Mixing
Probability of losing a game of chance.
$begingroup$
I am trying to work out the probabilities of losing a game of chance (1 in 2 chance per game) by sticking with the same answer, say heads on a coin toss.
I know there is a 1 in 2 chance of tails on the first game, then 1 of 2 possible outcomes again on the second game, and 3rd etc, but the chance of the same outcome repetitively is multiplied by the chance of each occurrence. So 2nd time 1/2 * 1/2 = 1/4, then 1/8 etc.
What I'm not sure about is, if that's the chance for each game, is the chance of losing all games multiplied together?
So to lose the first and second games you would need to multiply the 1/2 and 1/4 together giving 1/8, and to also lose the 3rd game you would also multiply by the 1/8 chance for that game. Giving 1/64. Chances for the 3rd game are a 1/8 chance of losing by sticking with heads following 2 tails outcomes, but are the chances of losing all 3 games 1 in 64?
It's 15 years since I did statistics in school and by the 4th game this gives 1/976, which seems too high to me, 1/16 seems more realistic, if anyone has a better understanding than me their input would be appreciated.
probability statistics
asked Sep 26 '15 at 22:33
ShaunShaun
11
$endgroup$
add a comment |
$begingroup$
I am trying to work out the probabilities of losing a game of chance (1 in 2 chance per game) by sticking with the same answer, say heads on a coin toss.
I know there is a 1 in 2 chance of tails on the first game, then 1 of 2 possible outcomes again on the second game, and 3rd etc, but the chance of the same outcome repetitively is multiplied by the chance of each occurrence. So 2nd time 1/2 * 1/2 = 1/4, then 1/8 etc.
What I'm not sure about is, if that's the chance for each game, is the chance of losing all games multiplied together?
So to lose the first and second games you would need to multiply the 1/2 and 1/4 together giving 1/8, and to also lose the 3rd game you would also multiply by the 1/8 chance for that game. Giving 1/64. Chances for the 3rd game are a 1/8 chance of losing by sticking with heads following 2 tails outcomes, but are the chances of losing all 3 games 1 in 64?
It's 15 years since I did statistics in school and by the 4th game this gives 1/976, which seems too high to me, 1/16 seems more realistic, if anyone has a better understanding than me their input would be appreciated.
probability statistics
asked Sep 26 '15 at 22:33
ShaunShaun
11
$endgroup$
$begingroup$
This may need clarification. Is not each game a single toss? But if so, then the probability of winning a given game is $frac 12$ regardless of what has come before (or what comes after). And where does $976$ come from? That isn't a power of $2$.
$endgroup$
– lulu
Sep 26 '15 at 22:36
$begingroup$
If i understand correctly you are wrong because every toss has $frac{1}{2}$ probability of being tail so the probability of losing two games (even if not sticking with the same choose) is $frac{1}{4}$ and in general the probability of losing all the first $n$ games is $frac{1}{2^n}$
$endgroup$
– karmalu
Sep 27 '15 at 12:06
add a comment |
$begingroup$
I am trying to work out the probabilities of losing a game of chance (1 in 2 chance per game) by sticking with the same answer, say heads on a coin toss.
I know there is a 1 in 2 chance of tails on the first game, then 1 of 2 possible outcomes again on the second game, and 3rd etc, but the chance of the same outcome repetitively is multiplied by the chance of each occurrence. So 2nd time 1/2 * 1/2 = 1/4, then 1/8 etc.
What I'm not sure about is, if that's the chance for each game, is the chance of losing all games multiplied together?
So to lose the first and second games you would need to multiply the 1/2 and 1/4 together giving 1/8, and to also lose the 3rd game you would also multiply by the 1/8 chance for that game. Giving 1/64. Chances for the 3rd game are a 1/8 chance of losing by sticking with heads following 2 tails outcomes, but are the chances of losing all 3 games 1 in 64?
It's 15 years since I did statistics in school and by the 4th game this gives 1/976, which seems too high to me, 1/16 seems more realistic, if anyone has a better understanding than me their input would be appreciated.
probability statistics
asked Sep 26 '15 at 22:33
ShaunShaun
11
$endgroup$
I am trying to work out the probabilities of losing a game of chance (1 in 2 chance per game) by sticking with the same answer, say heads on a coin toss.
I know there is a 1 in 2 chance of tails on the first game, then 1 of 2 possible outcomes again on the second game, and 3rd etc, but the chance of the same outcome repetitively is multiplied by the chance of each occurrence. So 2nd time 1/2 * 1/2 = 1/4, then 1/8 etc.
What I'm not sure about is, if that's the chance for each game, is the chance of losing all games multiplied together?
So to lose the first and second games you would need to multiply the 1/2 and 1/4 together giving 1/8, and to also lose the 3rd game you would also multiply by the 1/8 chance for that game. Giving 1/64. Chances for the 3rd game are a 1/8 chance of losing by sticking with heads following 2 tails outcomes, but are the chances of losing all 3 games 1 in 64?
