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Anomaly Probability - 0-99 vs 0.000-99.000
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I have a buddy who claims anomaly probability has a effect with decimal points here and will be 'less random' in a game dodge formula.
Right now his claim as that remove the decimal points would make the game less random, not in the overall result, but the short terms. To me this sounds absolutely crazy and wouldn't make a difference. The overall probability of it being 'less' random wouldn't be the case, because the end result is the same. If the probability changed, then the % average would change as well.
For example, if my hit rate is 40%, a dice roll of 0-99 would still be around 40%, and a dice roll of 0.000 to 99.000 would be the same thing.
So to try to put it into an example. He claims that more decimals 0.000000 to 99.000000 lets say will make a 50% hit rate chance more random and you'll get things like, 10 misses in a row and then 10 hits in a row which is still 50%, but then without decimals you'll have more 1 miss, 1 hit, 1 miss, 1 hit, etc, which again is 50%.
Is this true? I can't seem to wrap my head around it, because I feel like if I just chopped the decimal points, it would make 0 difference obviously, though maybe that's not the way around it either.
probability
asked Aug 3 at 23:32
Valleriani
11
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I have a buddy who claims anomaly probability has a effect with decimal points here and will be 'less random' in a game dodge formula.
Right now his claim as that remove the decimal points would make the game less random, not in the overall result, but the short terms. To me this sounds absolutely crazy and wouldn't make a difference. The overall probability of it being 'less' random wouldn't be the case, because the end result is the same. If the probability changed, then the % average would change as well.
For example, if my hit rate is 40%, a dice roll of 0-99 would still be around 40%, and a dice roll of 0.000 to 99.000 would be the same thing.
So to try to put it into an example. He claims that more decimals 0.000000 to 99.000000 lets say will make a 50% hit rate chance more random and you'll get things like, 10 misses in a row and then 10 hits in a row which is still 50%, but then without decimals you'll have more 1 miss, 1 hit, 1 miss, 1 hit, etc, which again is 50%.
Is this true? I can't seem to wrap my head around it, because I feel like if I just chopped the decimal points, it would make 0 difference obviously, though maybe that's not the way around it either.
probability
asked Aug 3 at 23:32
Valleriani
11
3
Stop thinking about it - your buddy is simply wrong!
– herb steinberg
Aug 3 at 23:34
Is there an issue that there are one hundred integers from $0$ through to $99$, so if you want more digits then perhaps you should be looking at $0.000$ through to $99.999$ rather than to $99.000$?
– Henry
Aug 3 at 23:46
Yeah I get the 0.000 to 99.999 bit too, was more of just an example, wouldn't really matter about that bit, but the logic was mostly to do with the way it worked. It wouldn't matter if there was 0 to 10, and we wanted to roll a 1 or less, it's the same as 0 to 100 and rolling a 10 or less, or a 0 to 1000 and rolling a 100 or less. The percent is always 10% in this case.
– Valleriani
Aug 3 at 23:55
If $Pr(A)=0.4$ and $Pr(B)=0.4000$ then of course $Pr(A)=Pr(B)$ since $0.4=0.4000$. Similarly, $frac410=frac400010000$, etc... Unless if your friend goes and very explicitly defines what he thinks will be changed, the apparent interpretation seems to have to do with looking at the chance of an event to occur and in both scenarios the event is equally likely to occur and increasing the size of the sample space effectively does nothing. Compare flipping a coin looking for heads to rolling a 20-sided die and looking for a number 11+.
– JMoravitz
Aug 3 at 23:57
There will be more variety in terms of specific outcomes seen, but if we were to replace the observed outcomes with simple yes/no's in terms of whether or not they were a part of the event in question, there would be absolutely no discernible difference between a sequence of outcomes from a smaller sample space to a sequence of outcomes from a larger sample space.
– JMoravitz
Aug 4 at 0:00
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a buddy who claims anomaly probability has a effect with decimal points here and will be 'less random' in a game dodge formula.
Right now his claim as that remove the decimal points would make the game less random, not in the overall result, but the short terms. To me this sounds absolutely crazy and wouldn't make a difference. The overall probability of it being 'less' random wouldn't be the case, because the end result is the same. If the probability changed, then the % average would change as well.
For example, if my hit rate is 40%, a dice roll of 0-99 would still be around 40%, and a dice roll of 0.000 to 99.000 would be the same thing.
So to try to put it into an example. He claims that more decimals 0.000000 to 99.000000 lets say will make a 50% hit rate chance more random and you'll get things like, 10 misses in a row and then 10 hits in a row which is still 50%, but then without decimals you'll have more 1 miss, 1 hit, 1 miss, 1 hit, etc, which again is 50%.
