Modelling Mafia: pick a ball without replacement while also removing a non-distinct ball
$begingroup$
I worded this question before quite poorly. This is a model of the party game mafia. Drawing without replacement represents killing off players.
Let's say I have $N$ balls (players), of which $k$ are red (mafia) and the rest are blue (innocent civilians). The probability that I pick all the red balls without replacement in $n$ attempts is:
$frac{binom{k}{k}*binom{N-k}{n-k}}{binom{N}{n}}$
right? But what if I peek into the bag and discard one blue ball after each random draw, regardless of what was chosen? This simulates the mafia players killing an innocent civilian each round. We're assuming that the mafia never kill other mafia members.
So yes, this means that you remove a blue ball after each red or blue draw. If you randomly draw a blue ball, then you draw and remove a second blue ball as well.
What would the probability be then?
probability
asked Jan 21 at 17:26
EliotEliot
112
$endgroup$
add a comment |
$begingroup$
I worded this question before quite poorly. This is a model of the party game mafia. Drawing without replacement represents killing off players.
Let's say I have $N$ balls (players), of which $k$ are red (mafia) and the rest are blue (innocent civilians). The probability that I pick all the red balls without replacement in $n$ attempts is:
$frac{binom{k}{k}*binom{N-k}{n-k}}{binom{N}{n}}$
right? But what if I peek into the bag and discard one blue ball after each random draw, regardless of what was chosen? This simulates the mafia players killing an innocent civilian each round. We're assuming that the mafia never kill other mafia members.
So yes, this means that you remove a blue ball after each red or blue draw. If you randomly draw a blue ball, then you draw and remove a second blue ball as well.
What would the probability be then?
probability
asked Jan 21 at 17:26
EliotEliot
112
$endgroup$
add a comment |
$begingroup$
I worded this question before quite poorly. This is a model of the party game mafia. Drawing without replacement represents killing off players.
Let's say I have $N$ balls (players), of which $k$ are red (mafia) and the rest are blue (innocent civilians). The probability that I pick all the red balls without replacement in $n$ attempts is:
$frac{binom{k}{k}*binom{N-k}{n-k}}{binom{N}{n}}$
right? But what if I peek into the bag and discard one blue ball after each random draw, regardless of what was chosen? This simulates the mafia players killing an innocent civilian each round. We're assuming that the mafia never kill other mafia members.
So yes, this means that you remove a blue ball after each red or blue draw. If you randomly draw a blue ball, then you draw and remove a second blue ball as well.
What would the probability be then?
probability
asked Jan 21 at 17:26
EliotEliot
112
$endgroup$
I worded this question before quite poorly. This is a model of the party game mafia. Drawing without replacement represents killing off players.
Let's say I have $N$ balls (players), of which $k$ are red (mafia) and the rest are blue (innocent civilians). The probability that I pick all the red balls without replacement in $n$ attempts is:
$frac{binom{k}{k}*binom{N-k}{n-k}}{binom{N}{n}}$
right? But what if I peek into the bag and discard one blue ball after each random draw, regardless of what was chosen? This simulates the mafia players killing an innocent civilian each round. We're assuming that the mafia never kill other mafia members.
So yes, this means that you remove a blue ball after each red or blue draw. If you randomly draw a blue ball, then you draw and remove a second blue ball as well.
What would the probability be then?
probability
probability
asked Jan 21 at 17:26
EliotEliot
112
asked Jan 21 at 17:26
EliotEliot
112
asked Jan 21 at 17:26
EliotEliot
112
asked Jan 21 at 17:26
EliotEliot
112
asked Jan 21 at 17:26
EliotEliot
112
112
add a comment |
add a comment |
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