Probability of get 3 cards out of 4

Question

How to calculate a probability of choose 3 off-suited and off-ranked cards out of 4 given cards from a standard 52-cards deck (the situation when all the cards off-suited/off-ranked should be avoid).

E.g. you're dealt 3 cards at a time.

P.S. By simulating the situation on a computer (1mln tries) the answer must me like 54.5%)

P.P.S. The same question but to get 2 cards (off-suited and off-ranked)

P.P.P.S. The same question byt 1 card (off-suited and off-ranked) or what is the same 3 cards either same suit or same rank

More about how I'm trying to do that (probably wrong): We can have 3 cards out of 4 with 4 different capabilities. Each time we draw a card, we reduce the chances to draw needed card: 52 - starting draw, 36 - rest suitable for 2nd card, 22 - rest suitable for 3rd card and $\bf(52 - 10 - 3)$ - rest suitable for last card. And each time we draw we have less cards in the deck, like 52,51,50,49.

$\frac{(52-52)\times36\times22\times10}{52\times51\times50\times49} + \frac{52\times(51-36)\times22\times10}{52\times51\times50\times49} + \frac{52\times36\times(50-22)\times10}{52\times51\times50\times49} + \frac{52\times36\times22\times(49-10)}{52\times51\times50\times49} = 35.428571\%$

but that is not $54.5\%$ I got from simulation. And I want to understand where I was wrong in calculation or in simulation program.

Answer 1

Case 1: Four cards of four different suits (one way to get suits).

Case 1A is edited out because it turns out not to be a desirable case ($4$ different suits and $4$ different ranks). 1A: Four different ranks: Pick a rank for spades ($13$ ways); hearts ($12$ ways); diamonds ($11$ ways); clubs (10 ways).

Total for case 1A: 17160 possibilities.

1B: Three different ranks: Pick three ranks ($13\choose 3$ ways); pick a rank to be used twice ($3$ ways); pick two suits to receive that same rank ($4\choose 2$ ways); assign remaining two ranks to remaining two suits ($2$ ways).

Total for case 1B: $10296$ possibilities.

Case 2: Three suits among the four cards. ($12$ ways to choose suits--$4$ choices for twice-appearing suit; $3$ choices for unused suit).

2A: Four different ranks occur: Pick two ranks for the twice-used suit ($13\choose 2$ ways); choose a rank for one of the other used suits ($11$ ways). Choose a rank for other used suit ($10$ ways).

Total for case 2A: $12\cdot 8580=102960$ possibilities (the factor of $12$ is from the $12$ choices for suit at the beginning of case 2).

2B: Three different ranks occur (but still an outcome of the desired type). This means one of the twice-used suit cards will have to match rank with one of the singly-used suit cards, and all other ranks different. Choose a rank for one of the singly-used suits ($13$ ways); choose a rank for the other singly-used suit ($12$ ways); for the twice-used suit cards, choose one already used rank and one new rand ($2\cdot 11$ ways).

Total for case 2B: $12\cdot 3432=41184$ possibilities (again the extra factor of $12$ is for suit selection at the beginning of case 2).

Grand total: (Add cases): There are $154440$ desirable card combinations.

The probability of getting a desirable combination is $\frac{171600}{52\choose 4}=\frac{154440}{270725}\approx .5705$.

This probability is different than the one suggested by OP based on computer simulation. But I think this value is correct--I also ran a program to check, looking at all $270725$ different $4$-card hands.

Answer 2

I try to tackle this question using the most elementary but tedious counting method, by first listing all possible mutually exclusive cases:

Case 1: The 2nd card is of a different rank and a different suit from the first card, while the 3rd card is of a third rank and a third suit. In this case, the 4th card is irrelevant. Since we are finding exactly 3 cards with different ranks and suits, The 4th card must NOT be (the 4th suit and a different rank).

e.g. $ {\spadesuit 8},{\heartsuit 7},{\clubsuit 6},{\clubsuit 5} $

So, the 4th card must not be ${\diamondsuit A}, {\diamondsuit 2}, {\diamondsuit 3}, {\diamondsuit 4}, {\diamondsuit 5}, {\diamondsuit 9}, {\diamondsuit 10}, {\diamondsuit J}, {\diamondsuit Q}$ or $ {\diamondsuit K}$, but any other will do.

Case 2: The 2nd card is of a different rank and a different suit from the first card, but the 3rd card is not a third rank or a third suit. This is further divided into the sub-cases below (note: the table is an image):

Case 3: The 2nd card is already the same rank or suit as the first card. This is further divided into the sub-cases below (note: the table is also an image):

So these are all the cases. What's remaining is to add up all the probabilities of all these cases:

