Eli5: If I flip a coin 10x what’s the chances that the next 10 flips will be the exact outcome of the first 10?
In: Mathematics
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.
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.
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%
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.
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%
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
(?)
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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.
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.
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.
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