, but only if the number on the card is one less or one more than the number on the revealed card (cards having values from 1 to 13, in 4 colours). What are the odds that out of the players' 16 face-up cards, not a single card can be placed on the revealed card?
So it's 40 cards out of 52, that's the binomial coefficient, then there's 16 out of 40 and 1 out of the remaining 12, so the question is, what are the odds that all the 16 revealed cards will be numbers 2 or more below or above, or the same as, the number of the revealed card? The revealed card is 1 out of 52; if it's a 1 or a 13 there are 4 cards that can be placed on it (2s in any color and 12s in any color, respectively), for the rest there are 8 that can be placed on it, What are the odds that out of the 16 face-up cards, none can be placed on the revealed card?
Can I ignore the order in which I draw the cards, and say I reveal the card first and then deal the rest (13 possible strengths)? In that case I have 16 out of 51 cards (binomial) face up, 4 or 8 of which can be played (depending on the first one). Do i separate the possibilities? I have 4 times 2 cases where only 4 cards can be played (if available), and 11 times 4 cases where 8 cards can be played, the odds of the first case are 2 against 13, the odds of the second case are 11 against 13, so i calculate conditional probability for the face-up cards for the two separate sets of cases and then add them, or do I multiply them by 2/13 and 11/13 respectively? I'm lost.
And an additional question, what about a deck of 32 cards (8 strengths, 4 colours), 6 cards dealt to each player (3 face up, 3 face down)?
Halp pls, thank you