Is this true? I can't seem to wrap my head around it, because I feel like if I just chopped the decimal points, it would make 0 difference obviously, though maybe that's not the way around it either.
probability
asked Aug 3 at 23:32
Valleriani
11
I have a buddy who claims anomaly probability has a effect with decimal points here and will be 'less random' in a game dodge formula.
Right now his claim as that remove the decimal points would make the game less random, not in the overall result, but the short terms. To me this sounds absolutely crazy and wouldn't make a difference. The overall probability of it being 'less' random wouldn't be the case, because the end result is the same. If the probability changed, then the % average would change as well.
For example, if my hit rate is 40%, a dice roll of 0-99 would still be around 40%, and a dice roll of 0.000 to 99.000 would be the same thing.
So to try to put it into an example. He claims that more decimals 0.000000 to 99.000000 lets say will make a 50% hit rate chance more random and you'll get things like, 10 misses in a row and then 10 hits in a row which is still 50%, but then without decimals you'll have more 1 miss, 1 hit, 1 miss, 1 hit, etc, which again is 50%.
Is this true? I can't seem to wrap my head around it, because I feel like if I just chopped the decimal points, it would make 0 difference obviously, though maybe that's not the way around it either.
probability
asked Aug 3 at 23:32
Valleriani
11
asked Aug 3 at 23:32
Valleriani
11
asked Aug 3 at 23:32
Valleriani
11
asked Aug 3 at 23:32
Valleriani
11
11
3
Stop thinking about it - your buddy is simply wrong!
– herb steinberg
Aug 3 at 23:34
Is there an issue that there are one hundred integers from $0$ through to $99$, so if you want more digits then perhaps you should be looking at $0.000$ through to $99.999$ rather than to $99.000$?
– Henry
Aug 3 at 23:46
Yeah I get the 0.000 to 99.999 bit too, was more of just an example, wouldn't really matter about that bit, but the logic was mostly to do with the way it worked. It wouldn't matter if there was 0 to 10, and we wanted to roll a 1 or less, it's the same as 0 to 100 and rolling a 10 or less, or a 0 to 1000 and rolling a 100 or less. The percent is always 10% in this case.
– Valleriani
Aug 3 at 23:55
If $Pr(A)=0.4$ and $Pr(B)=0.4000$ then of course $Pr(A)=Pr(B)$ since $0.4=0.4000$. Similarly, $frac410=frac400010000$, etc... Unless if your friend goes and very explicitly defines what he thinks will be changed, the apparent interpretation seems to have to do with looking at the chance of an event to occur and in both scenarios the event is equally likely to occur and increasing the size of the sample space effectively does nothing. Compare flipping a coin looking for heads to rolling a 20-sided die and looking for a number 11+.
– JMoravitz
Aug 3 at 23:57
There will be more variety in terms of specific outcomes seen, but if we were to replace the observed outcomes with simple yes/no's in terms of whether or not they were a part of the event in question, there would be absolutely no discernible difference between a sequence of outcomes from a smaller sample space to a sequence of outcomes from a larger sample space.
– JMoravitz
Aug 4 at 0:00
 |Â
show 1 more comment
3
Stop thinking about it - your buddy is simply wrong!
– herb steinberg
Aug 3 at 23:34
Is there an issue that there are one hundred integers from $0$ through to $99$, so if you want more digits then perhaps you should be looking at $0.000$ through to $99.999$ rather than to $99.000$?
– Henry
Aug 3 at 23:46
Yeah I get the 0.000 to 99.999 bit too, was more of just an example, wouldn't really matter about that bit, but the logic was mostly to do with the way it worked. It wouldn't matter if there was 0 to 10, and we wanted to roll a 1 or less, it's the same as 0 to 100 and rolling a 10 or less, or a 0 to 1000 and rolling a 100 or less. The percent is always 10% in this case.
– Valleriani
Aug 3 at 23:55
If $Pr(A)=0.4$ and $Pr(B)=0.4000$ then of course $Pr(A)=Pr(B)$ since $0.4=0.4000$. Similarly, $frac410=frac400010000$, etc... Unless if your friend goes and very explicitly defines what he thinks will be changed, the apparent interpretation seems to have to do with looking at the chance of an event to occur and in both scenarios the event is equally likely to occur and increasing the size of the sample space effectively does nothing. Compare flipping a coin looking for heads to rolling a 20-sided die and looking for a number 11+.
– JMoravitz
Aug 3 at 23:57
There will be more variety in terms of specific outcomes seen, but if we were to replace the observed outcomes with simple yes/no's in terms of whether or not they were a part of the event in question, there would be absolutely no discernible difference between a sequence of outcomes from a smaller sample space to a sequence of outcomes from a larger sample space.
– JMoravitz
Aug 4 at 0:00
3
3
Stop thinking about it - your buddy is simply wrong!
– herb steinberg
Aug 3 at 23:34
Stop thinking about it - your buddy is simply wrong!
