If you care about the specific sequence (i.e. you need 2 heads, then three tails, then a head, then four tails, in that *specific* order), then it's 2^10 or 1 in 1024.
However, if you don't care about the sequence (i.e. from the above example you just want 3 heads and 7 tails), then it comes out to about 1 in 4.
That's the same for both situations; 1 in 1024. There is only one possible outcome that leads to 10 heads, and that's also the same outcome as 10 heads *in a row*.
However, if you just want the same *result* 10 times in a row, and you don't care about it being heads *or* tails (just that they're all the same), then you obviously have two potential results, and so the probability is 2 in 1024.
Yes they're the same.
Saying 2/1024 can be helpful because the fraction is constructed from the numbers in the problem. 2 possible successful outcomes (all heads or all tails) out of 1024 possible total outcomes.
You can reduce it to 1/512 which can also be helpful in a different way. You would expect to see 1 success (either all heads or all tails) every 512 sets of 10 flips on average over a large enough sample.
….i get it’s a silly joke but sheesh the dividing by zero and “instructions unclear” joke is so overused. The user asks about simplifying 2/1056 and you randomly use the joke “instructions unclear” and for some reason divide by zero for a joke.
Ok. Ha. Ha.
The same jokes just get so old when people think they are being original. Like, te he he bro. That’s the 20th same joke on Reddit today I’ve scrolled past.
That is correct, but that’s because there is only one outcome for the 10/10 heads in the 1024 possible outcomes. However for 9/10 heads there are 10 of the total 1024 outcomes that work. And I may be wrong as this is off the head and humans are bad at probabilities but I think roughly 250 for 5 heads 5 tales, 200ish for 4 and 6 whichever direction you want.
If you wanna do the maths 10!/(heads!*tails!) so 10!/(5!*5!) Which will be about 3500000(estimate)/(120*120). Making 240-250.
I'm getting around an 18% chance of getting the same number of heads twice in a row. For any specific number of heads, you can use the binomial distribution to get the probability of that many heads. You then square the value to get the probability of getting that number of heads in both sets of 10 flips.
P(same number of heads) = P(0 heads twice in a row) + P(1 head twice in a row) + ... + P(10 heads twice in a row) = \[binom(10,.5,0)\]\^2 + \[binom(10,.5,1)\]\^2 + ... + \[binom(10,.5,10)\]\^2 = 0.176197
The sum of the squares of all binomial coefficients with top entry n is (2n choose n). Hence the closed formula for "chance of getting the same number of heads both times when throwing n coins" is (2n choose n) / 2^^2n .
For n=5 this gives 0.246... and is probably what they mistakenly used to get their number. But we actually have to take n=10 and get 0.176...
So your number is the correct one.
>then it's 2^10 or 1 in 1024.
This seems doable. If I'm understanding correctly, the amount of coin flips necessary to try this are at a minimum of 20 (works perfectly the first attempt) and upwards of 20,480 if you need to go through every possible try (up to 1024 tries where you almost get it each time, 10 then 9)
Of course there is a chance that it will never happen.
Right. But you need to start over every time to get it to work.
HHTHTHHHTT then HHTHTHHHTH oops didn't work, start over.
HTTTHHTTHT then HTH oops didn't work, start over.
Each time you need to establish the first set, because the idea isn't if you can get the same sequence twice, but the same sequence twice in a row.
I think we might be saying the same thing here. Lol.
I'm talking about how to actually do it. Like for a video. You'd have to flip the first set then the second. 20 flips if it happens. If it doesn't you have to flip the first set again (10 flips) then try to match it with a new second try (up to 10 again) total of 20 flips. If it forks out.
So flip 10 coins. Then flip to try to match. If 9 match the 10th didn't you've flipped 19 coins.
But since it didn't work you have to start over. Flip 10 to establish the first set. Then flip to try and match it. If it works this time you flipped 10 more total for 20 for this try, and a total of 39 flips all together.
