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
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!
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%
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.
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.
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)
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.
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.
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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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