Mixing
Randomly picking event outcome
$begingroup$
I had a thought today about randomly forecasting an event which is inherently random, which led to a lot of confusion.
Relating this to something trivial. If you flip a coin 10 times and you limit your choice to either heads only or tails only, you would obviously expect to be correct 5 times. But if you were to choose between heads/tails randomly, should this effect the expected value?
If you select heads or tails randomly, then are you dealing with two independent events? i.e. Is the probability of guessing correctly $P(A;and;B)$, where $P(A)$ is your pick of heads or tails and $P(B)$ is the outcome of the coin flip?
I'm dubious of my approach, but I applied the following to test my idea (in python):
import numpy as np
guess = np.random.choice(2, 1000000, .5)
flip_outcome = np.random.choice(2, 1000000, .5)
result = (guess * flip_outcome).mean()
Rounded to two decimal places, result = 0.25
If your selection of heads or tails is random, would this actually affect the outcome? Before writing this post, I experimented with this idea with an actual coin. Using python's numpy, I generated a random array (array([0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0]) to dictate my selection of either heads (1) or tails (0) in a 20 flip experiment. I then repeated this process 50 times and was able to predict the outcome with approximately 27% accuracy.
I feel like I'm missing something fundamental here, or accounting for something that doesn't need to be accounted for?
probability
asked Jan 24 at 13:47
EBBEBB
1012
$endgroup$
add a comment |
$begingroup$
I had a thought today about randomly forecasting an event which is inherently random, which led to a lot of confusion.
Relating this to something trivial. If you flip a coin 10 times and you limit your choice to either heads only or tails only, you would obviously expect to be correct 5 times. But if you were to choose between heads/tails randomly, should this effect the expected value?
If you select heads or tails randomly, then are you dealing with two independent events? i.e. Is the probability of guessing correctly $P(A;and;B)$, where $P(A)$ is your pick of heads or tails and $P(B)$ is the outcome of the coin flip?
I'm dubious of my approach, but I applied the following to test my idea (in python):
import numpy as np
guess = np.random.choice(2, 1000000, .5)
flip_outcome = np.random.choice(2, 1000000, .5)
result = (guess * flip_outcome).mean()
Rounded to two decimal places, result = 0.25
If your selection of heads or tails is random, would this actually affect the outcome? Before writing this post, I experimented with this idea with an actual coin. Using python's numpy, I generated a random array (array([0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0]) to dictate my selection of either heads (1) or tails (0) in a 20 flip experiment. I then repeated this process 50 times and was able to predict the outcome with approximately 27% accuracy.
I feel like I'm missing something fundamental here, or accounting for something that doesn't need to be accounted for?
probability
asked Jan 24 at 13:47
EBBEBB
1012
$endgroup$
$begingroup$
I don't understand how your simulation could have such a poor outcome. There are two scenarios in which you correctly identify the toss: It's $H$ and you guess $H$, it's $T$ and you guess $T$. Each has probability $.25$ and the events are mutually exclusive so the answer is $.5$
$endgroup$
– lulu
Jan 24 at 13:50
$begingroup$
Perhaps I have misunderstood your question. I read it as saying that you use the toss of a fair coin to guess the outcome of another coin. In truth, it doesn't even matter if the other coin is biased (so long as yours is fair). If the second coin is $H$ with probability $p$ then you guess $H$ when it is $H$ with probability $.5times p$ and you guess $T$ when it is $T$ with probability $.5times (1-p)$, and these always sum to $.5$ Or were you asking something different?
$endgroup$
– lulu
Jan 24 at 13:58
1
$begingroup$
Assume $1$ represents heads and $0$ tails. You are only counting it a success if you choose heads and the outcome is heads. If you choose tails and it lands tails, you are computingresult=0*0=0
$endgroup$
– saulspatz
Jan 24 at 14:00
$begingroup$
Thanks @lulu, as pointed out I was ignoring successful TT results.
$endgroup$
– EBB
Jan 24 at 14:36
$begingroup$
Aha thanks saulspatz, amended with(guess == flip_outcome).mean()which equates to .5 as expected. Thanks!
$endgroup$
– EBB
Jan 24 at 14:44
add a comment |
$begingroup$
I had a thought today about randomly forecasting an event which is inherently random, which led to a lot of confusion.
Relating this to something trivial. If you flip a coin 10 times and you limit your choice to either heads only or tails only, you would obviously expect to be correct 5 times. But if you were to choose between heads/tails randomly, should this effect the expected value?
If you select heads or tails randomly, then are you dealing with two independent events? i.e. Is the probability of guessing correctly $P(A;and;B)$, where $P(A)$ is your pick of heads or tails and $P(B)$ is the outcome of the coin flip?