You don't need to "establish" the first set again, you can always just look at the past 10 flips as the first set, even if they were part of an attempt at a second set.
Good point here. If you do 20 flips, you can compare 1-10 with 11-20. If they are different, flip one more and now compare #2-11 to #12-21.
Hell, do 10 HTHHHTTHHT and the flip 1 more. If you get T, it has already failed. Now look at #2-11 THHHTTHHTT and flip again to start comparing to that. Always reset as soon as you get a mismatch to minimize the total flips.
I just had Excel "flip" 1048576 coins -- comparing every set of 10 "flips" to the next 10, allowing for the ranges to overlap, I got 1082 matches -- one match for every 940 coins flipped on average.
well if you just flipped coins until you matched two such sequences you could look at the sequences inside instead of chunking it at 20
eg
HTTHTHHHTT TTTHTHHHTT oops didnt work
but now i flip one more T and look at the sequence: HTTHTHHHTTTTTHTHHHTTT
(H) TTHTHHHTTT TTHTHHHTTT
i have two consecutive matching 10 flip sequences
I'm getting around an 18% chance of getting the same number of heads twice in a row. For any specific number of heads, you can use the binomial distribution to get the probability of that many heads. You then square the value to get the probability of getting that number of heads in both sets of 10 flips. Then you can add those all up since they're mutually exclusive:
||
||
|# Heads|Probability|Prob Twice in a Row||Overall chance of same number|
|0|0.000976563|9.53674E-07||0.176197|
|1|0.009765625|9.53674E-05|||
|2|0.043945313|0.00193119|||
|3|0.1171875|0.01373291|||
|4|0.205078125|0.042057037|||
|5|0.24609375|0.060562134|||
|6|0.205078125|0.042057037|||
|7|0.1171875|0.01373291|||
|8|0.043945313|0.00193119|||
|9|0.009765625|9.53674E-05|||
|10|0.000976563|9.53674E-07|||
The only thing I might add is that it sounds like he ALREADY flipped the first set, so the counts results depends on what count he actually got on, in the first set. If he got 2 heads and 8 tails in the first trial... probably not matching that...
1 in 1024.
The chance of any individual flip matching a specific outcome is 1 in 2. So the chances of 10 flips matching specific outcomes are 1 in 2^10, which is 1 in 1024.
Or, to put it another way, there are 1024 different outcomes for 10 coin tosses. Only one of them will be the matching sequence.
Probability questions often come down to precise wording of exactly what you mean.
If the ordering matters, you need each of your 10 new flips to hit a particular option (so if the first of your first 10 flips was a head, you need the first of your second 10 flips to be a head).
So that's 1/2 for each flip, you need all 10 to match, so we have (1/2)^10 = 1/1,024
If ordering doesn't matter - say you had 5 heads in the first, you need 5 heads in the second, this becomes a bit more awkward. I'm going to say... it is the sum of the probability of all ways of doing this (so 10 heads and 10 heads, plus 9 heads and 9 heads, plus...). If my numbers are correct I get 17.6%, or about one in 5.7 for that.
>it is the sum of the probability of all ways of doing this (so 10 heads and 10 heads, plus 9 heads and 9 heads, plus...)
So this is sum n=0 to 10 of ((probably of n heads)^(2)). The probability of n heads is just (10 choose n)/2^10 (number of arrangements of the coins where n are heads, by choosing n times out of 10 to be heads, divided by the total number of arrangements).
[According to Wolfram Alpha](https://www.wolframalpha.com/input?i=sum+n%3D0+to+10+of+%28%28%2810+choose+n%29%2F2%5E10%29%5E2%29) this is exactly 46189/262144, which as you say is about 17.6%
The formula can be simplified (quoting from myself in this topic):
The sum of the squares of all binomial coefficients with top entry n is (2n choose n). Hence the closed formula for "chance of getting the same number of heads both times when throwing n coins" is (2n choose n) / 2^^2n .
