1 - x approach is the best while answering an 'at least' or 'at most' kind of probability questions, as it saves all the time wasted in imagining the various possibilities of a complex situation that could lead to careless mistakes on most of the occasions.

As in this case in particular, it is important for us to know what numerical values you took while using your so called 'conventional approach' as under:

The chance that John finds at least one pair of cards that have the same value

= Probability of getting one pair of cards that have the same value+ probability of getting both pair of cards that have the same value

I am more than 99 percent sure that you have missed a count here, which every second test taker is prone to while using your so called 'conventional approach' in this problem.

4 cards out of 12 can be dealt in 12C4 ways.

What is the chance that John finds at least one pair of cards that have the same value?

= 1 - What is the chance that John finds NO pair of cards that have the same value?

We have 6 different denominations in the shape of 12 cards

First card can be drawn in 12 ways, and in order to make certain that NO pair of cards is there that have the same value, the second draw could be made in 10 ways, subsequently, third in 8 ways, and the fourth in 6 ways. Hence, a total of 12 × 10 × 8 × 6 ÷ 4! ways are there to favor the translated condition.

1 - (12 × 10 × 8 × 6 ÷ 4!) ÷ 12C4

= 1 - (12 × 10 × 8 × 6) ÷ (12 × 11 × 10 × 9)

= 1 - 16/33

= [spoiler]17/33

C[/spoiler]

sanju09 , can u pls explain that why did u divide 12 * 10*8 *6 by 4 !.... i could figure out everything else ,but this one is not gng down my throat...

As Sanju said lets find the probability where all the four cards yield different value
Total cards= 2 (1's)+2 (2's)+.....+2 (6's)= 12 Cards

Out of this if i have to pick 4 different value cards it would be
1st card (any one card out of 12), 2nd card( any one card out of 10),3rd card (any one card out of 8), 4th card (any one card out of 6)
progressively we are decreasing it by 2 as the previous numbers selected + its compelementary suit should be avoided
Given Event = 12*10*8*6

Total no of selection possible is 12*11*10*9

There probability = 1-(12*10*8*6)/(12*11*10*9)

This gives you 1-16/63, hence the answer is answer is 17/63. Hope this explains.

Regard,
Mani

manishankar wrote:As Sanju said lets find the probability where all the four cards yield different value
Total cards= 2 (1's)+2 (2's)+.....+2 (6's)= 12 Cards

Out of this if i have to pick 4 different value cards it would be
1st card (any one card out of 12), 2nd card( any one card out of 10),3rd card (any one card out of 8), 4th card (any one card out of 6)
progressively we are decreasing it by 2 as the previous numbers selected + its compelementary suit should be avoided
Given Event = 12*10*8*6

Total no of selection possible is 12*11*10*9

There probability = 1-(12*10*8*6)/(12*11*10*9)

This gives you 1-16/63, hence the answer is answer is 17/63. Hope this explains.

Regard,
Mani

Total possible selections are 12C4, not 12*11*10*9 and the ways favoring otherwise are 12*10*8*6/4!, not just 12*10*8*6. By chance 4! is a common in denominators, it cancels out without effecting the overall result. Moreover, it's 33 not 63 in the denominator when (12*10*8*6)/(12*11*10*9) is worked out.

crackinggmat wrote:sanju09 , can u pls explain that why did u divide 12 * 10*8 *6 by 4 !.... i could figure out everything else ,but this one is not gng down my throat...

If 4 cards are the 4 free spaces to be filled by pairs of 1 through 6 under the given constraints, then 12 X 10 X 8 X 6 gives us the various permutations of filling the 4 free spaces, but since the cards are not picked one after the other, rather these are picked in a single go, order doesn't matter. It's just the selection of 4 things what is needed here, and nCr = nPr/r!. Our nPr is 12 X 10 X 8 X 6 and r is 4.