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11-15-2002, 07:28 PM
7 out of 10 hands, single deck, reshuffle after every hand.

meant to include predetermined, iow before the first hand say the king of spades.

Thanks

BruceZ
11-16-2002, 12:35 AM
sum[k=7,10]C(10,k)*(1/52)^k*(51/52)^(10-k) = 1.1 x 10^-10
= 1 in 9 billion (American billion).

11-18-2002, 04:13 AM
Probability of the dealer having the same up card in BJ

Hi BruceZ,

The formula you show:

sum[k=7,10]C(10,k)*(1/52)^k*(51/52)^(10-k) = 1.1 x 10^-10
= 1 in 9 billion (American billion).

Includes the probability for the same BJ predetermined card being the upcard for 7 out of 10 deals + 8 out of 10 deals + 9 out of 10 deals + 10 out of 10 deals. No big deal -- but did not the problem definition only ask for the chance of the same upcard coming up exactly 7 times out of 10. Your formula is for 7 or more. The problem definition should have said for 7 or more repeats for a sample of 10 deals.

Actually Bruce, I appreciate your expertise and learned a lot from your reply on this and other probability things. Thanks for your effort.

11-18-2002, 02:59 PM
Thank you both,

I had worked it out as (1/52)^7*(51/52)^3 for exactly 7 out of 10 times. That is actually what I was looking for, but I'm still waiting for a guy, to say yes that is correct.

THX again
AA

Homer
11-18-2002, 03:36 PM
The way you did it is for exactly 7/10 times with an exact required order. Do you care about the order, or do you simply want 7/10, in any order?

If you don't care about the order, you multiply (1/52)^7*(51/52)^3 by C(10,7). The C(10,7) is how many ways you can take 7 things out of 10 (120). Note that this equation is the same as Bruce's, with K = 7.

-- Homer

11-18-2002, 07:08 PM
Correct ans.: should be: 120 * (1/52)^7*(51/52)^3 for exactly 7 out of 10 times. Where combination of 10 things taken 7 at a time = the 120 in the answer.

CWH