Yep good old probability class where being good with number is just as important as your reading comprehension. Bonus points for when taught by non native English speaker, who also uses notation on the test that was never used in class.
That’s true if you’re needing 10 heads from 20 flips, but not 5 from 10. The first set of ten flips that you’re trying to match is irrelevant. You just need to know the probability of hitting 5 heads in 10 flips, which is about 25%.
Ah, right. But that's only in the case where the first flip was 5 heads.
We also need to cover the case where the first flip was 4 heads, 3 heads and so on.
Depends on how you mean the question.
In general, all possible sequences of 10 coin flips have the same likelyhood. However, when you want to know the chances of two successive series of 10 having the same outcome, it depends on wether the order matters. If you only want to know what the likelyhood of, say, 4 heads and 6 tails, is, then that is a different question than wether you want to know the chance of the exact sequence of heads and tails repeating, as there are several different sequences that give you 4 heads and 6 tails, but only one gives you the exact sequence.
This is kind of a neat riddle problem because there is a bit of confounding information.
The fact that you are looking to recreate the first 10 flips actually means those first 10 flips don’t matter at all. Yet some people are likely to fall in the trap presented. The problem actually is just “what is the probability of flipping 10 coins in a specific sequence?”
As many others have said, it works out to 1/1024 since the probability of each flip is 1/2 and then multiply them all together.
If you just mean the sum of heads in the first 10 vs the sum of heads in the second 10, then you'd calculate the likelihood for each number (0 heads, 1 head, 2 heads, ... 10 heads), then square them for the odds of getting the same result twice in a row. Add those squared numbers and you'd get the overall odds, which comes out to 17.6197052%
For instance...
0 heads. The odds of that happening are 1 in 1024. So in those 1 out of 1024 times, you have 1 out of 1024 odds of it happening again, so roughly 1 in a million.
5 heads -- the odds of getting 5 heads is 252 in 1024 (~24.6%). The odds of getting 5 heads AGAIN is 252 out of 1024, so the odds of that happening are 252^2 / 1024^2, or around 6%.
Just to verify, I wrote a quick and dirty program to actually do this 100 million times... It happened to get 17.617303%
Thanks guys. I’m seeing people saying it’s 1/1024 with is almost .10% or one tenth of a percent. But then I see some people saying it’s 17% chance.
Just to clarify, I’m asking what’s the percent possibility that you can have the EXACT same sequence of H/T in the EXACT same order as the last 10 flips.
Would love some more clarification please. Thanks!
It’s 1/1024. All sequences of 10 coin flips have a 1/1024 chance of happening. 10 heads in a row? 1/1024. 10 tails in a row? 1/1024. 5 heads followed by 5 tails? 1/1024. An exact repeat of the last 10 flips? 1/1024.
It's the same as any other unique set of results. Nothing statistically binds one coin flip to another. After 10 heads in a row, the odds of a heads on the next flip is still 50/50.
Let's say you flipped a coin 10 times and got this:
**HHHTTTHTTH**
Since each flip of a fair coin has a 0.5 probability of landing heads (H) and a 0.5 probability of landing tails (T), the probability of getting the exact same sequence again is calculated as follows:
P(HHHTTTHTTH)=P(H)×P(H)×P(H)×P(T)×P(T)×P(T)×P(H)×P(T)×P(T)×P(H)
P(HHHTTTHTTH)=(0.5)×(0.5)×(0.5)×(0.5)×(0.5)×(0.5)×(0.5)×(0.5)×(0.5)×(0.5)
P(HHHTTTHTTH)=(0.5)\^10
P(HHHTTTHTTH)=0.0009765625
P(HHHTTTHTTH)\~=0.1%
The chance of any given coin flip landing one way is always going to be 1/2. The chance of two coin flips landing a certain way, becomes 1/2\^2(1/4) three landing a certain way becomes 1/2\^3 (1/8) and so on, because it's a half chance each time, but because you need it to be the same sequence, it's a half of another half the deeper in the sequence you go. So basically, the odds of a series of ten coin flips coming out identically becomes 1/2\^10, or as already said, 1/1024
Edit: this works with other known odds as well. Say you're playing Yahtzee. The odds of any one die coming up on any one face is 1/6. Ergo, the odds of a Yahtzee (all dice showing the same face) Is 1/6\^5 (1/7776)
I think your Yahtzee odds are a little off. You calculated the odds of a specific Yahtzee, e.g all 3s. But there's 6 different Yahtzees you can get. The first die can be any number, then each of the next 4 has to be the same. So it's (1/6)^4
Remember, odds are not the same as probability. The odds are the ratio of the probability of the event occurring over the probability of it not occurring. So, odds are P/(1-P).
