Probability problem with 52-card deck.

[Guv]

There are 52 cards in a regular deck, 13 per suit. If you deal 13 cards to a person, what is the probability that he will be void in some suit? Id est: That he will have no cards in some suit?

[chrisf]

After 3 cards, the prop of having a void suit is 1.
On the 4th card, the prob. of having a void suit changes. If we pick any one suit not represented in the first three cards, the prob. of it not being represented in the first :
4 cards is 36/49
5 cards is (36/49) * (36/48)
13 cards is (36/49) * (36/48) * .... * (36/40)
= (36^10) / (49*48*47*46*45*44*43*42*41*40)
= (36^10)/(49! / 39!)
= 0.1226

[Guv]

Kedaman and Chrisf: Do not feel too bad. I got it wrong the first time and a few bridge playing mathematicians agreed with me.

This Thread was started because of my initial error, and the errors made by several others.

I initially calculated that C(39, 13) is the number of ways 13 cards can be selected from a 39 card deck (Id est: From a deck missing one suit). Multiply by four due to there being four suits. Then divide by C(52,13) to convert to a probability.

The above looks correct, and was agreed to by several of my friends, but is not correct. Later, when I discovered the correct answer, I posed the problem to a few more people, all of whom came up with incorrect analyses.

[kedaman]

I am almost surprised, well I guess this could be fun

[Starman]

Are you dealing four hands Guv, or just taking the top 13 cards from a shuffled deck?

[kedaman]

Hmm..Tell me if I'm wrong but, I don't think the order matters which you pick up the cards, so others picking card in between wouldn't differ from the case them picking cards afterwards.

[simonm]

I doubt if I can solve it but here's my idea:

If I start by calculating the odds of getting 13 cards and none of which is a, say, heart, here are the odds (I think):

39/52 * 38/51 * 37/50 ... 28/41 * 27/40 = 0.0128

Now since the odds we want are for any suit (i.e. hearts OR clubs OR diamonds OR spades), The probability is 0.0128 * 4 = 0.0511


Now, having said that, I'm almost certain my calculation is flawed because intuition tells me that it's much more likely to be devoid of one particular suit with a hand of 13 cards. Can't think why though...

[Starman]

I just thought that if you were the first of the four hands to be dealt then as your last card is dealt there is/could-be a 1/4 chance of an individual card, whereas if you were the fourth hand dealt the last card is rather predetermined. Maybe it doesn't change the total probabilities though.

[Starman]

Trying it from another direction:
If I have 13 cards, there are 39 possible distributions of suits.
13, 0, 0, 0
12, 1, 0, 0
11, 2, 0, 0
11, 1, 1, 0
10, 3, 0, 0
10, 2, 1, 0
10, 1, 1, 1
9, 4, 0, 0
9, 3, 1, 0
9, 2, 2, 0
9, 2, 1, 1
8, 5, 0, 0
8, 4, 1, 0
8, 3, 2, 0
8, 3, 1, 1
8, 2, 2, 1
7, 6, 0, 0
7, 5, 1, 0
7, 4, 2, 0
7, 4, 1, 1
7, 3, 3, 0
7, 3, 2, 1
7, 2, 2, 2
6, 6, 1, 0
6, 5, 2, 0
6, 5, 1, 1
6, 4, 3, 0
6, 4, 2, 1
6, 3, 3, 1
6, 3, 2, 2
5, 5, 3, 0
5, 5, 2, 1
5, 4, 4, 0
5, 4, 3, 1
5, 4, 2, 2
5, 3, 3, 2
4, 4, 4, 1
4, 4, 3, 2
4, 3, 3, 3

21 of these possibles are void in at least one suit.
So it may be 21/39 of holding a void.

but the probabilities of each distribution are not even and I suspect this needs to be taken into account. I can work out the prob of a 13, 0, 0, 0 hand, but how many ways can you have a 4,3,3,3 etc ?

[simonm]

Starman,

You must be way off, surely? How can there be a greater than 50/50 chance of being void in one particular suit?