It's 15 years since I did statistics in school and by the 4th game this gives 1/976, which seems too high to me, 1/16 seems more realistic, if anyone has a better understanding than me their input would be appreciated.
probability statistics
probability statistics
asked Sep 26 '15 at 22:33
ShaunShaun
11
asked Sep 26 '15 at 22:33
ShaunShaun
11
asked Sep 26 '15 at 22:33
ShaunShaun
11
asked Sep 26 '15 at 22:33
ShaunShaun
11
asked Sep 26 '15 at 22:33
ShaunShaun
11
11
$begingroup$
This may need clarification. Is not each game a single toss? But if so, then the probability of winning a given game is $frac 12$ regardless of what has come before (or what comes after). And where does $976$ come from? That isn't a power of $2$.
$endgroup$
– lulu
Sep 26 '15 at 22:36
$begingroup$
If i understand correctly you are wrong because every toss has $frac{1}{2}$ probability of being tail so the probability of losing two games (even if not sticking with the same choose) is $frac{1}{4}$ and in general the probability of losing all the first $n$ games is $frac{1}{2^n}$
$endgroup$
– karmalu
Sep 27 '15 at 12:06
add a comment |
$begingroup$
This may need clarification. Is not each game a single toss? But if so, then the probability of winning a given game is $frac 12$ regardless of what has come before (or what comes after). And where does $976$ come from? That isn't a power of $2$.
$endgroup$
– lulu
Sep 26 '15 at 22:36
$begingroup$
If i understand correctly you are wrong because every toss has $frac{1}{2}$ probability of being tail so the probability of losing two games (even if not sticking with the same choose) is $frac{1}{4}$ and in general the probability of losing all the first $n$ games is $frac{1}{2^n}$
$endgroup$
– karmalu
Sep 27 '15 at 12:06
$begingroup$
This may need clarification. Is not each game a single toss? But if so, then the probability of winning a given game is $frac 12$ regardless of what has come before (or what comes after). And where does $976$ come from? That isn't a power of $2$.
$endgroup$
– lulu
Sep 26 '15 at 22:36
$begingroup$
This may need clarification. Is not each game a single toss? But if so, then the probability of winning a given game is $frac 12$ regardless of what has come before (or what comes after). And where does $976$ come from? That isn't a power of $2$.
$endgroup$
– lulu
Sep 26 '15 at 22:36
$begingroup$
If i understand correctly you are wrong because every toss has $frac{1}{2}$ probability of being tail so the probability of losing two games (even if not sticking with the same choose) is $frac{1}{4}$ and in general the probability of losing all the first $n$ games is $frac{1}{2^n}$
$endgroup$
– karmalu
Sep 27 '15 at 12:06
$begingroup$
If i understand correctly you are wrong because every toss has $frac{1}{2}$ probability of being tail so the probability of losing two games (even if not sticking with the same choose) is $frac{1}{4}$ and in general the probability of losing all the first $n$ games is $frac{1}{2^n}$
$endgroup$
– karmalu
Sep 27 '15 at 12:06
add a comment |
1 Answer
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It comes down what you consider a win. Is it to get one of the tosses or all? I think you ask about getting everything right.
Lets say you stick to heads (H) and there are two games. Then all possible outcomes on your world are HH,HT,TH,TT and only HH is good for you. The probability you will win all games is $(1/2)^n$, for n=2 or 1/4. For n=3 it is $(1/2)^3=1/8$ etc
HOWEVER if you already are at the point that you have won the first game , then we are at a world that allows only HH and HT because the first turn is already played (this is called conditional: you are basing your estimations on the condition that something has already happened). Then your chances of winning this round conditional to the fact you already won all previous is 1/2. The same if you are asking about your chances of winning round n+1 conditional to the fact you won all previous rounds.
So every round, your chances are 50-50, but this is because you can only arrive to that round if you win all previous rounds. And for every player that arrived to round n, there will be $2^{n-1}$ players on average that never made it there. (for n>=2)
This means sticking makes no difference. Now, so that is not confused with a different question: if at every round you have more than one choices AND something is revealed to you (you get some additional information by say showing to you that there is nothing behind curtain 3, then this is a different problem)
answered Aug 23 '16 at 14:28
ntgntg
1236
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$begingroup$
It comes down what you consider a win. Is it to get one of the tosses or all? I think you ask about getting everything right.
Lets say you stick to heads (H) and there are two games. Then all possible outcomes on your world are HH,HT,TH,TT and only HH is good for you. The probability you will win all games is $(1/2)^n$, for n=2 or 1/4. For n=3 it is $(1/2)^3=1/8$ etc
HOWEVER if you already are at the point that you have won the first game , then we are at a world that allows only HH and HT because the first turn is already played (this is called conditional: you are basing your estimations on the condition that something has already happened). Then your chances of winning this round conditional to the fact you already won all previous is 1/2. The same if you are asking about your chances of winning round n+1 conditional to the fact you won all previous rounds.