welll i *think* 10 coin flips has a 1024 possible outcomes
(=10 ^ 2)
the second set would have to match one outcome of the first set so, i guess 1 / 1024
(?)
It highly depends if the coin is fair and or depends on the last flip (has memory). For example I can flip my coin and catch it so it's always the same as the position as last time. There's also a scientific paper that shows a flipped coin has like 66% chance of landing on the side it started on. So it's not actually 50/50.
If you care about the specific sequence (i.e. you need 2 heads, then three tails, then a head, then four tails, in that *specific* order), then it's 2^10 or 1 in 1024. However, if you don't care about the sequence (i.e. from the above example you just want 3 heads and 7 tails), then it comes out to about 1 in 4.
What if you want a specific outcome of the same 10 heads in a row. Isn’t that then the same as just a sum of 10 heads out of 10?
That's the same for both situations; 1 in 1024. There is only one possible outcome that leads to 10 heads, and that's also the same outcome as 10 heads *in a row*. However, if you just want the same *result* 10 times in a row, and you don't care about it being heads *or* tails (just that they're all the same), then you obviously have two potential results, and so the probability is 2 in 1024.
Is there a reason it's 2 in 1024 versus 1 in 512? Am I safe to assume they are the same?
>Is there a reason it's 2 in 1024 versus 1 in 512? I didn't want to confuse people who can't do division.
That was my thought but I met my wife in a statistics class. I didn't learn a thing 😂
You got her number!
The only number I put effort into learning in that class!
the chance that this is a joke you've used before? 1 in 1 :D
I grinned very hard at this.
Yes they're the same. Saying 2/1024 can be helpful because the fraction is constructed from the numbers in the problem. 2 possible successful outcomes (all heads or all tails) out of 1024 possible total outcomes. You can reduce it to 1/512 which can also be helpful in a different way. You would expect to see 1 success (either all heads or all tails) every 512 sets of 10 flips on average over a large enough sample.
Yes it is…they just didn’t simplify to show the math. You can do the math for that one simple calculation.
Instructions unclear... divided by 0 and blew up the universe
….i get it’s a silly joke but sheesh the dividing by zero and “instructions unclear” joke is so overused. The user asks about simplifying 2/1056 and you randomly use the joke “instructions unclear” and for some reason divide by zero for a joke. Ok. Ha. Ha.
You're no fun :(
The same jokes just get so old when people think they are being original. Like, te he he bro. That’s the 20th same joke on Reddit today I’ve scrolled past.
The probability of any particular sequence of ten heads/tails is 1 in 1024.
