Mixing
Probability of coin flip
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I was wondering about the probability of coin flips. How many times would i need to toss a coin to have at least 75% chance of getting a head.
I know the answer would be $1-left(left(frac{1}{2}right)^2right)$ = 2 throws but how would you work it out without trial and error.
probability
asked Dec 7 '16 at 3:00
mattmatt
103
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add a comment |
$begingroup$
I was wondering about the probability of coin flips. How many times would i need to toss a coin to have at least 75% chance of getting a head.
I know the answer would be $1-left(left(frac{1}{2}right)^2right)$ = 2 throws but how would you work it out without trial and error.
probability
asked Dec 7 '16 at 3:00
mattmatt
103
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One flip you gain .5, another flip you get half of that which makes .75 so 2
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– suomynonA
Dec 7 '16 at 3:01
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Log(1-3/4)/log(1/2)?
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– Kitter Catter
Dec 7 '16 at 3:06
add a comment |
$begingroup$
I was wondering about the probability of coin flips. How many times would i need to toss a coin to have at least 75% chance of getting a head.
I know the answer would be $1-left(left(frac{1}{2}right)^2right)$ = 2 throws but how would you work it out without trial and error.
probability
asked Dec 7 '16 at 3:00
mattmatt
103
$endgroup$
I was wondering about the probability of coin flips. How many times would i need to toss a coin to have at least 75% chance of getting a head.
I know the answer would be $1-left(left(frac{1}{2}right)^2right)$ = 2 throws but how would you work it out without trial and error.
probability
probability
asked Dec 7 '16 at 3:00
mattmatt
103
asked Dec 7 '16 at 3:00
mattmatt
103
asked Dec 7 '16 at 3:00
mattmatt
103
asked Dec 7 '16 at 3:00
mattmatt
103
asked Dec 7 '16 at 3:00
mattmatt
103
103
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One flip you gain .5, another flip you get half of that which makes .75 so 2
$endgroup$
– suomynonA
Dec 7 '16 at 3:01
$begingroup$
Log(1-3/4)/log(1/2)?
$endgroup$
– Kitter Catter
Dec 7 '16 at 3:06
add a comment |
$begingroup$
One flip you gain .5, another flip you get half of that which makes .75 so 2
$endgroup$
– suomynonA
Dec 7 '16 at 3:01
$begingroup$
Log(1-3/4)/log(1/2)?
$endgroup$
– Kitter Catter
Dec 7 '16 at 3:06
$begingroup$
One flip you gain .5, another flip you get half of that which makes .75 so 2
$endgroup$
– suomynonA
Dec 7 '16 at 3:01
$begingroup$
One flip you gain .5, another flip you get half of that which makes .75 so 2
$endgroup$
– suomynonA
Dec 7 '16 at 3:01
$begingroup$
Log(1-3/4)/log(1/2)?
$endgroup$
– Kitter Catter
Dec 7 '16 at 3:06
$begingroup$
Log(1-3/4)/log(1/2)?
$endgroup$
– Kitter Catter
Dec 7 '16 at 3:06
add a comment |
2 Answers
2
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You can use complementary counting.
Having at least a $75%$ chance of getting a head is equivalent to a $(100-75)%=25%$ or lower chance of getting no heads.
Getting no heads means flipping all tails. The probability of flipping $k$ consecutive tails is $frac{1}{2^k}$. We want this probability to be less than or equal to $25%$ (or $frac{1}{4}$).
$k=2$ is the smallest solution, so you would need to toss the coin twice.
answered Dec 7 '16 at 4:16
JedJed
719414
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add a comment |
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If you want to get a%, you need $log_{1over2}1-{aover100}$ times, then find the number that is closest to it that is greater than it.
answered Jan 22 at 0:47
Math LoverMath Lover
15910
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2 Answers
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2 Answers
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$begingroup$
You can use complementary counting.
Having at least a $75%$ chance of getting a head is equivalent to a $(100-75)%=25%$ or lower chance of getting no heads.
Getting no heads means flipping all tails. The probability of flipping $k$ consecutive tails is $frac{1}{2^k}$. We want this probability to be less than or equal to $25%$ (or $frac{1}{4}$).
$k=2$ is the smallest solution, so you would need to toss the coin twice.
answered Dec 7 '16 at 4:16
JedJed
719414
$endgroup$
add a comment |
$begingroup$
You can use complementary counting.
Having at least a $75%$ chance of getting a head is equivalent to a $(100-75)%=25%$ or lower chance of getting no heads.
Getting no heads means flipping all tails. The probability of flipping $k$ consecutive tails is $frac{1}{2^k}$. We want this probability to be less than or equal to $25%$ (or $frac{1}{4}$).
$k=2$ is the smallest solution, so you would need to toss the coin twice.
answered Dec 7 '16 at 4:16
JedJed
719414
$endgroup$
add a comment |
$begingroup$
You can use complementary counting.
Having at least a $75%$ chance of getting a head is equivalent to a $(100-75)%=25%$ or lower chance of getting no heads.
Getting no heads means flipping all tails. The probability of flipping $k$ consecutive tails is $frac{1}{2^k}$. We want this probability to be less than or equal to $25%$ (or $frac{1}{4}$).
$k=2$ is the smallest solution, so you would need to toss the coin twice.
answered Dec 7 '16 at 4:16
JedJed
719414
$endgroup$
You can use complementary counting.
Having at least a $75%$ chance of getting a head is equivalent to a $(100-75)%=25%$ or lower chance of getting no heads.
Getting no heads means flipping all tails. The probability of flipping $k$ consecutive tails is $frac{1}{2^k}$. We want this probability to be less than or equal to $25%$ (or $frac{1}{4}$).
$k=2$ is the smallest solution, so you would need to toss the coin twice.
answered Dec 7 '16 at 4:16
JedJed
719414
answered Dec 7 '16 at 4:16
JedJed
719414
answered Dec 7 '16 at 4:16
JedJed
719414
answered Dec 7 '16 at 4:16
JedJed
719414
719414
add a comment |
add a comment |
$begingroup$
If you want to get a%, you need $log_{1over2}1-{aover100}$ times, then find the number that is closest to it that is greater than it.
answered Jan 22 at 0:47
Math LoverMath Lover
15910
$endgroup$
add a comment |
$begingroup$
If you want to get a%, you need $log_{1over2}1-{aover100}$ times, then find the number that is closest to it that is greater than it.
answered Jan 22 at 0:47
Math LoverMath Lover
15910
$endgroup$
add a comment |
$begingroup$
If you want to get a%, you need $log_{1over2}1-{aover100}$ times, then find the number that is closest to it that is greater than it.
answered Jan 22 at 0:47
Math LoverMath Lover
15910
$endgroup$
If you want to get a%, you need $log_{1over2}1-{aover100}$ times, then find the number that is closest to it that is greater than it.
answered Jan 22 at 0:47
Math LoverMath Lover
15910
answered Jan 22 at 0:47
Math LoverMath Lover
15910
answered Jan 22 at 0:47
Math LoverMath Lover
15910
answered Jan 22 at 0:47
Math LoverMath Lover
15910
15910
add a comment |
add a comment |
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$begingroup$
One flip you gain .5, another flip you get half of that which makes .75 so 2
$endgroup$
– suomynonA
Dec 7 '16 at 3:01
$begingroup$
Log(1-3/4)/log(1/2)?
$endgroup$
– Kitter Catter
Dec 7 '16 at 3:06