Solving problems in Reverse

SheetWise · #1 10-10-2007, 11:36 AM

If I want to calculate the probability of having at least one diamond in a draw of two cards -- I will subtract from 1 the probability of having no diamonds in two cards,

1 - (39/52 * 38/51) = 1 - .5588 = .4412

What if I wanted to work the problem forward instead of backwards -- what would be the easiest way?

R Gibert · #2 10-10-2007, 12:56 PM

The answer depends on whether you would accept an approximation and how accurate you want the approximation to be e.g. my answer to all problems is 1 [img]/images/graemlins/wink.gif[/img]

The most obvious "forward" way as you call it would be:

13/52 + (39/52)*(13/51)

It is always possible to factor out the 13/52 in this type of problem i.e.

<font class="small">Code:</font><hr /><pre>x/y + ((y - x)/y)*(x/(y - 1)) = x/y + (y - x)*x/((y - 1)*y)
= x/y + ((y - x)/(y - 1))*(x/y)
= (x/y)*(1 + (y - x)/(y - 1))</pre><hr />

This would look like:

(13/52)*(1 + 39/51)

in the problem at hand. It is about as economical to calculate as the "reverse" way, but has drifted away from how most people think.

jay_shark · #3 10-10-2007, 01:00 PM

A is the event that the first card is a diamond but the second card isn't .

B is the event that the second card is a diamond but the first isn't .

C is the event that both cards are diamonds .

We simply add these 3 mutually exclusive events .

P(A)= 13/52* 39/51 = P(B)
P(C) = 13/52*12/51

P(A) + P(B) + P(C) = 0.4412

jay_shark · #4 10-10-2007, 02:13 PM

Here is a probabilistic explanation for R. Gilbert's answer .

13/52 is the probability the first card is a diamond . This may also include the probability when the second card is a diamond or not .

The probability the first card isn't a diamond but the second card is :

39/52*13/51 .

These two events are mutually exclusive and we can add them directly .

SheetWise · #5 10-11-2007, 08:14 AM

Thank you both for your answers. I know I asked for the easiest way (my bad) -- reviewing the two answers I should have said the most intuitive way. Thank you R Gilbert for the easiest -- Thank you jay_shark for the most intuitive and an explanation of the easiest.