Probability of a single trial within binomial experiment vs. stand-alone bernoulli experiment

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When a flip a coin several times, each throw is independent from another. In other words, my coin does not know what came out previous time. So, each next flip the result is unpredictable and random. Now, suppose I flipped a fair coin three times and got each time a head (head-head-head). Intuitively, the head cannot come out “head” all the time, so I can expect the on fourth throw to have higher chances to get finally a tail. But each flip is an independent event - the coin does not know what came out last time.



How this paradox is resolved?



Many thanks!










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  • 6




    I see no paradox, given trials are independent.
    – coffeemath
    Dec 27 '18 at 7:33















3














When a flip a coin several times, each throw is independent from another. In other words, my coin does not know what came out previous time. So, each next flip the result is unpredictable and random. Now, suppose I flipped a fair coin three times and got each time a head (head-head-head). Intuitively, the head cannot come out “head” all the time, so I can expect the on fourth throw to have higher chances to get finally a tail. But each flip is an independent event - the coin does not know what came out last time.



How this paradox is resolved?



Many thanks!










share|cite|improve this question



















  • 6




    I see no paradox, given trials are independent.
    – coffeemath
    Dec 27 '18 at 7:33













3












3








3


0





When a flip a coin several times, each throw is independent from another. In other words, my coin does not know what came out previous time. So, each next flip the result is unpredictable and random. Now, suppose I flipped a fair coin three times and got each time a head (head-head-head). Intuitively, the head cannot come out “head” all the time, so I can expect the on fourth throw to have higher chances to get finally a tail. But each flip is an independent event - the coin does not know what came out last time.



How this paradox is resolved?



Many thanks!










share|cite|improve this question















When a flip a coin several times, each throw is independent from another. In other words, my coin does not know what came out previous time. So, each next flip the result is unpredictable and random. Now, suppose I flipped a fair coin three times and got each time a head (head-head-head). Intuitively, the head cannot come out “head” all the time, so I can expect the on fourth throw to have higher chances to get finally a tail. But each flip is an independent event - the coin does not know what came out last time.



How this paradox is resolved?



Many thanks!







probability binomial-distribution paradoxes






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edited Dec 27 '18 at 8:57









Eevee Trainer

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5,0471734










asked Dec 27 '18 at 7:31









JohnJohn

1236




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  • 6




    I see no paradox, given trials are independent.
    – coffeemath
    Dec 27 '18 at 7:33












  • 6




    I see no paradox, given trials are independent.
    – coffeemath
    Dec 27 '18 at 7:33







6




6




I see no paradox, given trials are independent.
– coffeemath
Dec 27 '18 at 7:33




I see no paradox, given trials are independent.
– coffeemath
Dec 27 '18 at 7:33










6 Answers
6






active

oldest

votes


















7















Intuitively, the head cannot come out “head” all the time, so I can expect the on fourth throw to have higher chances to get finally a tail.




This marks the error.



Granted, there is no paradox - your error is known as the gambler's fallacy (or Monte Carlo fallacy). By assuming that tails is more likely, you assume that the trials are not independent even though they are. The fallacy is then that you feel the events are dependent, when they're not.



This was actually touched on in a recent Numberphile video somewhat: we think that streaks or heads of tails is nonrandom, i.e. would be a hint or sign of dependence or bias, when really they're not.



So there is no paradox, just bad intuition because of how human intuition tends to work.






share|cite|improve this answer






















  • Thanks a lot. The video is really illuminating.
    – John
    Dec 28 '18 at 11:18


















12














If you flipped a fair coin three times and got heads each time, then on the next toss your probability of getting heads is $1/2$. The fact that you just flipped three heads in a row is irrelevant. There is no paradox.






share|cite|improve this answer




















  • But what if I look retrospectively on the whole sequence? If I did get 4 gets, I would said that how lucky I was to get 4 out of 4 heads. But if I received fourth tail, I would say that 3 out of 4 heads is not so low chances.
    – John
    Dec 27 '18 at 8:09






