Probability to draw one set of marbles before another set
$begingroup$
I have a bag of 15 marbles - 5 white, 10 black.
White wins if 3 white marbles are drawn before 4 black marbles are drawn. Black wins if 4 black marbles are drawn first.
How do I calculate the odds white/black win this game?
probability combinatorics
$endgroup$
add a comment |
$begingroup$
I have a bag of 15 marbles - 5 white, 10 black.
White wins if 3 white marbles are drawn before 4 black marbles are drawn. Black wins if 4 black marbles are drawn first.
How do I calculate the odds white/black win this game?
probability combinatorics
$endgroup$
add a comment |
$begingroup$
I have a bag of 15 marbles - 5 white, 10 black.
White wins if 3 white marbles are drawn before 4 black marbles are drawn. Black wins if 4 black marbles are drawn first.
How do I calculate the odds white/black win this game?
probability combinatorics
$endgroup$
I have a bag of 15 marbles - 5 white, 10 black.
White wins if 3 white marbles are drawn before 4 black marbles are drawn. Black wins if 4 black marbles are drawn first.
How do I calculate the odds white/black win this game?
probability combinatorics
probability combinatorics
asked Dec 3 '18 at 17:38
Mike Mike
1
1
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1 Answer
1
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votes
$begingroup$
Some thoughts to get you started
White can win in as little as $3$ draws and at most $6$ draws, otherwise we would have had $4$ black marbles drawn before the third white
Black can win in as little as $4$ draws and at most $6$ draws, otherwise we would have had $3$ white marbles drawn before the fourth black.
Notice no game will ever go $7$ draws or more.
So all that needs to be done is to look at the four cases white can win and the three cases black can win and compare the two.
$endgroup$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Some thoughts to get you started
White can win in as little as $3$ draws and at most $6$ draws, otherwise we would have had $4$ black marbles drawn before the third white
Black can win in as little as $4$ draws and at most $6$ draws, otherwise we would have had $3$ white marbles drawn before the fourth black.
Notice no game will ever go $7$ draws or more.
So all that needs to be done is to look at the four cases white can win and the three cases black can win and compare the two.
$endgroup$
add a comment |
$begingroup$
Some thoughts to get you started
White can win in as little as $3$ draws and at most $6$ draws, otherwise we would have had $4$ black marbles drawn before the third white
Black can win in as little as $4$ draws and at most $6$ draws, otherwise we would have had $3$ white marbles drawn before the fourth black.
Notice no game will ever go $7$ draws or more.
So all that needs to be done is to look at the four cases white can win and the three cases black can win and compare the two.
$endgroup$
add a comment |
$begingroup$
Some thoughts to get you started
White can win in as little as $3$ draws and at most $6$ draws, otherwise we would have had $4$ black marbles drawn before the third white
Black can win in as little as $4$ draws and at most $6$ draws, otherwise we would have had $3$ white marbles drawn before the fourth black.
Notice no game will ever go $7$ draws or more.
So all that needs to be done is to look at the four cases white can win and the three cases black can win and compare the two.
$endgroup$
Some thoughts to get you started
White can win in as little as $3$ draws and at most $6$ draws, otherwise we would have had $4$ black marbles drawn before the third white
Black can win in as little as $4$ draws and at most $6$ draws, otherwise we would have had $3$ white marbles drawn before the fourth black.
Notice no game will ever go $7$ draws or more.
So all that needs to be done is to look at the four cases white can win and the three cases black can win and compare the two.
answered Dec 3 '18 at 17:48
WaveXWaveX
2,6922722
2,6922722
add a comment |
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