Does any formula exists with which one can reckon the probability that x cards of any demonination will be dealt after h number of cards have been dealt depending on n number of decks.
Example: How great is the probability that three aces will be dealt after fifteen cards are dealt out of a two deck game. Since this also has great relevance in canasta, I'd be happy to receive an answer.
Thanks in advance.
Regards, Robert V. Lux
> Does any formula exists with which one can
> reckon the probability that x cards of any
> demonination will be dealt after h number of
> cards have been dealt depending on n number
> of decks.
> Example: How great is the probability that
> three aces will be dealt after fifteen cards
> are dealt out of a two deck game. Since this
> also has great relevance in canasta, I'd be
> happy to receive an answer.
You can use Peter Griffin's formula for the probability of a RC at a particular depth. Of course, you would need a counting system for tracking aces.
Sincerely,
Cacarulo
> You can use Peter Griffin's formula for the
> probability of a RC at a particular depth.
> Of course, you would need a counting system
> for tracking aces.
Well, I am not looking for the TC. I am looking for how great the probability is that any number of a specific card (not only aces that is) will be dealt within various ranges, such as 10, 20, 30 cards, etc.
Regards, Robert V. Lux
> Well, I am not looking for the TC. I am
> looking for how great the probability is
> that any number of a specific card (not only
> aces that is) will be dealt within various
> ranges, such as 10, 20, 30 cards, etc.
I'm not talking about TC. The formula gives you the probability of getting a particular RC at an specific depth (10, 20, 30 cards). Say your counting system only count the aces (or any other card) then you can ask: what is the probability of getting 3 aces (RC = +3) if there are 20 cards left? See what I mean?
Sincerely,
Cacarulo
> Example: How great is the probability that
> three aces will be dealt after fifteen cards
> are dealt out of a two deck game. Since this
> also has great relevance in canasta, I'd be
> happy to receive an answer.
This can be answered with a straightforward calculation, which I'll provide once I'm sure of one thing: do you want the probability of precisely three aces in the first 15 cards of 104, or do you want the probability of at least three aces?
Don
> This can be answered with a straightforward
> calculation, which I'll provide once I'm
> sure of one thing: do you want the
> probability of precisely three aces in the
> first 15 cards of 104, or do you want the
> probability of at least three aces?
I decided I'd just assume exactly three aces and provide the math for you.
There are 8 aces available from 104 cards. So, we have 96 "others." The idea is to simply specify, for all 15 cards, whether each card is an ace or an "other." Then, one final trick at the end gives the answer.
So, we begin like this: ace, ace, ace, followed by 12 "others."
That gives: (8/104) x (7/103) x (6/102) x (96/101) x (95/100) x ... x (85/90). I don't know what that equals yet, so let's hold it aside for a while.
Now, of course, the aces don't have to be the first three cards. They can appear spread around anywhere in the 15 positions. So, we need to multiply the above answer by the number of ways that the three aces can appear in the 15 positions. This is expressed by C15,3, or the combination of 15 things taken three at a time.
The above value is simply (15x14x13)/(3x2x1) = 455.
So, the final answer is whatever the long fraction value is, multiplied by 455.
I get for the top, 1/16,115. So that would make the answer: 455/16,115 = 0.0282345, or one chance in 35.4.
Don
> I decided I'd just assume exactly three aces
> and provide the math for you.
For this type of problem your formula is exact. The one that I suggested to Robert is a normal approximation.
Sincerely,
Cacarulo
> I get for the top, 1/16,115. So that would
> make the answer: 455/16,115 = 0.0282345, or
> one chance in 35.4.
Here is my answer using the Distribution of a Point Count (TOB, page 92):
For monitoring the presence of aces I'll use the following balanced count (A-T):
-12 1 1 1 1 1 1 1 1 1
If we wanted to monitor a different card we would put -12 in the corresponding position.
ss = 12 * 1^2 + (-12)^2 = 156
N = 104
n = 89 cards remaining
b = sqrt((156*15*89)/(13*103)) = 12.47
Now, we want exactly 3 aces within the first 15 cards. This corresponds to a RC of -24
r1 = -24 + 0.5 = -23.5
r2 = -24 - 0.5 = -24.5
z1 = |-24.5| / 12.47 = 1.96
z2 = |-23.5| / 12.47 = 1.88
Looking into the Normal Distribution table we found that:
n1 (z1) = 0.4750
n2 (z2) = 0.4699
The probability is equal to the difference of these values (n1 - n2)
Prob = 0.0051
Don, could you check your values again? It seems like there is a big difference.