  • 3




    @John The probability on getting $4$ heads by $4$ throws is exactly the same as getting at first $3$ heads and then as last a tail. Also in that case you can say: "how lucky I was to get $3$ heads at first and then a tail as last."
    – drhab
    Dec 27 '18 at 8:55










  • Or with more runs "how lucky I was to get heads, tails, tails, heads, tails, heads, tails, tails, tails, heads".
    – immibis
    Dec 27 '18 at 12:15










  • Thanks, it is now clear to me!
    – John
    Dec 28 '18 at 11:17


















5















"...so I can expect the on fourth throw to have higher chances to get finally a tail"




By independence there is no reason at all for expecting that, so there is no paradox.






share|cite|improve this answer




















  • Thanks a lot for help!
    – John
    Dec 28 '18 at 11:18


















5















Intuitively, the head cannot come out “head” all the time, so I can expect the on fourth throw to have higher chances to get finally a tail




Not quite.



"In four tries, it's more probable to have 3 heads and 1 tail than to have 4 heads" => TRUE



"In four tries, the result HHHT is more probable than HHHH " => FALSE



This is a common confusion.
See for example.






share|cite|improve this answer




















  • Thanks a lot for help! We should not confuse between chances of getting specific order and the number of tails (or tails)
    – John
    Dec 28 '18 at 11:18



















3














If you already flipped three heads you already did the "hard part" of getting 4 heads in a row. Mathematically, if you look at the probability of four heads in a row with already three heads you should use conditional proababilities



$$P(textfour heads in a row | textI already got three heads in a row)
$$



(read as : "Given that I already got three heads in a row, what is the probability that I get four heads in a row") which is then, by independence of events, exactly $frac12$.



This might get a bit more intuitive if you look at more extreme cases. The probability of apocalypse in the next 5 minutes is (hopefully) very low. But if you see a nuclear warhead in the sky the probability of a nuclear apocalypse is high. The "hard part" (= unprobable events) already happened, so there is no paradox when $P(textapocalypse)$ is low but $P(textapocalypse|textseeing a nuke)$ is high.






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  • Thank a lot for your help!
    – John
    Dec 28 '18 at 11:19


















2














I find that numbers clear things up. For a fair coin,



$$beginalign
&Pr(mathrmHHH) = frac18\
&Pr(mathrmHHHH) = frac116 \
&Pr(mathrmHHHT) = frac116 \
endalign$$



That is, both the following are true:



  • getting four heads in four coin tosses is somewhat rare (probability $1/16$: let's call it a “rarity of $16$”) before you start,


  • but “half of the rarity” (i.e. a probability $1/8$ event) has already happened by the time you get three heads in a row, so that getting a further heads or tails are now equally likely again.



A point here is that all possible sequences of $mathrmH$s and $mathrmT$s of the same length have the same probability. That is, although it is true that:



$$beginalign
Pr(text$4$ Hs) &= frac116 \
Pr(text$3$ Hs, $1$ T) &= frac416 = frac28 = frac14 \
Pr(text$2$ Hs, $2$ Ts) &= frac616 = frac38 \
Pr(text$1$ H, $3$ Ts) &= frac416 = frac28 = frac14 \
Pr(text$4$ Ts) &= frac116
endalign$$



it is still the case that $mathrmHHHH$ and (say) $mathrmHTTH$ are both equally likely (probability of $1/16$ each), and the “rarity” of “four heads in a row” comes from there being fewer ways for that to happen (or conversely, the higher probability of “two heads and two tails” comes from there being more ways for that to happen), rather than any individual sequence being less or more rare.



Richard Feynman in a lecture:




You know, the most amazing thing happened to me tonight. I was coming here, on the way to the lecture, and I came in through the parking lot. And you won’t believe what happened. I saw a car with the license plate ARW 357. Can you imagine? Of all the millions of license plates in the state, what was the chance that I would see that particular one tonight? Amazing!