I'm dubious of my approach, but I applied the following to test my idea (in python):
import numpy as np
guess = np.random.choice(2, 1000000, .5)
flip_outcome = np.random.choice(2, 1000000, .5)
result = (guess * flip_outcome).mean()
Rounded to two decimal places, result = 0.25
If your selection of heads or tails is random, would this actually affect the outcome? Before writing this post, I experimented with this idea with an actual coin. Using python's numpy, I generated a random array (array([0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0]) to dictate my selection of either heads (1) or tails (0) in a 20 flip experiment. I then repeated this process 50 times and was able to predict the outcome with approximately 27% accuracy.
I feel like I'm missing something fundamental here, or accounting for something that doesn't need to be accounted for?
probability
asked Jan 24 at 13:47
EBBEBB
1012
$endgroup$
I had a thought today about randomly forecasting an event which is inherently random, which led to a lot of confusion.
Relating this to something trivial. If you flip a coin 10 times and you limit your choice to either heads only or tails only, you would obviously expect to be correct 5 times. But if you were to choose between heads/tails randomly, should this effect the expected value?
If you select heads or tails randomly, then are you dealing with two independent events? i.e. Is the probability of guessing correctly $P(A;and;B)$, where $P(A)$ is your pick of heads or tails and $P(B)$ is the outcome of the coin flip?
I'm dubious of my approach, but I applied the following to test my idea (in python):
import numpy as np
guess = np.random.choice(2, 1000000, .5)
flip_outcome = np.random.choice(2, 1000000, .5)
result = (guess * flip_outcome).mean()
Rounded to two decimal places, result = 0.25
If your selection of heads or tails is random, would this actually affect the outcome? Before writing this post, I experimented with this idea with an actual coin. Using python's numpy, I generated a random array (array([0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0]) to dictate my selection of either heads (1) or tails (0) in a 20 flip experiment. I then repeated this process 50 times and was able to predict the outcome with approximately 27% accuracy.
I feel like I'm missing something fundamental here, or accounting for something that doesn't need to be accounted for?
probability
probability
asked Jan 24 at 13:47
EBBEBB
1012
asked Jan 24 at 13:47
EBBEBB
1012
asked Jan 24 at 13:47
EBBEBB
1012
asked Jan 24 at 13:47
EBBEBB
1012
asked Jan 24 at 13:47
EBBEBB
1012
1012
$begingroup$
I don't understand how your simulation could have such a poor outcome. There are two scenarios in which you correctly identify the toss: It's $H$ and you guess $H$, it's $T$ and you guess $T$. Each has probability $.25$ and the events are mutually exclusive so the answer is $.5$
$endgroup$
– lulu
Jan 24 at 13:50
$begingroup$
Perhaps I have misunderstood your question. I read it as saying that you use the toss of a fair coin to guess the outcome of another coin. In truth, it doesn't even matter if the other coin is biased (so long as yours is fair). If the second coin is $H$ with probability $p$ then you guess $H$ when it is $H$ with probability $.5times p$ and you guess $T$ when it is $T$ with probability $.5times (1-p)$, and these always sum to $.5$ Or were you asking something different?
$endgroup$
– lulu
Jan 24 at 13:58
1
$begingroup$
Assume $1$ represents heads and $0$ tails. You are only counting it a success if you choose heads and the outcome is heads. If you choose tails and it lands tails, you are computingresult=0*0=0
$endgroup$
– saulspatz
Jan 24 at 14:00
$begingroup$
Thanks @lulu, as pointed out I was ignoring successful TT results.
$endgroup$
– EBB
Jan 24 at 14:36
$begingroup$
Aha thanks saulspatz, amended with(guess == flip_outcome).mean()which equates to .5 as expected. Thanks!
$endgroup$
– EBB
Jan 24 at 14:44
add a comment |
$begingroup$
I don't understand how your simulation could have such a poor outcome. There are two scenarios in which you correctly identify the toss: It's $H$ and you guess $H$, it's $T$ and you guess $T$. Each has probability $.25$ and the events are mutually exclusive so the answer is $.5$
$endgroup$
– lulu
Jan 24 at 13:50
$begingroup$
Perhaps I have misunderstood your question. I read it as saying that you use the toss of a fair coin to guess the outcome of another coin. In truth, it doesn't even matter if the other coin is biased (so long as yours is fair). If the second coin is $H$ with probability $p$ then you guess $H$ when it is $H$ with probability $.5times p$ and you guess $T$ when it is $T$ with probability $.5times (1-p)$, and these always sum to $.5$ Or were you asking something different?