Case 1: $\frac{36}{51} \times \frac{22}{50} \times \frac{39}{49}$

Case 2a1: $\frac{36}{51} \times \frac{2}{50} \times \frac{11}{49}$

Case 2a2: $\frac{36}{51} \times \frac{2}{50} \times \frac{11}{49}$

Case 2a3: $\frac{36}{51} \times \frac{2}{50} \times \frac{11}{49}$

Case 2b1: $\frac{36}{51} \times \frac{11}{50} \times \frac{2}{49}$

Case 2b2: $\frac{36}{51} \times \frac{11}{50} \times \frac{2}{49}$

Case 2b3: $\frac{36}{51} \times \frac{11}{50} \times \frac{20}{49}$

Case 2c1: $\frac{36}{51} \times \frac{2}{50} \times \frac{11}{49}$

Case 2c2: $\frac{36}{51} \times \frac{2}{50} \times \frac{11}{49}$

Case 2c3: $\frac{36}{51} \times \frac{2}{50} \times \frac{11}{49}$

Case 2d1: $\frac{36}{51} \times \frac{11}{50} \times \frac{2}{49}$

Case 2d2: $\frac{36}{51} \times \frac{11}{50} \times \frac{2}{49}$

Case 2d3: $\frac{36}{51} \times \frac{11}{50} \times \frac{20}{49}$

Case 2e: $\frac{36}{51} \times \frac{1}{50} \times \frac{22}{49}$

Case 2f: $\frac{36}{51} \times \frac{1}{50} \times \frac{22}{49}$

Case 3a: $\frac{3}{51} \times \frac{12}{50} \times \frac{22}{49}$

Case 3b: $\frac{3}{51} \times \frac{12}{50} \times \frac{22}{49}$

Case 3c1: $\frac{3}{51} \times \frac{24}{50} \times \frac{11}{49}$

Case 3c2: $\frac{3}{51} \times \frac{24}{50} \times \frac{11}{49}$

Case 3c3: $\frac{3}{51} \times \frac{24}{50} \times \frac{11}{49}$

Case 3d: $\frac{12}{51} \times \frac{3}{50} \times \frac{22}{49}$

Case 3e: $\frac{12}{51} \times \frac{3}{50} \times \frac{22}{49}$

Case 3f1: $\frac{12}{51} \times \frac{33}{50} \times \frac{2}{49}$

Case 3f2: $\frac{12}{51} \times \frac{33}{50} \times \frac{2}{49}$

Case 3f3: $\frac{12}{51} \times \frac{33}{50} \times \frac{20}{49}$

I suppose these are all the possible cases being dealt with. Surely this method is clumsy and there are likely ways to generalize some of the above sub-cases.

Edit: The above sum is 0.63385 , which is quite different from the result obtained by the original poster 0.57047.

Answer 3

For a 3-card hand here is an explanation that was stolen from another site (courtesy BruceZ of twoplustwo.com):

$\frac{52}{52}\times\frac{36}{51}\times\frac{22}{50}\times(\frac{6}{49}\times4+\frac{33}{49}\times2)≈ 57.05\%$

Of the remaining 39 cards that aren't part of the Badugi, there are 6 that match the suit of one of the other cards and a rank of a different card, and the other 33 match just the rank or a suit but not both. For the first 6, we multiply by 4 because there are 4 positions that non-Badugi card can come in. For the 33, we multiply by 2 because there are 6 possible positions that the 2 matching cards can come, and only 3 of those have one of them as the last card (either one of them can be part of the Badugi).

We can do it a little more cleanly with combinations as $\binom{4}{3}\times13\times12\times11\times(6+\frac{33}{2}) = 154,440$

3-card hands and $\binom{52}{4} = 270,725$ possible hands for $\frac{154,440}{270,725} ≈ 57.05\%.$ That's $\binom{4}{3} = 4$ ways to choose the 3 suits of the Badugi, times $13\times12\times11$ ways to choose the ranks for the 3 suits, times 6 ways to choose the non-Badugi card that matches a rank and a suit, and 33 ways to choose the non-Badugi card that matches a rank or a suit but not both, but then divide that by 2 since otherwise we would be counting each Badugi twice since there would be 2 choices for the non-Badugi card.

You could work out the possibility of a 2-card hand in a similar manner to the 3-card situation, but since we know a given hand must be either 1, 2, 3, or 4 cards we can just subtract all the other probabilities from 1 to find the 2-card hand:

$1 - (6.34 + 57.05 + 1.06) ≈ 35.55$

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Here is another answer from David Sklanski:

There are three categories:

1. Four suited with a pair: $A\spadesuit A\clubsuit 2\diamondsuit 3\heartsuit$

2. Three suited with no pair: $A\spadesuit 2\diamondsuit 3\heartsuit 4\spadesuit$

3. Three suited with a pair: $A\spadesuit A\clubsuit 2\spadesuit 3\diamondsuit$

   1. The chances that the second card pairs the first is $\frac{3}{51}$. If so, the chances that the third card is a different rank and suit is $\frac{24}{51}$ and if so the chances that the last card is different is $\frac{11}{49}$. Multiply these fractions and then multiply by 6 because the pair can be in six different places and you get $3.8\%$

   2. The chances that the second card matches the suit of the first is $\frac{12}{51}$. If it does the chances that the third is a different suit and rank of the first two is $\frac{33}{50}$. If so the chances the last card is different is $\frac{20}{49}$. Multiply these fractions and then multiply by 6 because the "suit pair" can be in six different spots and you get $38.03\%$

   3. The chances that the second card pairs the first is $\frac{3}{51}$. If it does the chances that the third card is of one of those two suits is $\frac{24}{51}$. If so the chances that the last card is a different rank and a different suit is $\frac{22}{49}$. Multiply those fractions and then multiply by 12 because the pair can be in six spots and the card that matches the suit of one of those pair cards can be in in either one of the remaining two spots. That's $15.21\%$

$38.03\%+3.80\%+15.21\%=57.04\%$