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  • Thank a lot for your help!
    – John
    Dec 28 '18 at 11:19










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6 Answers
6






active

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6 Answers
6






active

oldest

votes









active

oldest

votes






active

oldest

votes









7















Intuitively, the head cannot come out “head” all the time, so I can expect the on fourth throw to have higher chances to get finally a tail.




This marks the error.



Granted, there is no paradox - your error is known as the gambler's fallacy (or Monte Carlo fallacy). By assuming that tails is more likely, you assume that the trials are not independent even though they are. The fallacy is then that you feel the events are dependent, when they're not.



This was actually touched on in a recent Numberphile video somewhat: we think that streaks or heads of tails is nonrandom, i.e. would be a hint or sign of dependence or bias, when really they're not.



So there is no paradox, just bad intuition because of how human intuition tends to work.






share|cite|improve this answer






















  • Thanks a lot. The video is really illuminating.
    – John
    Dec 28 '18 at 11:18















7















Intuitively, the head cannot come out “head” all the time, so I can expect the on fourth throw to have higher chances to get finally a tail.




This marks the error.



Granted, there is no paradox - your error is known as the gambler's fallacy (or Monte Carlo fallacy). By assuming that tails is more likely, you assume that the trials are not independent even though they are. The fallacy is then that you feel the events are dependent, when they're not.



This was actually touched on in a recent Numberphile video somewhat: we think that streaks or heads of tails is nonrandom, i.e. would be a hint or sign of dependence or bias, when really they're not.



So there is no paradox, just bad intuition because of how human intuition tends to work.






share|cite|improve this answer






















  • Thanks a lot. The video is really illuminating.
    – John
    Dec 28 '18 at 11:18













7












7








7







Intuitively, the head cannot come out “head” all the time, so I can expect the on fourth throw to have higher chances to get finally a tail.




This marks the error.



Granted, there is no paradox - your error is known as the gambler's fallacy (or Monte Carlo fallacy). By assuming that tails is more likely, you assume that the trials are not independent even though they are. The fallacy is then that you feel the events are dependent, when they're not.



This was actually touched on in a recent Numberphile video somewhat: we think that streaks or heads of tails is nonrandom, i.e. would be a hint or sign of dependence or bias, when really they're not.



So there is no paradox, just bad intuition because of how human intuition tends to work.






share|cite|improve this answer















Intuitively, the head cannot come out “head” all the time, so I can expect the on fourth throw to have higher chances to get finally a tail.




This marks the error.



Granted, there is no paradox - your error is known as the gambler's fallacy (or Monte Carlo fallacy). By assuming that tails is more likely, you assume that the trials are not independent even though they are. The fallacy is then that you feel the events are dependent, when they're not.



This was actually touched on in a recent Numberphile video somewhat: we think that streaks or heads of tails is nonrandom, i.e. would be a hint or sign of dependence or bias, when really they're not.



So there is no paradox, just bad intuition because of how human intuition tends to work.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 27 '18 at 8:56

























answered Dec 27 '18 at 7:46









Eevee TrainerEevee Trainer

5,0471734




5,0471734











  • Thanks a lot. The video is really illuminating.
    – John
    Dec 28 '18 at 11:18
















  • Thanks a lot. The video is really illuminating.
    – John
    Dec 28 '18 at 11:18















Thanks a lot. The video is really illuminating.
– John
Dec 28 '18 at 11:18




Thanks a lot. The video is really illuminating.
– John
Dec 28 '18 at 11:18











12














If you flipped a fair coin three times and got heads each time, then on the next toss your probability of getting heads is $1/2$. The fact that you just flipped three heads in a row is irrelevant. There is no paradox.






share|cite|improve this answer




















  • But what if I look retrospectively on the whole sequence? If I did get 4 gets, I would said that how lucky I was to get 4 out of 4 heads. But if I received fourth tail, I would say that 3 out of 4 heads is not so low chances.
    – John
    Dec 27 '18 at 8:09






  • 3




    @John The probability on getting $4$ heads by $4$ throws is exactly the same as getting at first $3$ heads and then as last a tail. Also in that case you can say: "how lucky I was to get $3$ heads at first and then a tail as last."
    – drhab
    Dec 27 '18 at 8:55










