Re: A card game probability

Faton Berisha wrote:
On Jan 28, 12:30 am, matt271829-n...@xxxxxxxxxxx wrote:
[...]
Right, I figured out how you get the value of the expression to be
12/13 for n = 1. I didn't realise that you were assuming binom{n-1}i =
0 when i > n-1. I just looked at it and thought "that must be wrong".

I also assume sum_{i=a}^b=0 and prod_{i=a}^b=1 for a>b.

There are several other things I don't understand in your reply, but
rather than labouring over those, let's cut to the chase and look at
your final result, which I quote below:
[...]

I agree with the first result (n = 1). For n = 2 I calculate the
probability as the sum of the following cases:

First card = 1 (prob 4/52); second card = anything (prob 1)
First card = 2 (prob 4/52); second card = 2 (prob 3/51)
First card not 1 or 2 (prob 44/52); second card = 2 (prob 4/51)

Total probability = 4/52*1 + 4/52*3/51 + 44/52*4/51 = 98/663 =
0.14781...

And for n = 3, using an analogous method, I get a probability of
3533/16575 = 0.21315...

These are rather different from your results.

If you take into the consideratio the remark above,
you should get those results;
after all, as I mentioned, I got them by a computer.

However, as I had also mantioned,
the formula does'n behave properly for values of n near 52,
and I found out this is not due to roundoff error.
So it needs a modification for such values,
and this is the one that i propose:

Denote by c(n) a "correction" function

c(n) := 1+floor((n-14)/13),

for floor(x) being the greatest integer not greater than x.

Then, the probability bar p_n = Pr(bar A_n | bar A_1 ... bar A_n)
is approximated by the following formula

bar p_n = sum_{i=0}^4 binom{n-1-c(n)} i
times prod_{j=0}^{i-1}(4 - j)/(48 - j)
times prod_{j=0}^{n-i-2-c(n)} (44-j-c(n)+c(j))/(48-i-j-c(n)
+c(j))
times (52-n+i-3)/(52-n+1).

Now, for the probability of getting at least one match
in n trial

p_n = 1 - prod_{i=1}^n bar p_i

the following table of results is obtained

( 0.0769, 0.1493, 0.2160, 0.2773, 0.3337, 0.3856,
0.4333, 0.4772, 0.5176, 0.5548, 0.5890, 0.6204,
0.6494, 0.6837, 0.7151, 0.7437, 0.7697, 0.7929,
0.8136, 0.8321, 0.8486, 0.8633, 0.8764, 0.8881,
0.8986, 0.9081, 0.9204, 0.9311, 0.9404, 0.9485,
0.9555, 0.9615, 0.9666, 0.9709, 0.9745, 0.9776,
0.9802, 0.9825, 0.9844, 0.9870, 0.9892, 0.9909,
0.9924, 0.9936, 0.9946, 0.9954, 0.9960, 0.9965,
0.9968, 0.9971, 0.9972, 0.9972 )

The results don't quite agree with the ones obtained by David
in his experiment using a computer simulation.
I guess, on of our two approximations is less approximate :)

I've been thinking about stopping the simulation. Further
pseudo-random trials will lower the pseudo-uncertainty
only very slowly (days of tests...) .

A recent line of output:

9000 million tests; approx. prob. is: 0.9837662454
[ of 1 or more matches ].

David Bernier


Regards,
Faton Berisha

.

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