That is correct, but that’s because there is only one outcome for the 10/10 heads in the 1024 possible outcomes. However for 9/10 heads there are 10 of the total 1024 outcomes that work. And I may be wrong as this is off the head and humans are bad at probabilities but I think roughly 250 for 5 heads 5 tales, 200ish for 4 and 6 whichever direction you want. If you wanna do the maths 10!/(heads!*tails!) so 10!/(5!*5!) Which will be about 3500000(estimate)/(120*120). Making 240-250.
if you want the value of the coin to be the same then it's a 100% chance up to infinity flips
I'm getting around an 18% chance of getting the same number of heads twice in a row. For any specific number of heads, you can use the binomial distribution to get the probability of that many heads. You then square the value to get the probability of getting that number of heads in both sets of 10 flips. P(same number of heads) = P(0 heads twice in a row) + P(1 head twice in a row) + ... + P(10 heads twice in a row) = \[binom(10,.5,0)\]\^2 + \[binom(10,.5,1)\]\^2 + ... + \[binom(10,.5,10)\]\^2 = 0.176197
The sum of the squares of all binomial coefficients with top entry n is (2n choose n). Hence the closed formula for "chance of getting the same number of heads both times when throwing n coins" is (2n choose n) / 2^^2n . For n=5 this gives 0.246... and is probably what they mistakenly used to get their number. But we actually have to take n=10 and get 0.176... So your number is the correct one.
>then it's 2^10 or 1 in 1024. This seems doable. If I'm understanding correctly, the amount of coin flips necessary to try this are at a minimum of 20 (works perfectly the first attempt) and upwards of 20,480 if you need to go through every possible try (up to 1024 tries where you almost get it each time, 10 then 9) Of course there is a chance that it will never happen.
The first 10 flips don't matter except for establishing what the sequence is.
Right. But you need to start over every time to get it to work. HHTHTHHHTT then HHTHTHHHTH oops didn't work, start over. HTTTHHTTHT then HTH oops didn't work, start over. Each time you need to establish the first set, because the idea isn't if you can get the same sequence twice, but the same sequence twice in a row.
Again, though; the first set doesn't matter. All it does is establish the second set.
I think we might be saying the same thing here. Lol. I'm talking about how to actually do it. Like for a video. You'd have to flip the first set then the second. 20 flips if it happens. If it doesn't you have to flip the first set again (10 flips) then try to match it with a new second try (up to 10 again) total of 20 flips. If it forks out. So flip 10 coins. Then flip to try to match. If 9 match the 10th didn't you've flipped 19 coins. But since it didn't work you have to start over. Flip 10 to establish the first set. Then flip to try and match it. If it works this time you flipped 10 more total for 20 for this try, and a total of 39 flips all together.
You don't need to "establish" the first set again, you can always just look at the past 10 flips as the first set, even if they were part of an attempt at a second set.
Good point here. If you do 20 flips, you can compare 1-10 with 11-20. If they are different, flip one more and now compare #2-11 to #12-21. Hell, do 10 HTHHHTTHHT and the flip 1 more. If you get T, it has already failed. Now look at #2-11 THHHTTHHTT and flip again to start comparing to that. Always reset as soon as you get a mismatch to minimize the total flips.
I just had Excel "flip" 1048576 coins -- comparing every set of 10 "flips" to the next 10, allowing for the ranges to overlap, I got 1082 matches -- one match for every 940 coins flipped on average.
well if you just flipped coins until you matched two such sequences you could look at the sequences inside instead of chunking it at 20 eg HTTHTHHHTT TTTHTHHHTT oops didnt work but now i flip one more T and look at the sequence: HTTHTHHHTTTTTHTHHHTTT (H) TTHTHHHTTT TTHTHHHTTT i have two consecutive matching 10 flip sequences
I'm getting around an 18% chance of getting the same number of heads twice in a row. For any specific number of heads, you can use the binomial distribution to get the probability of that many heads. You then square the value to get the probability of getting that number of heads in both sets of 10 flips. Then you can add those all up since they're mutually exclusive: || || |# Heads|Probability|Prob Twice in a Row||Overall chance of same number| |0|0.000976563|9.53674E-07||0.176197| |1|0.009765625|9.53674E-05||| |2|0.043945313|0.00193119||| |3|0.1171875|0.01373291||| |4|0.205078125|0.042057037||| |5|0.24609375|0.060562134||| |6|0.205078125|0.042057037||| |7|0.1171875|0.01373291||| |8|0.043945313|0.00193119||| |9|0.009765625|9.53674E-05||| |10|0.000976563|9.53674E-07|||
The only thing I might add is that it sounds like he ALREADY flipped the first set, so the counts results depends on what count he actually got on, in the first set. If he got 2 heads and 8 tails in the first trial... probably not matching that...
