4 ms·
The most interesting part of this paper to me is that "HHT" is more likely than "HHH" to appear in a sub-set of tosses of length 4 or 12 (for example), and that
by ccvannorman 8y ago
The most interesting part of this paper to me is that "HHT" is more likely than "HHH" to appear in a sub-set of tosses of length 4 or 12 (for example), and that any sequence like HHH or HHT has a "wait time" calculated by mathematicians. The wait times are how many flips on average you need to produce the sequence and they are not equal -- HHHH has a wait time of 30 vs HHHT has a wait time of 16 (not sure if I got the numbers exactly right.)
So the answer to the question "Which is more likely to be seen in a sequence of 10 coin flips, HHH or HHT", the answer is surprisingly, "HHT".
This is the key point of the paper. People are shown a sequence and they answer the question as if the sequence was pulled from a longer sequence (though that isn't strictly what the question asks).
- johnsimer 8y agoThis is fascinating to me. Do you have a link to the paper? I can't quickly wrap my mind around why HHT would be have a shorter wait time than HHH
- fpalmans 8y agoNot having read the paper to which to parent referred, but I might be able to shed some light on this... Let us describe the state of the previous two tosses as XX, with X being either H or T. If --(X)--> describes the next toss, we can describe the transition between states as follows: HH --(H)--> HH HH --(T)--> HT and HT --(H)--> TH HT --(T)--> TT etc. If we are waiting for HHT, starting from state HH HH --(H)--> HH : or, we get back to state HH HH --(T)--> "HHT" : HHT is found and we stop searching as opposed to waiting for HHH HH --(H)--> "HHH" : HHH is found, and we stop HH --(T)--> HT : and we are in a new state which requires at least two more --(H)--> in sequence before we return to state HH. I hope this helps. EDIT: reformatting & typo