So every round, your chances are 50-50, but this is because you can only arrive to that round if you win all previous rounds. And for every player that arrived to round n, there will be $2^{n-1}$ players on average that never made it there. (for n>=2)
This means sticking makes no difference. Now, so that is not confused with a different question: if at every round you have more than one choices AND something is revealed to you (you get some additional information by say showing to you that there is nothing behind curtain 3, then this is a different problem)
answered Aug 23 '16 at 14:28
ntgntg
1236
$endgroup$
add a comment |
$begingroup$
It comes down what you consider a win. Is it to get one of the tosses or all? I think you ask about getting everything right.
Lets say you stick to heads (H) and there are two games. Then all possible outcomes on your world are HH,HT,TH,TT and only HH is good for you. The probability you will win all games is $(1/2)^n$, for n=2 or 1/4. For n=3 it is $(1/2)^3=1/8$ etc
HOWEVER if you already are at the point that you have won the first game , then we are at a world that allows only HH and HT because the first turn is already played (this is called conditional: you are basing your estimations on the condition that something has already happened). Then your chances of winning this round conditional to the fact you already won all previous is 1/2. The same if you are asking about your chances of winning round n+1 conditional to the fact you won all previous rounds.
So every round, your chances are 50-50, but this is because you can only arrive to that round if you win all previous rounds. And for every player that arrived to round n, there will be $2^{n-1}$ players on average that never made it there. (for n>=2)
This means sticking makes no difference. Now, so that is not confused with a different question: if at every round you have more than one choices AND something is revealed to you (you get some additional information by say showing to you that there is nothing behind curtain 3, then this is a different problem)
answered Aug 23 '16 at 14:28
ntgntg
1236
$endgroup$
add a comment |
$begingroup$
It comes down what you consider a win. Is it to get one of the tosses or all? I think you ask about getting everything right.
Lets say you stick to heads (H) and there are two games. Then all possible outcomes on your world are HH,HT,TH,TT and only HH is good for you. The probability you will win all games is $(1/2)^n$, for n=2 or 1/4. For n=3 it is $(1/2)^3=1/8$ etc
HOWEVER if you already are at the point that you have won the first game , then we are at a world that allows only HH and HT because the first turn is already played (this is called conditional: you are basing your estimations on the condition that something has already happened). Then your chances of winning this round conditional to the fact you already won all previous is 1/2. The same if you are asking about your chances of winning round n+1 conditional to the fact you won all previous rounds.
So every round, your chances are 50-50, but this is because you can only arrive to that round if you win all previous rounds. And for every player that arrived to round n, there will be $2^{n-1}$ players on average that never made it there. (for n>=2)
This means sticking makes no difference. Now, so that is not confused with a different question: if at every round you have more than one choices AND something is revealed to you (you get some additional information by say showing to you that there is nothing behind curtain 3, then this is a different problem)
answered Aug 23 '16 at 14:28
ntgntg
1236
$endgroup$
It comes down what you consider a win. Is it to get one of the tosses or all? I think you ask about getting everything right.
Lets say you stick to heads (H) and there are two games. Then all possible outcomes on your world are HH,HT,TH,TT and only HH is good for you. The probability you will win all games is $(1/2)^n$, for n=2 or 1/4. For n=3 it is $(1/2)^3=1/8$ etc
HOWEVER if you already are at the point that you have won the first game , then we are at a world that allows only HH and HT because the first turn is already played (this is called conditional: you are basing your estimations on the condition that something has already happened). Then your chances of winning this round conditional to the fact you already won all previous is 1/2. The same if you are asking about your chances of winning round n+1 conditional to the fact you won all previous rounds.
So every round, your chances are 50-50, but this is because you can only arrive to that round if you win all previous rounds. And for every player that arrived to round n, there will be $2^{n-1}$ players on average that never made it there. (for n>=2)
This means sticking makes no difference. Now, so that is not confused with a different question: if at every round you have more than one choices AND something is revealed to you (you get some additional information by say showing to you that there is nothing behind curtain 3, then this is a different problem)
answered Aug 23 '16 at 14:28
ntgntg
1236
edited Aug 24 '16 at 11:51
answered Aug 23 '16 at 14:28
ntgntg
1236
answered Aug 23 '16 at 14:28
ntgntg
1236
answered Aug 23 '16 at 14:28
ntgntg
1236
1236
add a comment |
add a comment |
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$begingroup$
This may need clarification. Is not each game a single toss? But if so, then the probability of winning a given game is $frac 12$ regardless of what has come before (or what comes after). And where does $976$ come from? That isn't a power of $2$.
$endgroup$
– lulu
Sep 26 '15 at 22:36
$begingroup$
If i understand correctly you are wrong because every toss has $frac{1}{2}$ probability of being tail so the probability of losing two games (even if not sticking with the same choose) is $frac{1}{4}$ and in general the probability of losing all the first $n$ games is $frac{1}{2^n}$
$endgroup$
– karmalu
Sep 27 '15 at 12:06