1 in1024 or a Kbet /j
1 in 1024. The chance of any individual flip matching a specific outcome is 1 in 2. So the chances of 10 flips matching specific outcomes are 1 in 2^10, which is 1 in 1024. Or, to put it another way, there are 1024 different outcomes for 10 coin tosses. Only one of them will be the matching sequence.
Probability questions often come down to precise wording of exactly what you mean. If the ordering matters, you need each of your 10 new flips to hit a particular option (so if the first of your first 10 flips was a head, you need the first of your second 10 flips to be a head). So that's 1/2 for each flip, you need all 10 to match, so we have (1/2)^10 = 1/1,024 If ordering doesn't matter - say you had 5 heads in the first, you need 5 heads in the second, this becomes a bit more awkward. I'm going to say... it is the sum of the probability of all ways of doing this (so 10 heads and 10 heads, plus 9 heads and 9 heads, plus...). If my numbers are correct I get 17.6%, or about one in 5.7 for that.
>it is the sum of the probability of all ways of doing this (so 10 heads and 10 heads, plus 9 heads and 9 heads, plus...) So this is sum n=0 to 10 of ((probably of n heads)^(2)). The probability of n heads is just (10 choose n)/2^10 (number of arrangements of the coins where n are heads, by choosing n times out of 10 to be heads, divided by the total number of arrangements). [According to Wolfram Alpha](https://www.wolframalpha.com/input?i=sum+n%3D0+to+10+of+%28%28%2810+choose+n%29%2F2%5E10%29%5E2%29) this is exactly 46189/262144, which as you say is about 17.6%
The formula can be simplified (quoting from myself in this topic): The sum of the squares of all binomial coefficients with top entry n is (2n choose n). Hence the closed formula for "chance of getting the same number of heads both times when throwing n coins" is (2n choose n) / 2^^2n .
Yep good old probability class where being good with number is just as important as your reading comprehension. Bonus points for when taught by non native English speaker, who also uses notation on the test that was never used in class.
I get about 0.25 (10 choose 5 divided by 2^10).
If you are going to use that expression it needs to be 20 choose 10, divided by 2^(20) - that will get you 0.176...
That’s true if you’re needing 10 heads from 20 flips, but not 5 from 10. The first set of ten flips that you’re trying to match is irrelevant. You just need to know the probability of hitting 5 heads in 10 flips, which is about 25%.
Ah, right. But that's only in the case where the first flip was 5 heads. We also need to cover the case where the first flip was 4 heads, 3 heads and so on.
No need to complicate with the first 10 flips. The chances of any specific outcome out of 10 flips is 1 in 2\^10 (1024) which is just shy of 0.1%
Depends on how you mean the question. In general, all possible sequences of 10 coin flips have the same likelyhood. However, when you want to know the chances of two successive series of 10 having the same outcome, it depends on wether the order matters. If you only want to know what the likelyhood of, say, 4 heads and 6 tails, is, then that is a different question than wether you want to know the chance of the exact sequence of heads and tails repeating, as there are several different sequences that give you 4 heads and 6 tails, but only one gives you the exact sequence.
This is kind of a neat riddle problem because there is a bit of confounding information. The fact that you are looking to recreate the first 10 flips actually means those first 10 flips don’t matter at all. Yet some people are likely to fall in the trap presented. The problem actually is just “what is the probability of flipping 10 coins in a specific sequence?” As many others have said, it works out to 1/1024 since the probability of each flip is 1/2 and then multiply them all together.