But, following your logic and counting the number of combinations of each distribution:

13, 0, 0, 0 x4 * 4
12, 1, 0, 0 x12 * 12
11, 2, 0, 0 x12 * 12
11, 1, 1, 0 x12 * 12
10, 3, 0, 0 x12 * 12
10, 2, 1, 0 x24 * 24
10, 1, 1, 1 x4
9, 4, 0, 0 x12 * 12
9, 3, 1, 0 x24 * 24
9, 2, 2, 0 x12 * 12
9, 2, 1, 1 x12
8, 5, 0, 0 x12 * 12
8, 4, 1, 0 x24 * 24
8, 3, 2, 0 x24 * 24
8, 3, 1, 1 x12
8, 2, 2, 1 x12
7, 6, 0, 0 x12 * 12
7, 5, 1, 0 x24 * 24
7, 4, 2, 0 x24 * 24
7, 4, 1, 1 x12
7, 3, 3, 0 x12 * 12
7, 3, 2, 1 x24
7, 2, 2, 2 x4
6, 6, 1, 0 x12 * 12
6, 5, 2, 0 x24 * 24
6, 5, 1, 1 x12
6, 4, 3, 0 x24 * 24
6, 4, 2, 1 x24
6, 3, 3, 1 x12
6, 3, 2, 2 x12
5, 5, 3, 0 x12 * 12
5, 5, 2, 1 x12
5, 4, 4, 0 x12 * 12
5, 4, 3, 1 x24
5, 4, 2, 2 x12
5, 3, 3, 2 x12
4, 4, 4, 1 x4
4, 4, 3, 2 x12
4, 3, 3, 3 x4

---- ---

560 340

Prob. = 340 / 560 = 0.607


And that is just ridiculous!?!

[Starman]

simonm, I'm not sure I agree with the number of combinations you have offered

for 13,0,0,0 - four ways to do this- one for each suit

for 12,1,0,0 -

pick one suit, there are 13 ways to take 12 cards

multiplied by 39 for the remaining cards, multiplied by 4 to try all the suits

making 2028 possible 12,1,0,0 hands

It gets a lot worse as you go down the list

but I think that these totals for the voids against the total possible hands will be the answer.

[kedaman]

39/52 * 38/51 * 37/50 ... 28/41 * 27/40 = 0.0128
...
The probability is 0.0128 * 4 = 0.0511

This sounds good, and I tested it empirically, and it seems to be correct

    int in[4],out[4],intot,temp;
    unsigned int case_=0,total=0;

    for (;; ){
        in[3]=in[2]=in[1]=in[0]=13;
        out[3]=out[2]=out[1]=out[0]=0;
        for(intot=52;intot>39; ){
            temp=rand()%intot--;
            for (int* i=in;temp>=0;++i)temp-=*i;
            --(*--i);
            ++out[i-in];
            if (!(out[3]==0||out[2]==0||out[1]==0||out[0]==0))break;
        }
        if (intot==39&&(out[3]==0||out[2]==0||out[1]==0||out[0]==0))case_++;
        total++;
        if (!(total&1048575))cout << (double)case_/(double)total << " ( " << case_ << " : " << total << ")" << endl;
    } 

[simonm]

Starman, I see what you are saying and it does get very complex.

So, I worked out the probabilities with a much simpler example (deck of 9 cards, 3 suits and a hand of three cards).

I first assigned probabilities your way:

3,0,0 1
0,3,0 1
0,0,3 1
2,1,0 9
2,0,1 9
1,2,0 9
1,0,2 9
0,2,1 9
0,1,2 9
1,1,1 27

Therefore Probability of not getting a suit = 67/94 = 0.7128

Then I used my method:

(6/9 * 5/8 * 4/7) * 3 = 0.7143

Very close but not quite the same. Curious...

[simonm]

Kedaman

I know you prefer C++ but this is (generally) a VB forum.

Unfortunately, not being familiar with C++ I don't understand your example program.

