Toss a coin 10 times.

[calorific]

Same principle http://www.mathsisfun.com/data/quincunx.html

 

[TriggerSlab]

2 heads = 45 possible sequences = 45/1024 (about 4 and a bit %)
3 heads = 120 possible sequences = 120/1024 (nearly 12%)
4 heads = 210 possible sequences = 210/1024 (shade over 20%)
5 heads = 252 possible sequences = 252/1024 (nearly 25%)

And no heads is 1/1024 which is not even close to 1%.

Yup jeds is spot on. Just to expand his workings for anyone who's interested - how many possible sequences (of 10) have exactly 4 heads in them? Well, if we have 10 positions in the sequence then the first head has 10 possibilities, the second has 9 etc giving us 10 x 9 x 8 x 7 possible sequences. But because all heads are the same this will 'double count' many possibilities - if I allocate the first head to position one and the second head to position two then it's the same as doing it the other way round. So we need to remove this 'double counting' by dividing by the number of ways of arranging our four heads in a row. Using the same logic as above, this is 4 x 3 x 2 x 1. So our answer is:
10 x 9 x 8 x 7 / (4 x 3 x 2 x 1) giving 210.

 

[jeds]

Same principle http://www.mathsisfun.com/data/quincunx.html[/QUOTE]

What a great website. One of the better uses of t'internet.