![]() On average we will have to wait longer for the monkey to to type ‘abracadabra’ than ‘abracadabrx’ In other words, the monkey needs to type the word ‘abracadabra’ completely, and that counts as one appearance, and then the monkey needs to type it completely again for the next appearance.Īnswer: a) is greater. ![]() Likewise, ‘abracadabrabracadabra’ is only one ‘abracadabra’. Second, if the monkey types ‘abracadabracadabra’ this only counts as one ‘abracadabra’. Another way of phrasing the question would be: over the long run, which of ‘abracadabra’ or ‘abracadabrx’ appears more frequently? The one that is more frequent is the one it takes, on average, less time to get to. When I say ‘the average time it will take the monkey to type abracadabra’, I do not mean how long it takes to type out the word ‘abracadabra’ on its own, which is always 11 seconds (or 10 seconds since the first letter is typed on zero seconds and the 11th letter is typed on the 10th second.) I mean the average of the time it takes to get to an ‘abracadabra’, either from the beginning of the experiment or from a previous appearance of ‘abracadabra’. ![]() It favours no letters: all letters at any second have a 1/26 probability of being typed.Ī) the average time it will take the monkey to type “abracadabra”ī) the average time it will take the monkey to type “abracadabrx”īefore I get to the answer, some clarifications. The monkey types at random, with a constant speed of one letter per second. A monkey is sitting at a typewriter that has only 26 keys, one per letter of the alphabet.
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