  • Or with more runs "how lucky I was to get heads, tails, tails, heads, tails, heads, tails, tails, tails, heads".
    – immibis
    Dec 27 '18 at 12:15










  • Thanks, it is now clear to me!
    – John
    Dec 28 '18 at 11:17















12














If you flipped a fair coin three times and got heads each time, then on the next toss your probability of getting heads is $1/2$. The fact that you just flipped three heads in a row is irrelevant. There is no paradox.






share|cite|improve this answer




















  • But what if I look retrospectively on the whole sequence? If I did get 4 gets, I would said that how lucky I was to get 4 out of 4 heads. But if I received fourth tail, I would say that 3 out of 4 heads is not so low chances.
    – John
    Dec 27 '18 at 8:09






  • 3




    @John The probability on getting $4$ heads by $4$ throws is exactly the same as getting at first $3$ heads and then as last a tail. Also in that case you can say: "how lucky I was to get $3$ heads at first and then a tail as last."
    – drhab
    Dec 27 '18 at 8:55










  • Or with more runs "how lucky I was to get heads, tails, tails, heads, tails, heads, tails, tails, tails, heads".
    – immibis
    Dec 27 '18 at 12:15










  • Thanks, it is now clear to me!
    – John
    Dec 28 '18 at 11:17













12












12








12






If you flipped a fair coin three times and got heads each time, then on the next toss your probability of getting heads is $1/2$. The fact that you just flipped three heads in a row is irrelevant. There is no paradox.






share|cite|improve this answer












If you flipped a fair coin three times and got heads each time, then on the next toss your probability of getting heads is $1/2$. The fact that you just flipped three heads in a row is irrelevant. There is no paradox.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 27 '18 at 7:43









littleOlittleO

29.2k644109




29.2k644109











  • But what if I look retrospectively on the whole sequence? If I did get 4 gets, I would said that how lucky I was to get 4 out of 4 heads. But if I received fourth tail, I would say that 3 out of 4 heads is not so low chances.
    – John
    Dec 27 '18 at 8:09






  • 3




    @John The probability on getting $4$ heads by $4$ throws is exactly the same as getting at first $3$ heads and then as last a tail. Also in that case you can say: "how lucky I was to get $3$ heads at first and then a tail as last."
    – drhab
    Dec 27 '18 at 8:55










  • Or with more runs "how lucky I was to get heads, tails, tails, heads, tails, heads, tails, tails, tails, heads".
    – immibis
    Dec 27 '18 at 12:15










  • Thanks, it is now clear to me!
    – John
    Dec 28 '18 at 11:17
















  • But what if I look retrospectively on the whole sequence? If I did get 4 gets, I would said that how lucky I was to get 4 out of 4 heads. But if I received fourth tail, I would say that 3 out of 4 heads is not so low chances.
    – John
    Dec 27 '18 at 8:09






  • 3




    @John The probability on getting $4$ heads by $4$ throws is exactly the same as getting at first $3$ heads and then as last a tail. Also in that case you can say: "how lucky I was to get $3$ heads at first and then a tail as last."
    – drhab
    Dec 27 '18 at 8:55










  • Or with more runs "how lucky I was to get heads, tails, tails, heads, tails, heads, tails, tails, tails, heads".
    – immibis
    Dec 27 '18 at 12:15










  • Thanks, it is now clear to me!
    – John
    Dec 28 '18 at 11:17















But what if I look retrospectively on the whole sequence? If I did get 4 gets, I would said that how lucky I was to get 4 out of 4 heads. But if I received fourth tail, I would say that 3 out of 4 heads is not so low chances.
– John
Dec 27 '18 at 8:09




But what if I look retrospectively on the whole sequence? If I did get 4 gets, I would said that how lucky I was to get 4 out of 4 heads. But if I received fourth tail, I would say that 3 out of 4 heads is not so low chances.
– John
Dec 27 '18 at 8:09




