Two Plus Two Older Archives Probability of the dealer having the same up card in BJ
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#1
11-15-2002, 07:28 PM
 Am Alert Member Join Date: Nov 2002 Posts: 80
Probability of the dealer having the same up card in BJ

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
#2
11-16-2002, 12:35 AM
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636
Re: Probability of the dealer having the same up card in BJ

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).
#3
11-18-2002, 04:13 AM
 Guest Posts: n/a
Re: Probability of the dealer having the same up card in BJ

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.

#4
11-18-2002, 02:59 PM
 Am Alert Member Join Date: Nov 2002 Posts: 80
Re: Probability of the dealer having the same up card in BJ

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
#5
11-18-2002, 03:36 PM
 Homer Senior Member Join Date: Sep 2002 Posts: 5,909
Re: Probability of the dealer having the same up card in BJ

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
#6
11-18-2002, 07:08 PM
 Guest Posts: n/a
Re: Probability of the dealer having the same up card in BJ

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
#7
11-19-2002, 03:26 PM
 Am Alert Member Join Date: Nov 2002 Posts: 80
Re: Probability of the dealer having the same up card in BJ

Thx guys! yes, that was what I was missing.
#8
11-21-2002, 03:02 AM
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636
Re: Probability of the dealer having the same up card in BJ

No, because if you have 8,9, or 10 you certainly have 7. The problem didn't restrict the other hands, which is why I answered as I did without further explanation.

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