Mindbenders # 18

Last week, we featured a sophisticated New Age hustle. This week one offers a cruder, lowlife flimflam which demands applied physics and an exercise in pure logic.

1. The one-coin con

Could you drop a coin any normal coin from a height of around eight inches and make it stand on edge? Could anyone of your acquaintance? Try this one in a bar and thats a hint. If you work it out, count yourself in exalted company. Sir Isaac Newton proposed an ancestor.

2. Loon Tunes

Suppose you went to an insane asylum where the inhabitants consisted of staffers and pat- ients and strict segregation was not practised since everyone wore coveralls, and everyone knew everyone in that closed community.

 

Some of the staff had cracked under the strain of managing their charges and could now be certified insane. Some of the patients had meanwhile regained their balance, and could be considered sane. Moreover this was an asylum full of honest folk they all believed in telling the truth as they saw it. Naturally the insane made false statements in full faith, while the sane made true statements with a certitude that was almost pathological.

As a visiting inspector to an insane asylum, you have to judge which cases deserve review. One inhabitant makes a statement that leads to the deduction that he is a sane patient. Furnish some examples of such statements. Someone else makes a statement that leads to the deduction that he is an insane staffer. What could this be?

Answer to last weeks puzzle:

Just the one problem last time. A party, six random people, 12 astrological signs. Would you expect to gain by an even money bet that any two of them will share the same sign? The problem boils down to calculating the probability of such an event occurring.

Lets assume the first person has any given sign. The second person has a 1/12 chance of matching. Assuming he/she does not, the third party has a 2/12 chance of hitting one of the first two signs.

Similarly, the sixth party has a 5/12 chance of matching one of the earlier five signs, assuming no matches have already occurred.

If you remember your probability theory, you can use the mutually exclusive outcomes formula to calculate the proba-bility of (11/12x10/12x9/12x8/ 12x7/12) and come to the number 0.2228 which is the probability that none out of six will have a sign in common. Subtracting from 1, there is a 0.7772 chance of at least one sign matching.

A more painful if logical enough way of calculating this is via an iteration. The second chaps chance of matching being 1/12 of 1. If it doesnt match, the third chap has a 2/12 of 11/12 chance of matching. If this doesnt match, the fourth chap has a 3/12 chance of 1 minus the first plus second persons chances. As you carry on, add each chance up, or look at the remainder. It will give you the same answer.

The real odds for any two out of six persons sharing the same starsign is 0.7772 to 0.2228 or nearly 3.5 to 1 in favour. Thus, at even odds you stand to gain substantially, if you can con someone into taking the bet. Given the number of New Age Innumerates who believe in astrology, your chance of getting a taker or three is pretty high.