– herb steinberg
Aug 3 at 23:34
Is there an issue that there are one hundred integers from $0$ through to $99$, so if you want more digits then perhaps you should be looking at $0.000$ through to $99.999$ rather than to $99.000$?
– Henry
Aug 3 at 23:46
Is there an issue that there are one hundred integers from $0$ through to $99$, so if you want more digits then perhaps you should be looking at $0.000$ through to $99.999$ rather than to $99.000$?
– Henry
Aug 3 at 23:46
Yeah I get the 0.000 to 99.999 bit too, was more of just an example, wouldn't really matter about that bit, but the logic was mostly to do with the way it worked. It wouldn't matter if there was 0 to 10, and we wanted to roll a 1 or less, it's the same as 0 to 100 and rolling a 10 or less, or a 0 to 1000 and rolling a 100 or less. The percent is always 10% in this case.
– Valleriani
Aug 3 at 23:55
Yeah I get the 0.000 to 99.999 bit too, was more of just an example, wouldn't really matter about that bit, but the logic was mostly to do with the way it worked. It wouldn't matter if there was 0 to 10, and we wanted to roll a 1 or less, it's the same as 0 to 100 and rolling a 10 or less, or a 0 to 1000 and rolling a 100 or less. The percent is always 10% in this case.
– Valleriani
Aug 3 at 23:55
If $Pr(A)=0.4$ and $Pr(B)=0.4000$ then of course $Pr(A)=Pr(B)$ since $0.4=0.4000$. Similarly, $frac410=frac400010000$, etc... Unless if your friend goes and very explicitly defines what he thinks will be changed, the apparent interpretation seems to have to do with looking at the chance of an event to occur and in both scenarios the event is equally likely to occur and increasing the size of the sample space effectively does nothing. Compare flipping a coin looking for heads to rolling a 20-sided die and looking for a number 11+.
– JMoravitz
Aug 3 at 23:57
If $Pr(A)=0.4$ and $Pr(B)=0.4000$ then of course $Pr(A)=Pr(B)$ since $0.4=0.4000$. Similarly, $frac410=frac400010000$, etc... Unless if your friend goes and very explicitly defines what he thinks will be changed, the apparent interpretation seems to have to do with looking at the chance of an event to occur and in both scenarios the event is equally likely to occur and increasing the size of the sample space effectively does nothing. Compare flipping a coin looking for heads to rolling a 20-sided die and looking for a number 11+.
– JMoravitz
Aug 3 at 23:57
There will be more variety in terms of specific outcomes seen, but if we were to replace the observed outcomes with simple yes/no's in terms of whether or not they were a part of the event in question, there would be absolutely no discernible difference between a sequence of outcomes from a smaller sample space to a sequence of outcomes from a larger sample space.
– JMoravitz
Aug 4 at 0:00
There will be more variety in terms of specific outcomes seen, but if we were to replace the observed outcomes with simple yes/no's in terms of whether or not they were a part of the event in question, there would be absolutely no discernible difference between a sequence of outcomes from a smaller sample space to a sequence of outcomes from a larger sample space.
– JMoravitz
Aug 4 at 0:00
 |Â
show 1 more comment
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3
Stop thinking about it - your buddy is simply wrong!
– herb steinberg
Aug 3 at 23:34
Is there an issue that there are one hundred integers from $0$ through to $99$, so if you want more digits then perhaps you should be looking at $0.000$ through to $99.999$ rather than to $99.000$?
– Henry
Aug 3 at 23:46
Yeah I get the 0.000 to 99.999 bit too, was more of just an example, wouldn't really matter about that bit, but the logic was mostly to do with the way it worked. It wouldn't matter if there was 0 to 10, and we wanted to roll a 1 or less, it's the same as 0 to 100 and rolling a 10 or less, or a 0 to 1000 and rolling a 100 or less. The percent is always 10% in this case.
– Valleriani
Aug 3 at 23:55
If $Pr(A)=0.4$ and $Pr(B)=0.4000$ then of course $Pr(A)=Pr(B)$ since $0.4=0.4000$. Similarly, $frac410=frac400010000$, etc... Unless if your friend goes and very explicitly defines what he thinks will be changed, the apparent interpretation seems to have to do with looking at the chance of an event to occur and in both scenarios the event is equally likely to occur and increasing the size of the sample space effectively does nothing. Compare flipping a coin looking for heads to rolling a 20-sided die and looking for a number 11+.
– JMoravitz
Aug 3 at 23:57
There will be more variety in terms of specific outcomes seen, but if we were to replace the observed outcomes with simple yes/no's in terms of whether or not they were a part of the event in question, there would be absolutely no discernible difference between a sequence of outcomes from a smaller sample space to a sequence of outcomes from a larger sample space.
– JMoravitz
Aug 4 at 0:00