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Discussion in 'Psychology' started by sittingduck, Oct 16, 2008.

1. ### sittingduck

1- Four cards are dealt off the top of a well shuffled deck. You have a choice:
(a) To win \$1 if the first card is a club, and the second a diamond, and the third is a heart, and the fourth is a spade, in this order.
(b) To win \$1 if the four cards are of four different suits, in any order.

Which option is better? Or are they equivalent?
Justify by showing what is the probability in each case.

2. ### goldenarm

Option b is better but I don't feel like doing the math.

3. ### Fractals 'R Us

in option A you have 1/52 ways to fill the first card, 1/51, 1/50, 1/49 for the rest, multiply it all out you have a chance in 6.5 million. Option B, you have 4 ways of 52 to fill the first one, 3 of 51 to fill the second, 2/50, 1/49... multiply it all out you have a chance in 270,000 or so. Maybe I'm wrong but that's what I think.

Why am I doing this math?

4. ### Gringinho

Of course it is B...
Like Fractals said - do the maths - it's discrete maths.

5. ### axehawk

I'm pretty sure your math is wrong here.

You don't have a 1/52 chance in the first example, you have a 13/52 chance.

6. ### telozo

Case (a) is a subset of case (b). So (b) is the best choice, no matter the probabilites.

8. ### Bond

ok my guess
A- 13/52 x 13/51 x 13/50 x 13/49 = .0043%
B-52/52 x 39/51 x 26/50 x 13/49 = .105%
of course B is better

This is how I see it:

Option (a) and option (b) are very similar. They both require the player to draw 4 cards in different suits. The only different is that for option (a), you need to get four different suits in a defined order.

So, option (b) would be less difficult given one less criteria to fulfill.

PA

10. ### AMT4SWA

YOU WIN!!!!!!

#10     Oct 17, 2008
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