3




3




@John The probability on getting $4$ heads by $4$ throws is exactly the same as getting at first $3$ heads and then as last a tail. Also in that case you can say: "how lucky I was to get $3$ heads at first and then a tail as last."
– drhab
Dec 27 '18 at 8:55




@John The probability on getting $4$ heads by $4$ throws is exactly the same as getting at first $3$ heads and then as last a tail. Also in that case you can say: "how lucky I was to get $3$ heads at first and then a tail as last."
– drhab
Dec 27 '18 at 8:55












Or with more runs "how lucky I was to get heads, tails, tails, heads, tails, heads, tails, tails, tails, heads".
– immibis
Dec 27 '18 at 12:15




Or with more runs "how lucky I was to get heads, tails, tails, heads, tails, heads, tails, tails, tails, heads".
– immibis
Dec 27 '18 at 12:15












Thanks, it is now clear to me!
– John
Dec 28 '18 at 11:17




Thanks, it is now clear to me!
– John
Dec 28 '18 at 11:17











5















"...so I can expect the on fourth throw to have higher chances to get finally a tail"




By independence there is no reason at all for expecting that, so there is no paradox.






share|cite|improve this answer




















  • Thanks a lot for help!
    – John
    Dec 28 '18 at 11:18















5















"...so I can expect the on fourth throw to have higher chances to get finally a tail"




By independence there is no reason at all for expecting that, so there is no paradox.






share|cite|improve this answer




















  • Thanks a lot for help!
    – John
    Dec 28 '18 at 11:18













5












5








5







"...so I can expect the on fourth throw to have higher chances to get finally a tail"




By independence there is no reason at all for expecting that, so there is no paradox.






share|cite|improve this answer













"...so I can expect the on fourth throw to have higher chances to get finally a tail"




By independence there is no reason at all for expecting that, so there is no paradox.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 27 '18 at 8:50









drhabdrhab

98.5k544129




98.5k544129











  • Thanks a lot for help!
    – John
    Dec 28 '18 at 11:18
















  • Thanks a lot for help!
    – John
    Dec 28 '18 at 11:18















Thanks a lot for help!
– John
Dec 28 '18 at 11:18




Thanks a lot for help!
– John
Dec 28 '18 at 11:18











5















Intuitively, the head cannot come out “head” all the time, so I can expect the on fourth throw to have higher chances to get finally a tail




Not quite.



"In four tries, it's more probable to have 3 heads and 1 tail than to have 4 heads" => TRUE



"In four tries, the result HHHT is more probable than HHHH " => FALSE



This is a common confusion.
See for example.






share|cite|improve this answer




















  • Thanks a lot for help! We should not confuse between chances of getting specific order and the number of tails (or tails)
    – John
    Dec 28 '18 at 11:18
















5















Intuitively, the head cannot come out “head” all the time, so I can expect the on fourth throw to have higher chances to get finally a tail




Not quite.



"In four tries, it's more probable to have 3 heads and 1 tail than to have 4 heads" => TRUE



"In four tries, the result HHHT is more probable than HHHH " => FALSE



This is a common confusion.
See for example.






share|cite|improve this answer




















  • Thanks a lot for help! We should not confuse between chances of getting specific order and the number of tails (or tails)
    – John
    Dec 28 '18 at 11:18














5












5








5







Intuitively, the head cannot come out “head” all the time, so I can expect the on fourth throw to have higher chances to get finally a tail




Not quite.



"In four tries, it's more probable to have 3 heads and 1 tail than to have 4 heads" => TRUE



"In four tries, the result HHHT is more probable than HHHH " => FALSE



This is a common confusion.
See for example.






share|cite|improve this answer













Intuitively, the head cannot come out “head” all the time, so I can expect the on fourth throw to have higher chances to get finally a tail




Not quite.