[kedaman]

Simon
well i didn't code that for just anyone to understand, I code in C++ because it's easier for me and i posted it for anyone to find bugs in it.

not going into too much detail, this is what it does in an infinite loop:

1. reset amount per suit picked (out) and left(in)
2. pick 13 random cards by index of 0 to amount of cards left
- if no suit is void, exit the loop
3. total is incremented and case_ is incremented in case there is some void suit.
4. each 1048576'th iteration, show current statistics

[Starman]

simonm,

I make that 57 hands with a void out of 84 possible hands

=0.67857

it's not as close to your answer now - either right or wrong :7).

I've been looking along the lines of:
(using your 9 card pack)

1st - any card will do = 9/9
2nd - any card will do = 8/8
3rd - must not be the third suit = 4/7
=.5714

but this doesn't agree with your answer either although it's nearly an anagram.

back to excel.......

[simonm]

OK, I can't add up!

I stand by the result of my method though. Besides, it's been verified by Kedaman's cryptic function.

1st - any card will do = 9/9
2nd - any card will do = 8/8
3rd - must not be the third suit = 4/7
=.5714

I think this is flawed becasue you are not taking account of the possibility that the first two cards you drew might have been the same suit. In that case, the last card you draw could be anything (it doesn't matter).

[Starman]

simonm

yes, that's where it all goes horribly confusing

1st - any will do = 9/9
2nd
a) 2/8 of matching the 1st
b) 6/8 of not

3a - 6/7 'safe' cards
3b - 4/7 'safe' cards

a) 9/9 * 2/8 * 6/7 = 12/56 = .21428

b) 9/9 * 6/8 * 4/7 = 24/56 = .42857

but I think this just gives the chances of having a pair instead of a triple and has nothing to do with the actual question

combin(52,13)=635013559600 possible bridge hands.

some are void and I'm still trying to guess how many

[kedaman]

I still wonder what's wrong with my analogy, is anyone finds the flaw i'd be glad to know

[Guv]

If you shuffle the deck thoroughly, it does not matter whether you deal 13 cards off the top or deal four hands and examine one of them. The odds are the same for 13 cards chosen by any random process. Of course, if you have information about one hand at the table, it will affect the odds relating to any of the other three.

Kedaman and Simon are getting close. The formula for the combination of 39 cards taken 13 at a time is.

C(39,13) = 39!/13!*26!

The above is the number of hands which can be made up using only 39 cards from 3 suits. If you divide by C(52,13) you get the probability of a hand containing a void in the fourth suit.

The above is the approach I took. Then I multiplied by four to get the probability of a void in any suit. On my calculator, I got the following.

C(39,13) / C(52,13) = .012 790 948, which is the probability of a 13-card hand with no Spades.

Multiply by 4 = .051 163 792, which is not the probability of a hand with a void !!! ??

Permutations, combinations, and probability have subtle pitfalls for the unwary.

I, and several bridge playing mathematicians, accepted the above as correct. Then I wrote a program for my hand calculator to calculate various bridge table probabilities. There are 21 patterns with void suits. When I added the 21 probabilities, I did not get the above, and thought I had a bug in the program.

Then I realized the subtle error in the above analysis. Does anybody else see it?

[chrisf]

Guv,
Some of the combinations which have a void in the fourth suit also have a void in one or two other suits, so are 'counted' more than once when you multiply by 4.

[Guv]

ChrisF: You got it.

All hands with distributions like 0 0 6 7 are counted twice: Once with the Spade void hands and once with the Heart void hands.

The four hands with three voids (0 0 0 13) are counted three times.

Hence the correct calculation is as follows.

[ 4 * C(39, 13) - 6 * C(26, 13) + 4 * C(13, 13 ) ] / C(52, 13)

this results in .051 065 521, which is close to the incorect value I originally computed.

You must add 4 * C(13, 13 ), which equals 4, because subtracting hands with two voids overcorrects slightly by subtracting out all the hands with three voids.

[Starman]

My way Works!!!! - it's just takes a while

Now I know the answer it's easier.

