You play a coin flipping game with 5 coins. On round 1 you flip all of them. On round 2, you pick up all the ones that came up tails (leaving all the heads alone) and flip them again. You continue to do this until all the coins are heads. For example:
Round 1: H T T H T
Round 2: - H T - H
Round 3: - - T - -
Round 4: - - T - -
Round 5: - - H - -
Done in 5 Rounds.
What is the expected number of rounds you'll need to finish the game?
What is the probability you will finish the game in 3 rounds or less?
(In reply to
re: About the expected value by nikki)
"Expected value" is a technical statistical term, and just means the average value that you'd expect after a large number of trials. The results of any large collection of trials, being numbers, has several types of average, or "typicality" that can be specified. The expected value (or mean of the values) is never expected to be an integer. There are other measures of the "typical" number or result, such as the mode and the median.
For trials in which only integers make sense, it is only the mode that is expected to be integral. That is the value that comes up most commonly. The following table shows this is 3, as that has a 27.56% probability of being the number of rounds needed.
1 0.031250
2 0.206055
3 0.275604
4 0.211287
5 0.129019
6 0.071063
7 0.037265
8 0.019078
9 0.009652
10 0.004854
11 0.002434
12 0.001219
The median is the value at which you'd expect half the results to be more than, and half less than. Of course in this integral case, the best you can do is some number between 3 and 4. Regardless of what number you choose, slightly over 51% of trials will be less and under 49% more.
BTW, it would usually be said that the average family has 1.5 children, rather than that most families have 1.5 children. The former is a mean, while the latter would imply a mode. Actually the latter claims even more than a mode; it claims a majority, rather than the mere plurality that is needed for a mode.
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Posted by Charlie
on 2004-10-15 15:17:35 |