Is there a statistician in the house?
#1
Boost Pope
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Is there a statistician in the house?
I need some help visualizing this.
Assume that two fair coins are tossed. For each toss, there is a 25% chance that both will be heads, a 50% chance of one heads one tails, and a 25% chance that both will be tails.
Now assume that, outside of your sight, I toss both coins and reveal to you that they are not both tails. What is the probability of one heads and one tails? Is it still 50%, or does the elimination of one of the possible outcomes make it 66.6% heads/tails and 33.3% both heads?
Assume that two fair coins are tossed. For each toss, there is a 25% chance that both will be heads, a 50% chance of one heads one tails, and a 25% chance that both will be tails.
Now assume that, outside of your sight, I toss both coins and reveal to you that they are not both tails. What is the probability of one heads and one tails? Is it still 50%, or does the elimination of one of the possible outcomes make it 66.6% heads/tails and 33.3% both heads?
#3
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I say the 2nd, 66.7% and 33.3%.
Usually 4 possible outcomes.
coin 1, coin 2
H, H
H, T
T, H
T, T
Thats how you get the 25/50/25.... eliminate the double tails and you have...
H, H
H, T
T, H
1 of the 3 (33.3%) times it is two heads.
2 of the 3 (66.7%) times it is one heads and one tails.
Usually 4 possible outcomes.
coin 1, coin 2
H, H
H, T
T, H
T, T
Thats how you get the 25/50/25.... eliminate the double tails and you have...
H, H
H, T
T, H
1 of the 3 (33.3%) times it is two heads.
2 of the 3 (66.7%) times it is one heads and one tails.
#4
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If it were 50% odds of one heads/one tails, it would have to be equal odds to have the opposite (one tails/one heads), and that leaves no room for the heads/heads possibility. It's 66/33.
#7
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Agreed on pretty much everything that's been posted, except I'm not quite so certain of the outcome.
Before I toss the coins, each of the four possible outcomes (HH, HT, TH, TT) has a 25% chance of occurring. In fact, even after I toss the coins, the odds are the same. The question becomes- after I look at the coins and reveal to you that they are not TT, do the odds change?
Bear in mind that if we did this enough times, in one out of every four tosses I'd have produced TT, and thus the remainder of the experiment could not have taken place. Or put another way, the fact that I am able to ask you the question on three out of four executions (having not tossed TT) seems to validate the fact that the 25% TT probability did exist, thus maintaining the balance of probabilities.
To me, it's almost a Schroedinger problem. Do the odds change simply because I look at the coins and describe the result? I'm sure that there is a solid, concrete answer, I just don't know what it is.
Before I toss the coins, each of the four possible outcomes (HH, HT, TH, TT) has a 25% chance of occurring. In fact, even after I toss the coins, the odds are the same. The question becomes- after I look at the coins and reveal to you that they are not TT, do the odds change?
Bear in mind that if we did this enough times, in one out of every four tosses I'd have produced TT, and thus the remainder of the experiment could not have taken place. Or put another way, the fact that I am able to ask you the question on three out of four executions (having not tossed TT) seems to validate the fact that the 25% TT probability did exist, thus maintaining the balance of probabilities.
To me, it's almost a Schroedinger problem. Do the odds change simply because I look at the coins and describe the result? I'm sure that there is a solid, concrete answer, I just don't know what it is.
#8
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joe I refuse to answer the question until you make a video demonstrating the procedure. Please include at minimum 30 coin tosses that are not tails-tails results and a spreadsheet showing the occurrences of the remaining outcomes.
ok i'll answer a little: if you have four possible states (HT, TH, TT, HH) and the chances are each 25% out of a possible 100% of tosses, removing one state makes the total possible chances 75%. hence overall you are still 1 in 4 but for the subset it becomes 25%/75% or 33.333%. For HT/TH you get 66.667% chance.
ok i'll answer a little: if you have four possible states (HT, TH, TT, HH) and the chances are each 25% out of a possible 100% of tosses, removing one state makes the total possible chances 75%. hence overall you are still 1 in 4 but for the subset it becomes 25%/75% or 33.333%. For HT/TH you get 66.667% chance.
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Before I toss the coins, each of the four possible outcomes (HH, HT, TH, TT) has a 25% chance of occurring. In fact, even after I toss the coins, the odds are the same. The question becomes- after I look at the coins and reveal to you that they are not TT, do the odds change?
#14
I guess I wasn't explicit enough in post #5. The first case is an experiment of unconditional probability, the second being conditional probability. P[xx=HH]=0.5 (probability of heads-heads is 0.5)regardless of the experiment, however P[xx=HH|xxTT]=0.33 (probability of heads-heads given that the result is NOT tails-tails). The conditional probability is a subset of the total probability. Chapter 1 of any probability textbook; "Law of Total Probability."
#16
Are we taking wind speed/direction, air temp. , humidity and altitude into consideration? Don't forget you'll need to also use a 20% correction factor.
So if we were to flip these coins at let's say, 6,000 ft with an air temp. of 55* and 30% humidity and no wind, then i'd have to say there is a 100% chance that you'll get a result and that those results will be disputeable because not everyone is going to agree with your correction factor.
But since you're asking the question after clearly omitting one of the probabilities, then you must use a correction factor of 33% and 66%.
So if we were to flip these coins at let's say, 6,000 ft with an air temp. of 55* and 30% humidity and no wind, then i'd have to say there is a 100% chance that you'll get a result and that those results will be disputeable because not everyone is going to agree with your correction factor.
But since you're asking the question after clearly omitting one of the probabilities, then you must use a correction factor of 33% and 66%.
#17
Boost Pope
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I guess I wasn't explicit enough in post #5. The first case is an experiment of unconditional probability, the second being conditional probability. P[xx=HH]=0.5 (probability of heads-heads is 0.5)regardless of the experiment, however P[xx=HH|xxTT]=0.33 (probability of heads-heads given that the result is NOT tails-tails). The conditional probability is a subset of the total probability. Chapter 1 of any probability textbook; "Law of Total Probability."
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