Sincerely,
Cacarulo
> Don, could you check your values again? It
> seems like there is a big difference.
Anyone can take a calculator and double-check my figures, but the math is straightforward. It would be better if someone else tried to use a calculator to get my answer, I think.
Don
> Here is my answer using the Distribution of
> a Point Count (TOB, page 92):
> For monitoring the presence of aces I'll use
> the following balanced count (A-T):
> -12 1 1 1 1 1 1 1 1 1
> If we wanted to monitor a different card we
> would put -12 in the corresponding position.
> ss = 12 * 1^2 + (-12)^2 = 156
> N = 104
> n = 89 cards remaining
> b = sqrt((156*15*89)/(13*103)) = 12.47
> Now, we want exactly 3 aces within the first
> 15 cards. This corresponds to a RC of -24
> r1 = -24 + 0.5 = -23.5
> r2 = -24 - 0.5 = -24.5
> z1 = |-24.5| / 12.47 = 1.96
> z2 = |-23.5| / 12.47 = 1.88
> Looking into the Normal Distribution table
> we found that:
> n1 (z1) = 0.4750
> n2 (z2) = 0.4699
> The probability is equal to the difference
> of these values (n1 - n2)
> Prob = 0.0051
> Don, could you check your values again? It
> seems like there is a big difference.
> Sincerely,
> Cacarulo
For the present problem I've got these:
N = 104 cards
n = 15 cards
As drawn = 3
Average As drawn = 1.1538
For finding the variance of the expected number
of As by drawing 15 cards from a pack of 104
variance = 15*(1/13)*(12/13)*(89/103)= .9203
Now, to find the std we have to standardized the
differences in aces dividing by the sqr of the variance:
std = (3 aces - 1.1538 aces)/sqr variance
Solving we get: std = 1.9245
And now looking at the probability under the
Normal curve we get:
0.0271
That's 1 chance in 36.9, a figure very close
to Don's calculations.
I wouldn't bet money for the accuracy of my calculations,on the other hand. Just my two cents.
Regards
Z
> For the present problem I've got these:
> N = 104 cards
> n = 15 cards
> As drawn = 3
> Average As drawn = 1.1538
> For finding the variance of the expected
> number
> of As by drawing 15 cards from a pack of 104
> variance = 15*(1/13)*(12/13)*(89/103)= .9203
> Now, to find the std we have to standardized
> the
> differences in aces dividing by the sqr of
> the variance:
> std = (3 aces - 1.1538 aces)/sqr variance
> Solving we get: std = 1.9245
> And now looking at the probability under the
> Normal curve we get:
> 0.0271
> That's 1 chance in 36.9, a figure very close
> to Don's calculations.
> I wouldn't bet money for the accuracy of my
> calculations,on the other hand. Just my two
> cents.
I don't understand what you are doing here. I'm not saying that you're wrong but if I did the calculations correctly then Don's number should be 0.0733 which is still quite different from my 0.0051. Could you please verify Don's calculations? Don't know what I'm doing wrong.
Sincerely,
Cacarulo
> Could you please
> verify Don's calculations? Don't know what
> I'm doing wrong.
My math is simple and certainly correct. So, use a calculator to verify that the value I obtained was correct.
Don
> My math is simple and certainly correct. So,
> use a calculator to verify that the value I
> obtained was correct.
That's what I did with my abacus
(8/104)*(7/103)*(6/102)*(96/101)*(95/100)*(94/99)*
(93/98)*(92/97)*(91/96)*(90/95)*(89/94)*(88/93)*
(87/92)*(86/91)*(85/90)*455=
=0.07332209549650789440
Sincerely,
Cacarulo
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