"In four tries, it's more probable to have 3 heads and 1 tail than to have 4 heads" => TRUE



"In four tries, the result HHHT is more probable than HHHH " => FALSE



This is a common confusion.
See for example.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 27 '18 at 13:28









leonbloyleonbloy

40.3k645107




40.3k645107











  • Thanks a lot for help! We should not confuse between chances of getting specific order and the number of tails (or tails)
    – John
    Dec 28 '18 at 11:18

















  • Thanks a lot for help! We should not confuse between chances of getting specific order and the number of tails (or tails)
    – John
    Dec 28 '18 at 11:18
















Thanks a lot for help! We should not confuse between chances of getting specific order and the number of tails (or tails)
– John
Dec 28 '18 at 11:18





Thanks a lot for help! We should not confuse between chances of getting specific order and the number of tails (or tails)
– John
Dec 28 '18 at 11:18












3














If you already flipped three heads you already did the "hard part" of getting 4 heads in a row. Mathematically, if you look at the probability of four heads in a row with already three heads you should use conditional proababilities



$$P(textfour heads in a row | textI already got three heads in a row)
$$



(read as : "Given that I already got three heads in a row, what is the probability that I get four heads in a row") which is then, by independence of events, exactly $frac12$.



This might get a bit more intuitive if you look at more extreme cases. The probability of apocalypse in the next 5 minutes is (hopefully) very low. But if you see a nuclear warhead in the sky the probability of a nuclear apocalypse is high. The "hard part" (= unprobable events) already happened, so there is no paradox when $P(textapocalypse)$ is low but $P(textapocalypse|textseeing a nuke)$ is high.






share|cite|improve this answer




















  • Thank a lot for your help!
    – John
    Dec 28 '18 at 11:19















3














If you already flipped three heads you already did the "hard part" of getting 4 heads in a row. Mathematically, if you look at the probability of four heads in a row with already three heads you should use conditional proababilities



$$P(textfour heads in a row | textI already got three heads in a row)
$$



(read as : "Given that I already got three heads in a row, what is the probability that I get four heads in a row") which is then, by independence of events, exactly $frac12$.



This might get a bit more intuitive if you look at more extreme cases. The probability of apocalypse in the next 5 minutes is (hopefully) very low. But if you see a nuclear warhead in the sky the probability of a nuclear apocalypse is high. The "hard part" (= unprobable events) already happened, so there is no paradox when $P(textapocalypse)$ is low but $P(textapocalypse|textseeing a nuke)$ is high.






share|cite|improve this answer




















  • Thank a lot for your help!
    – John
    Dec 28 '18 at 11:19













3












3








3






If you already flipped three heads you already did the "hard part" of getting 4 heads in a row. Mathematically, if you look at the probability of four heads in a row with already three heads you should use conditional proababilities



$$P(textfour heads in a row | textI already got three heads in a row)
$$



(read as : "Given that I already got three heads in a row, what is the probability that I get four heads in a row") which is then, by independence of events, exactly $frac12$.



This might get a bit more intuitive if you look at more extreme cases. The probability of apocalypse in the next 5 minutes is (hopefully) very low. But if you see a nuclear warhead in the sky the probability of a nuclear apocalypse is high. The "hard part" (= unprobable events) already happened, so there is no paradox when $P(textapocalypse)$ is low but $P(textapocalypse|textseeing a nuke)$ is high.






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If you already flipped three heads you already did the "hard part" of getting 4 heads in a row. Mathematically, if you look at the probability of four heads in a row with already three heads you should use conditional proababilities



$$P(textfour heads in a row | textI already got three heads in a row)
$$



(read as : "Given that I already got three heads in a row, what is the probability that I get four heads in a row") which is then, by independence of events, exactly $frac12$.



This might get a bit more intuitive if you look at more extreme cases. The probability of apocalypse in the next 5 minutes is (hopefully) very low. But if you see a nuclear warhead in the sky the probability of a nuclear apocalypse is high. The "hard part" (= unprobable events) already happened, so there is no paradox when $P(textapocalypse)$ is low but $P(textapocalypse|textseeing a nuke)$ is high.







