Prob 2 red aces appear on the flop

Kazlic · #1 09-10-2006, 10:01 PM

What is the probability that both red aces appear on the flop?

This is an interesting question because there are 3 slots for an event involving only 2 cards. Normally I would just multiply the probabilities together (2/52) * (1/51) * (?/?) , but as you can see I have no idea what to do for the third slot.

Anybody know?

Thanks,
Kazlic

BruceZ · #2 09-10-2006, 11:29 PM

[ QUOTE ]
What is the probability that both red aces appear on the flop?

This is an interesting question because there are 3 slots for an event involving only 2 cards. Normally I would just multiply the probabilities together (2/52) * (1/51) * (?/?) , but as you can see I have no idea what to do for the third slot.

Anybody know?

[/ QUOTE ]

2/52 * 1/51 * 3 = 1 in 442. Multiply by 3 since the non-red ace can come in any of 3 positions.

Or, 50/C(52,3) = 50/22100 = 1 in 442. There are 50 flops with both red aces since there are 50 ways to choose the 3rd card. The total number of flops is C(52,3) = 52*51*50/(3*2*1) = 22100.

Kazlic · #3 09-11-2006, 12:03 AM

Amazing. Not only did you answer my question, you explained it so well.

Thank you so much!

Kazlic · #4 09-12-2006, 02:09 PM

Aren't there 6 arrangements of the 2-red aces and the non-red-ace?

Ad Ah X
Ah Ad X
Ah X Ad
Ad X Ah
X Ad Ah
X Ah Ad

So would the calculation be (2/52) * (1/51) * 6 = 1 in 221.

Or am I doing something wrong here?

rjp · #5 09-12-2006, 02:29 PM

Easiest way for me is to look at it like this (where C is a red ace):

CCx, CxC, xCC

CCx = (2/52)*(1/51)*(50/50)
CxC = (2/52)*(50/51)*(1/50)
xCC = (50/52)*(2/51)*(1/50)

Add this up and it's ~00.23% or 1 in 442 [img]/images/graemlins/grin.gif[/img]

BruceZ · #6 09-12-2006, 03:10 PM

[ QUOTE ]
Aren't there 6 arrangements of the 2-red aces and the non-red-ace?

Ad Ah X
Ah Ad X
Ah X Ad
Ad X Ah
X Ad Ah
X Ah Ad

So would the calculation be (2/52) * (1/51) * 6 = 1 in 221.

Or am I doing something wrong here?

[/ QUOTE ]

There are 6 arrangements of the 3 cards, but when you do (2/52) *(1/51), this is actually the probability of 2 of these 6 arrangements, such as Ah Ad X and Ad Ah X, since the first 2/52 can be either red ace. So you only multiply by 3 to account for the 3 positions of the non-red ace, and this will count all 6 arrangements.

If you want to think about all 6 arrangements, then you would say that the probability of each one of these is 1/52 * 1/51 * 50/50, and then multiply that by 6 since each of the 6 arrangements are equally likely, 1/52 * 1/51 * 50/50 * 6 = 1 in 442.