4 ms·
I don't have time to put numbers on it, but consider the following with 3 tosses only. There is 8 different enumerations: * HHH * THH * HTH * TTH * HHT * THT *
by hvidgaard 11y ago
I don't have time to put numbers on it, but consider the following with 3 tosses only. There is 8 different enumerations:
* HHH
* THH
* HTH
* TTH
* HHT
* THT
* HTT
* TTT
Alice is looking for HT, so she will succeed in HTH, HHT, THT, HTT, that is 4 out of 8 possible outcomes. Bob on the other hand is looking for HH, that is only in HHH, THH, HHT, 3 out of 8 possible outcomes. So while HH and HT are equal in probability when you consider 2 coin flips, the combination of HT happens more often than HH. This is the case with 3 coin flips - there is no guarantee it translate to the same with more coin flips, but that is my bet.
- calebm 11y agoI came across this scenario while interviewing at a HFT firm about a year ago - caught me off guard :)
- deleted 11y ago[deleted]
- cousin_it 11y agoAmusingly, the question "which substring occurs earlier on average" is different from the question "which substring is more likely to occur before the other". In fact the second question sometimes has a circular answer! For example, THH typically (with >50% probability) occurs before HHT, which typically occurs before HTT, which typically occurs before TTH, which typically occurs before THH. Also the question "which substring occurs earlier on average" is intimately connected with algorithms for substring search. For example, if you want to check that a string doesn't contain HHH, you need to look at every third character, but for THH that's not enough. Fascinating stuff.
- akavi 11y agoThe key "insight" that helped me intuit this is that "HH" can self overlap, while "HT" can't.