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answered Dec 27 '18 at 13:25









syntonymsyntonym

1312




1312











  • Thank a lot for your help!
    – John
    Dec 28 '18 at 11:19
















  • Thank a lot for your help!
    – John
    Dec 28 '18 at 11:19















Thank a lot for your help!
– John
Dec 28 '18 at 11:19




Thank a lot for your help!
– John
Dec 28 '18 at 11:19











2














I find that numbers clear things up. For a fair coin,



$$beginalign
&Pr(mathrmHHH) = frac18\
&Pr(mathrmHHHH) = frac116 \
&Pr(mathrmHHHT) = frac116 \
endalign$$



That is, both the following are true:



  • getting four heads in four coin tosses is somewhat rare (probability $1/16$: let's call it a “rarity of $16$”) before you start,


  • but “half of the rarity” (i.e. a probability $1/8$ event) has already happened by the time you get three heads in a row, so that getting a further heads or tails are now equally likely again.



A point here is that all possible sequences of $mathrmH$s and $mathrmT$s of the same length have the same probability. That is, although it is true that:



$$beginalign
Pr(text$4$ Hs) &= frac116 \
Pr(text$3$ Hs, $1$ T) &= frac416 = frac28 = frac14 \
Pr(text$2$ Hs, $2$ Ts) &= frac616 = frac38 \
Pr(text$1$ H, $3$ Ts) &= frac416 = frac28 = frac14 \
Pr(text$4$ Ts) &= frac116
endalign$$



it is still the case that $mathrmHHHH$ and (say) $mathrmHTTH$ are both equally likely (probability of $1/16$ each), and the “rarity” of “four heads in a row” comes from there being fewer ways for that to happen (or conversely, the higher probability of “two heads and two tails” comes from there being more ways for that to happen), rather than any individual sequence being less or more rare.



Richard Feynman in a lecture:




You know, the most amazing thing happened to me tonight. I was coming here, on the way to the lecture, and I came in through the parking lot. And you won’t believe what happened. I saw a car with the license plate ARW 357. Can you imagine? Of all the millions of license plates in the state, what was the chance that I would see that particular one tonight? Amazing!







share|cite|improve this answer




















  • Thank a lot for your help!
    – John
    Dec 28 '18 at 11:19















2














I find that numbers clear things up. For a fair coin,



$$beginalign
&Pr(mathrmHHH) = frac18\
&Pr(mathrmHHHH) = frac116 \
&Pr(mathrmHHHT) = frac116 \
endalign$$



That is, both the following are true:



  • getting four heads in four coin tosses is somewhat rare (probability $1/16$: let's call it a “rarity of $16$”) before you start,


  • but “half of the rarity” (i.e. a probability $1/8$ event) has already happened by the time you get three heads in a row, so that getting a further heads or tails are now equally likely again.



A point here is that all possible sequences of $mathrmH$s and $mathrmT$s of the same length have the same probability. That is, although it is true that:



$$beginalign
Pr(text$4$ Hs) &= frac116 \
Pr(text$3$ Hs, $1$ T) &= frac416 = frac28 = frac14 \
Pr(text$2$ Hs, $2$ Ts) &= frac616 = frac38 \
Pr(text$1$ H, $3$ Ts) &= frac416 = frac28 = frac14 \
Pr(text$4$ Ts) &= frac116
endalign$$



it is still the case that $mathrmHHHH$ and (say) $mathrmHTTH$ are both equally likely (probability of $1/16$ each), and the “rarity” of “four heads in a row” comes from there being fewer ways for that to happen (or conversely, the higher probability of “two heads and two tails” comes from there being more ways for that to happen), rather than any individual sequence being less or more rare.