I was calculating the different numbers of possible hands but couldn't get the totals to add up to the total possible hands.
The problem appears when duplicate amounts of two suits are present - I've fixed that now.

continuing from 11,2,0,0

11,1,1,0 - 78 (ways of 11) X 13 (ways of 1) X 13 (ways of 1) X 3 (positions for the 0) X 4(suits) = 158184

10,3,0,0 - 286 ways for 10 X 286 ways for 3 X 3 X 4=981552
etc. etc. etc.
To save a very overlong post, the totals for the voids are:
4
2028
73008
158184
981552
6960096
6134700
63800880
52200720
19876428
287103960
689049504
35335872
689049504
2296831680
1684343232
459366336
4134297024
8421716160
5684658408
7895358900


Add this lot together to get: 32427298180

Divide by the total possible hands : 635013559600

Result: 0.051065521

(full calculation available on request)

I think I prefer the look of
4 * C(39, 13) - 6 * C(26, 13) + 4 * C(13, 13 ) ] / C(52, 13)
It's obviously quicker.

Nice question Guv.

[kedaman]

Nothing wrong with my algoritm either, It approached the correct value but as Simon's value was close I thought his was the correct analytical solution.

This listing shows that it doesn't converge towards .05116 but rather towards .05106. I expect if I wait enough the 5'th digit and later the 6'th will stabilize and so on..
0.0507059 ( 53169 : 1048576)
0.0508876 ( 106719 : 2097152)
0.0509485 ( 160270 : 3145728)
0.0508549 ( 213301 : 4194304)
0.0508642 ( 266675 : 5242880)
0.0509036 ( 320258 : 6291456)
0.0508945 ( 373567 : 7340032)
0.0508931 ( 426922 : 8388608)
0.0509138 ( 480483 : 9437184)
0.0508969 ( 533693 : 10485760)
0.0509373 ( 587528 : 11534336)
0.0509635 ( 641269 : 12582912)
0.0509798 ( 694931 : 13631488)
0.0509904 ( 748542 : 14680064)
0.051 ( 802161 : 15728640)
0.0510134 ( 855863 : 16777216)
0.0510173 ( 909423 : 17825792)
0.0510189 ( 962950 : 18874368)
0.051018 ( 1016428 : 19922944)
0.0510449 ( 1070490 : 20971520)
0.0510505 ( 1124137 : 22020096)
0.0510491 ( 1177634 : 23068672)
0.051055 ( 1231306 : 24117248)
0.0510482 ( 1284671 : 25165824)
0.0510595 ( 1338495 : 26214400)
0.0510725 ( 1392389 : 27262976)
0.0510765 ( 1446056 : 28311552)
0.0510786 ( 1499674 : 29360128)
0.0510803 ( 1553285 : 30408704)
0.0510829 ( 1606929 : 31457280)
0.0510835 ( 1660513 : 32505856)
0.0510744 ( 1713773 : 33554432)
0.0510753 ( 1767358 : 34603008)
0.051078 ( 1821011 : 35651584)
0.0510827 ( 1874742 : 36700160)
0.051084 ( 1928358 : 37748736)
0.051076 ( 1981613 : 38797312)
0.0510798 ( 2035318 : 39845888)
0.051078 ( 2088806 : 40894464)
0.0510714 ( 2142090 : 41943040)
0.0510788 ( 2195962 : 42991616)
0.0510924 ( 2250120 : 44040192)
0.051089 ( 2303538 : 45088768)
0.05109 ( 2357155 : 46137344)
0.0510914 ( 2410796 : 47185920)
0.0510895 ( 2464278 : 48234496)
0.0510855 ( 2517648 : 49283072)
0.0510867 ( 2571279 : 50331648)
0.0510873 ( 2624876 : 51380224)
0.051085 ( 2678324 : 52428800)
0.0510903 ( 2732175 : 53477376)
0.0510858 ( 2785500 : 54525952)
0.0510861 ( 2839088 : 55574528)
0.0510857 ( 2892630 : 56623104)
0.0510822 ( 2945999 : 57671680)
0.0510818 ( 2999535 : 58720256)