Wikipedia:Reference desk/Archives/Mathematics/2014 November 3
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November 3
editSuper task and probability
editIf the probability of obtaining head on a coin is p and I am allow to flip the coin infinity about of times. Then as long as p is greater than zero then I will get a head eventually.
But if the above is true then if the probability of obtaining head is v and v>p then I would still get a head eventually.
but if the probabity of obtaining head for the n toss is f(n) and f(n) > 0 for all n, then I can possibly get a situation where I will not get a head eventually. And this is driving me crazy.
For example:
f(n)=1/(10^n)
- First toss: f(1)= 1/10 = 0.1
- Second toss: f(2)= 1/100 = 0.01
- Third toss: f(3)= 1/1000 = 0.001
So the probabity of me eventually getting a head on a coin toss is
Pr(eventually getting a head) = 0.1 + 0.9 * 0.01 + 0.9 * 0.99 * 0.001 + 0.9 * 0.99 * 0.999 * 0.0001 + ...
I noticed that
Pr(eventually getting a head) < 0.1 + 0.01 + 0.001 + 0.0001 + ...
Pr(eventually getting a head) < 0.1111...
So what can I conclude about this? That flipping a coin infinity number of times may not get you a head eventually if the probability is becoming smaller and smaller for each toss even if you have an infinity number of tosses. 202.177.218.59 (talk) 00:36, 3 November 2014 (UTC)
- It seem unbelievable to me that even if I have an infinite amount of opportunity to perform a task, and each opportunity has a positive finite probability of success. It is still possible for me to never accomplish the task successfully despite an infinite number of attempts. 202.177.218.59 (talk) 00:54, 3 November 2014 (UTC)
- The probability of getting a head on the each individual flip is still 0.1 which is what you need to use when constructing the probability chain
- Pr(eventually getting a head) = 0.1 + 0.9*0.1 + 0.9*0.9*0.1 + 0.9*0.9*0.9*0.1 + ... -> 1
- Dragons flight (talk) 00:58, 3 November 2014 (UTC)
- I have already stated that the probability of getting a head is dependent on the toss number. This is not an ordinary coin where the probability of head is fixed. So what you said is incorrect. Stop thinking in terms of coin flips. This about an attempt to perform a task where each subsequent attempt is much harder than the last attempt. 202.177.218.59 (talk) 03:18, 3 November 2014 (UTC)
- My bad, I misunderstood where you were going. Yes, if the probability of getting a head on the n-th toss is some arbitrary function f(n) > 0 then you can choose f(n) in such a way that the probability of getting a head in infinite tries is less than 1. It will require that f(n) -> 0 as n -> infinity. Actually you can set the final probability to be any number > 0. In general, Pr(eventually getting a head) < Sum(f(n)), which is basically the property you used above. Dragons flight (talk) 03:55, 3 November 2014 (UTC)
- To expand on this a bit, the initial statement is wrong in two ways. First it's assuming a constant (>0) probability of success. Second, the conclusion is that the probability of eventually getting a success is 1, but that's not the same as saying you must eventually get a success. There are many examples of events that occur with probability 0 but are still possible. For one, there is probability of 0 that a random number between 0 and 1 will be 1/2, but it's still possible that the number will be 1/2. Ignoring the second point, which is getting off-topic, if you don't assume that the probability of success is constant then the conclusion doesn't hold. For example if the probability of success for the first roll is 1/4, the second 1/9, the third 1/16, etc., then the probability of eventually getting success is only 1/2. In other words you have a equal chance of eventually getting a success and never getting a success no matter how long you continue. In general, if the probability of success is p(n)<1 then the critical factor is whether the series ∑p(n) converges. If it does, as in the case p(n)=1/(n+1)2 or p(n)=.1n then the probability of going forever without success is positive. If it doesn't, as in p(n)=p or p(n)=1/(n+1) then the probability is 1 that you eventually get success. The p(n)=1/(n+1) case is a bit odd in that even though you eventually get success with probability 1, the expected number of tries needed is infinite. --RDBury (talk) 13:59, 3 November 2014 (UTC)
- My bad, I misunderstood where you were going. Yes, if the probability of getting a head on the n-th toss is some arbitrary function f(n) > 0 then you can choose f(n) in such a way that the probability of getting a head in infinite tries is less than 1. It will require that f(n) -> 0 as n -> infinity. Actually you can set the final probability to be any number > 0. In general, Pr(eventually getting a head) < Sum(f(n)), which is basically the property you used above. Dragons flight (talk) 03:55, 3 November 2014 (UTC)
- Thank you, I didn't realized how dangerous playing around with infinity and supertasks are. Ordinary common sense does not apply and being given an infinite number of attempts is not an automatic "get out of jail free card". 220.239.43.253 (talk) 03:13, 4 November 2014 (UTC)