Richard Feynman in a lecture:




You know, the most amazing thing happened to me tonight. I was coming here, on the way to the lecture, and I came in through the parking lot. And you won’t believe what happened. I saw a car with the license plate ARW 357. Can you imagine? Of all the millions of license plates in the state, what was the chance that I would see that particular one tonight? Amazing!







share|cite|improve this answer




















  • Thank a lot for your help!
    – John
    Dec 28 '18 at 11:19













2












2








2






I find that numbers clear things up. For a fair coin,



$$beginalign
&Pr(mathrmHHH) = frac18\
&Pr(mathrmHHHH) = frac116 \
&Pr(mathrmHHHT) = frac116 \
endalign$$



That is, both the following are true:



  • getting four heads in four coin tosses is somewhat rare (probability $1/16$: let's call it a “rarity of $16$”) before you start,


  • but “half of the rarity” (i.e. a probability $1/8$ event) has already happened by the time you get three heads in a row, so that getting a further heads or tails are now equally likely again.



A point here is that all possible sequences of $mathrmH$s and $mathrmT$s of the same length have the same probability. That is, although it is true that:



$$beginalign
Pr(text$4$ Hs) &= frac116 \
Pr(text$3$ Hs, $1$ T) &= frac416 = frac28 = frac14 \
Pr(text$2$ Hs, $2$ Ts) &= frac616 = frac38 \
Pr(text$1$ H, $3$ Ts) &= frac416 = frac28 = frac14 \
Pr(text$4$ Ts) &= frac116
endalign$$



it is still the case that $mathrmHHHH$ and (say) $mathrmHTTH$ are both equally likely (probability of $1/16$ each), and the “rarity” of “four heads in a row” comes from there being fewer ways for that to happen (or conversely, the higher probability of “two heads and two tails” comes from there being more ways for that to happen), rather than any individual sequence being less or more rare.



Richard Feynman in a lecture:




You know, the most amazing thing happened to me tonight. I was coming here, on the way to the lecture, and I came in through the parking lot. And you won’t believe what happened. I saw a car with the license plate ARW 357. Can you imagine? Of all the millions of license plates in the state, what was the chance that I would see that particular one tonight? Amazing!







share|cite|improve this answer












I find that numbers clear things up. For a fair coin,



$$beginalign
&Pr(mathrmHHH) = frac18\
&Pr(mathrmHHHH) = frac116 \
&Pr(mathrmHHHT) = frac116 \
endalign$$



That is, both the following are true:



  • getting four heads in four coin tosses is somewhat rare (probability $1/16$: let's call it a “rarity of $16$”) before you start,


  • but “half of the rarity” (i.e. a probability $1/8$ event) has already happened by the time you get three heads in a row, so that getting a further heads or tails are now equally likely again.



A point here is that all possible sequences of $mathrmH$s and $mathrmT$s of the same length have the same probability. That is, although it is true that:



$$beginalign
Pr(text$4$ Hs) &= frac116 \
Pr(text$3$ Hs, $1$ T) &= frac416 = frac28 = frac14 \
Pr(text$2$ Hs, $2$ Ts) &= frac616 = frac38 \
Pr(text$1$ H, $3$ Ts) &= frac416 = frac28 = frac14 \
Pr(text$4$ Ts) &= frac116
endalign$$



it is still the case that $mathrmHHHH$ and (say) $mathrmHTTH$ are both equally likely (probability of $1/16$ each), and the “rarity” of “four heads in a row” comes from there being fewer ways for that to happen (or conversely, the higher probability of “two heads and two tails” comes from there being more ways for that to happen), rather than any individual sequence being less or more rare.



Richard Feynman in a lecture:




You know, the most amazing thing happened to me tonight. I was coming here, on the way to the lecture, and I came in through the parking lot. And you won’t believe what happened. I saw a car with the license plate ARW 357. Can you imagine? Of all the millions of license plates in the state, what was the chance that I would see that particular one tonight? Amazing!








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answered Dec 28 '18 at 2:17









ShreevatsaRShreevatsaR

34.3k668105




34.3k668105











  • Thank a lot for your help!
    – John
    Dec 28 '18 at 11:19
















  • Thank a lot for your help!
    – John
    Dec 28 '18 at 11:19















Thank a lot for your help!
– John
Dec 28 '18 at 11:19




Thank a lot for your help!
– John
Dec 28 '18 at 11:19

















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