$endgroup$
– lulu
Jan 24 at 13:58
1
$begingroup$
Assume $1$ represents heads and $0$ tails. You are only counting it a success if you choose heads and the outcome is heads. If you choose tails and it lands tails, you are computingresult=0*0=0
$endgroup$
– saulspatz
Jan 24 at 14:00
$begingroup$
Thanks @lulu, as pointed out I was ignoring successful TT results.
$endgroup$
– EBB
Jan 24 at 14:36
$begingroup$
Aha thanks saulspatz, amended with(guess == flip_outcome).mean()which equates to .5 as expected. Thanks!
$endgroup$
– EBB
Jan 24 at 14:44
$begingroup$
I don't understand how your simulation could have such a poor outcome. There are two scenarios in which you correctly identify the toss: It's $H$ and you guess $H$, it's $T$ and you guess $T$. Each has probability $.25$ and the events are mutually exclusive so the answer is $.5$
$endgroup$
– lulu
Jan 24 at 13:50
$begingroup$
I don't understand how your simulation could have such a poor outcome. There are two scenarios in which you correctly identify the toss: It's $H$ and you guess $H$, it's $T$ and you guess $T$. Each has probability $.25$ and the events are mutually exclusive so the answer is $.5$
$endgroup$
– lulu
Jan 24 at 13:50
$begingroup$
Perhaps I have misunderstood your question. I read it as saying that you use the toss of a fair coin to guess the outcome of another coin. In truth, it doesn't even matter if the other coin is biased (so long as yours is fair). If the second coin is $H$ with probability $p$ then you guess $H$ when it is $H$ with probability $.5times p$ and you guess $T$ when it is $T$ with probability $.5times (1-p)$, and these always sum to $.5$ Or were you asking something different?
$endgroup$
– lulu
Jan 24 at 13:58
$begingroup$
Perhaps I have misunderstood your question. I read it as saying that you use the toss of a fair coin to guess the outcome of another coin. In truth, it doesn't even matter if the other coin is biased (so long as yours is fair). If the second coin is $H$ with probability $p$ then you guess $H$ when it is $H$ with probability $.5times p$ and you guess $T$ when it is $T$ with probability $.5times (1-p)$, and these always sum to $.5$ Or were you asking something different?
$endgroup$
– lulu
Jan 24 at 13:58
1
1
$begingroup$
Assume $1$ represents heads and $0$ tails. You are only counting it a success if you choose heads and the outcome is heads. If you choose tails and it lands tails, you are computing
result=0*0=0$endgroup$
– saulspatz
Jan 24 at 14:00
$begingroup$
Assume $1$ represents heads and $0$ tails. You are only counting it a success if you choose heads and the outcome is heads. If you choose tails and it lands tails, you are computing
result=0*0=0$endgroup$
– saulspatz
Jan 24 at 14:00
$begingroup$
Thanks @lulu, as pointed out I was ignoring successful TT results.
$endgroup$
– EBB
Jan 24 at 14:36
$begingroup$
Thanks @lulu, as pointed out I was ignoring successful TT results.
$endgroup$
– EBB
Jan 24 at 14:36
$begingroup$
Aha thanks saulspatz, amended with
(guess == flip_outcome).mean() which equates to .5 as expected. Thanks!$endgroup$
– EBB
Jan 24 at 14:44
$begingroup$
Aha thanks saulspatz, amended with
(guess == flip_outcome).mean() which equates to .5 as expected. Thanks!$endgroup$
– EBB
Jan 24 at 14:44
add a comment |
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$begingroup$
I don't understand how your simulation could have such a poor outcome. There are two scenarios in which you correctly identify the toss: It's $H$ and you guess $H$, it's $T$ and you guess $T$. Each has probability $.25$ and the events are mutually exclusive so the answer is $.5$
$endgroup$
– lulu
Jan 24 at 13:50
$begingroup$
Perhaps I have misunderstood your question. I read it as saying that you use the toss of a fair coin to guess the outcome of another coin. In truth, it doesn't even matter if the other coin is biased (so long as yours is fair). If the second coin is $H$ with probability $p$ then you guess $H$ when it is $H$ with probability $.5times p$ and you guess $T$ when it is $T$ with probability $.5times (1-p)$, and these always sum to $.5$ Or were you asking something different?
$endgroup$
– lulu
Jan 24 at 13:58
1
$begingroup$
Assume $1$ represents heads and $0$ tails. You are only counting it a success if you choose heads and the outcome is heads. If you choose tails and it lands tails, you are computing
result=0*0=0$endgroup$
– saulspatz
Jan 24 at 14:00
$begingroup$
Thanks @lulu, as pointed out I was ignoring successful TT results.
$endgroup$
– EBB
Jan 24 at 14:36
$begingroup$
Aha thanks saulspatz, amended with
(guess == flip_outcome).mean()which equates to .5 as expected. Thanks!$endgroup$
– EBB
Jan 24 at 14:44