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 Don't try this barefoot. (Posted on 2011-12-01)
You have a box of tacks for which it has been determined that if you toss them on the floor precisely 36% will land point up.

How many of these tacks could you toss to make the most likely result be that 10 of them will land point up.

 No Solution Yet Submitted by Jer Rating: 5.0000 (1 votes)

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 computer-assisted solution | Comment 1 of 3

Assuming that 36% is the expected number of pointed-up tacks, it's implied that each tack has probability 0.36 of coming up heads. Only an infinite-size box would assure the 36% ratio, but even a 100-tack box would have an expected number being 36.

As a start one can estimate 10/0.36 = 27.777... as being near the required answer.

The following program evaluates for given numbers of tacks tossed between 26 and 32, the probability of 8, 9, 10, 11 or 12 tacks landing upward:

`   10   for N=26 to 32   15    print N;   20    for Succ=8 to 12   30      Prob=combi(N,Succ)*0.36^Succ*0.64^(N-Succ)   40      print using(3,5),Prob;   50    next Succ   60    print   70   next N`
`It finds:`
`  N       8        9       10       11       12 26   0.14303  0.16091  0.15387  0.12589  0.08852 27   0.13008  0.15447  0.15640  0.13596  0.10197 28   0.11655  0.14569  0.15570  0.14332  0.11421 29   0.10301  0.13520  0.15210  0.14778  0.12469 30   0.08990  0.12361  0.14601  0.14933  0.13300 31   0.07755  0.11147  0.13795  0.14814  0.13888 32   0.06617  0.09926  0.12842  0.14447  0.14221 `

So, if 27, 28 or 29 tacks were tossed, 10 of them pointing up would be the most likely outcome, at probabilities of 0.15640, 0.15570 and 0.15210 respectively. The 27 count maximizes this probability, but all three values satisfy the requirement that 10 be the most likely outcome.

 Posted by Charlie on 2011-12-01 13:32:42

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