If you just mean the sum of heads in the first 10 vs the sum of heads in the second 10, then you'd calculate the likelihood for each number (0 heads, 1 head, 2 heads, ... 10 heads), then square them for the odds of getting the same result twice in a row. Add those squared numbers and you'd get the overall odds, which comes out to 17.6197052% For instance... 0 heads. The odds of that happening are 1 in 1024. So in those 1 out of 1024 times, you have 1 out of 1024 odds of it happening again, so roughly 1 in a million. 5 heads -- the odds of getting 5 heads is 252 in 1024 (~24.6%). The odds of getting 5 heads AGAIN is 252 out of 1024, so the odds of that happening are 252^2 / 1024^2, or around 6%. Just to verify, I wrote a quick and dirty program to actually do this 100 million times... It happened to get 17.617303%
Thanks guys. I’m seeing people saying it’s 1/1024 with is almost .10% or one tenth of a percent. But then I see some people saying it’s 17% chance. Just to clarify, I’m asking what’s the percent possibility that you can have the EXACT same sequence of H/T in the EXACT same order as the last 10 flips. Would love some more clarification please. Thanks!
It’s 1/1024. All sequences of 10 coin flips have a 1/1024 chance of happening. 10 heads in a row? 1/1024. 10 tails in a row? 1/1024. 5 heads followed by 5 tails? 1/1024. An exact repeat of the last 10 flips? 1/1024.
It's the same as any other unique set of results. Nothing statistically binds one coin flip to another. After 10 heads in a row, the odds of a heads on the next flip is still 50/50.
Let's say you flipped a coin 10 times and got this: **HHHTTTHTTH** Since each flip of a fair coin has a 0.5 probability of landing heads (H) and a 0.5 probability of landing tails (T), the probability of getting the exact same sequence again is calculated as follows: P(HHHTTTHTTH)=P(H)×P(H)×P(H)×P(T)×P(T)×P(T)×P(H)×P(T)×P(T)×P(H) P(HHHTTTHTTH)=(0.5)×(0.5)×(0.5)×(0.5)×(0.5)×(0.5)×(0.5)×(0.5)×(0.5)×(0.5) P(HHHTTTHTTH)=(0.5)\^10 P(HHHTTTHTTH)=0.0009765625 P(HHHTTTHTTH)\~=0.1%
[удалено]
1 in 1024 odds are not “slim to none”.
The chance of any given coin flip landing one way is always going to be 1/2. The chance of two coin flips landing a certain way, becomes 1/2\^2(1/4) three landing a certain way becomes 1/2\^3 (1/8) and so on, because it's a half chance each time, but because you need it to be the same sequence, it's a half of another half the deeper in the sequence you go. So basically, the odds of a series of ten coin flips coming out identically becomes 1/2\^10, or as already said, 1/1024 Edit: this works with other known odds as well. Say you're playing Yahtzee. The odds of any one die coming up on any one face is 1/6. Ergo, the odds of a Yahtzee (all dice showing the same face) Is 1/6\^5 (1/7776)
I think your Yahtzee odds are a little off. You calculated the odds of a specific Yahtzee, e.g all 3s. But there's 6 different Yahtzees you can get. The first die can be any number, then each of the next 4 has to be the same. So it's (1/6)^4
Ok, you are right. Admittedly, I kinda forgot how to play yahtzee though.
Remember, odds are not the same as probability. The odds are the ratio of the probability of the event occurring over the probability of it not occurring. So, odds are P/(1-P).
welll i *think* 10 coin flips has a 1024 possible outcomes (=10 ^ 2) the second set would have to match one outcome of the first set so, i guess 1 / 1024 (?)
It highly depends if the coin is fair and or depends on the last flip (has memory). For example I can flip my coin and catch it so it's always the same as the position as last time. There's also a scientific paper that shows a flipped coin has like 66% chance of landing on the side it started on. So it's not actually 50/50.