Interpreting probability over 100% and math check

Question

I'm in no way good at probabilities, so I might have done something terribly wrong. Please bear with me!

We are looking at the probability of drawing exactly 3 cards of any suit when you draw 13 cards simultaneously from a full deck of 52 with no replacement. I have calculated as follows: $$ 4\cdot \frac{ {13 \choose 3} {39 \choose 10 } }{ {52 \choose 13} } )= 114~\% $$

1) Does my math seem ok?

2) If my math checks out, how do I interpret a probability of more than 100%? That shouldn't even be possible as far as my non-mathematical mind is concerned.

I really hope one of you guys have a minute to take a look at it. It would be truly appreciated.

Have a nice day.

Answer 1

Your math is not okay and calculated probabilities that exceed $1$ are enough already to draw this conclusion.

Let $C$ denote the event that you will draw exactly $3$ cards of the suite of clubs and similarly define $S,D,H$.

Then your "calculation" looks like: $$P(C\cup S\cup D\cup H)=P(C)+P(S)+P(D)+P(H)=4P(C)=4\frac{\binom{13}3\binom{39}{10}}{\binom{52}{13}}$$

But such a calculation is only legal if $C,S,D,H$ are events that are mutually exclusive in the sense that at most one of these events can occur.

That is not the case here. It is e.g. quite well possible to draw exactly $3$ clubs and exactly $3$ spades.

If you would have been talking about $7$ instead of $3$ then indeed the events would exclude eachother, so a sortlike computation would have been okay then.

Let me end with a correct calculation.

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Then to be found is $P(C\cup S\cup D\cup H)$ which can be found by applying the principle of inclusion/exclusion.

Also symmetry can be applied and we arrive at:$$P(C\cup S\cup D\cup H)=4P(C)-6P(C\cap S)+4P(C\cap S\cap D)=$$$$\binom{52}{13}^{-1}\left[4\binom{13}3\binom{39}{10}-6\binom{13}3^2\binom{26}{7}+4\binom{13}3^3\binom{13}{4}\right]$$ Observe that $P(C\cap S\cap D\cap H)$ is left out here. This for the simple reason that this probability is $0$. By a draw of $13$ cards it is not possible to draw exactly $3$ cards of each suite.

Answer 2

Probabilities are never outside of the interval $[0,1]$.

You have misapplied the hypergeometric distribution. I have interpreted your question as asking for the probability that, if $13$ cards are drawn without replacement from a standard deck of $52$ cards, that among these cards, there exists at least one suit for which there are exactly $3$ cards of that suit. If we denote the random number of cards drawn of each suit by $C, D, H, S$ for clubs, diamonds, hearts, and spades, respectively, then $$C + D + H + S = 13, \\ 0 \le C, D, H, S \le 13,$$ and we want $$\Pr[(C = 3) \cup (D = 3) \cup (H = 3) \cup (S = 3)].$$ This is not a hypergeometric model, although the sampling occurs without replacement.

We can perform the calculation without inclusion/exclusion, but it involves more tedious work. Consider the complement; that is to say, what is the probability that of all $13$ cards drawn, none of the suits have exactly three of that type? There are exactly $25$ partitions of $13$ into exactly four nonnegative integers excluding $3$:
$$13 + 0 + 0 + 0, \\ 12 + 1 + 0 + 0, \\ 11 + 2 + 0 + 0, \\ 11 + 1 + 1 + 0, \\ 10 + 2 + 1 + 0, \\ 10 + 1 + 1 + 1, \\ 9 + 4 + 0 + 0, \\ 9 + 2 + 2 + 0, \\ 9 + 2 + 1 + 1, \\ 8 + 5 + 0 + 0, \\ 8 + 4 + 1 + 0, \\ 8 + 2 + 2 + 1, \\ 7 + 6 + 0 + 0, \\ 7 + 5 + 1 + 0, \\ 7 + 4 + 2 + 0, \\ 7 + 4 + 1 + 1, \\ 7 + 2 + 2 + 2, \\ 6 + 6 + 1 + 0, \\ 6 + 5 + 2 + 0, \\ 6 + 5 + 1 + 1, \\ 6 + 4 + 2 + 1, \\ 5 + 5 + 2 + 1, \\ 5 + 4 + 4 + 0, \\ 5 + 4 + 2 + 2, \\ 4 + 4 + 4 + 1.$$ For such a partition, there are $$\binom{13}{C}\binom{13}{D}\binom{13}{H}\binom{13}{S}$$ ways to select the cards, but the number of such outcomes must also be multiplied by the number of permutations of suits comprising the partition; e.g., there are $\frac{4!}{3!1!} = 4$ ways to identify $(C,D,H,S)$ with the partition $13+0+0+0$, but there are $\frac{4!}{2!1!1!} = 12$ ways for the partition $5+4+2+2$. After doing the computation, we get the desired probability $$\frac{29460383524}{39688347475}.$$

In Mathematica, the code to produce this result is:

1 - Total[4!/(Times @@ (Last[Transpose[Tally[#]]]!)) (Times @@ Binomial[13, #]) 
    & /@ (PadRight[#, 4] & /@ Select[IntegerPartitions[13, 4], ! MemberQ[#, 3] &])] 
    / Binomial[52, 13]

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Rather, it seems that counting the desired outcomes is better than what I did above in counting the complement, since there are only $14$ partitions of $13$ containing at least one instance of $3$: $$10+3+0+0 \\ 9+3+1+0 \\ 8+3+2+0 \\ 8+3+1+1 \\ 7+3+3+0 \\ 7+3+2+1 \\ 6+4+3+0 \\ 6+3+3+1 \\ 6+3+2+2 \\ 5+5+3+0 \\ 5+4+3+1 \\ 5+3+2+2 \\ 4+4+3+2 \\ 4+3+3+3$$ and the rest of the computation is analogous as above. The modified code is nearly identical:

Total[4!/(Times @@ (Last[Transpose[Tally[#]]]!)) (Times @@ Binomial[13, #]) & /@
(PadRight[#, 4] & /@ Select[IntegerPartitions[13, 4], MemberQ[#, 3] &])]/
Binomial[52, 13]

Answer 3

Recall that $\,P(A \cup B)=P(A)+P(B)\,$ only if $\,A,B\,$ are mutually exclusive. In general, $$\,P(A \cup B)=P(A)+P(B)-P(A\cap B).\,$$

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Even though you are correct that

$$P(3 \,{\small ♠})=P(3 \,\color{red}{♥})=P(3 \,\color{red}{♦})=P(3 \,{\small ♣})= \frac{\binom{13}{3}\binom{39}{10}}{\binom{52}{13}}$$

these events are not mutually exclusive, so you cannot just multiply that value by 4.

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Unfortunately, to properly calculate this probability, you'd have to use inclusion/exclusion and that would take some effort. For the four events, the formula would look like this:

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$$ \begin{align} & P(3 \,{\small ♠} \, \cup \, 3 \,\color{red}{♥} \cup 3 \,\color{red}{♦} \cup 3 \,{\small ♣})\\[1ex] &= P(3 \,{\small ♠})+P(3 \,\color{red}{♥}) +P(3 \,\color{red}{♦})+P(3 \,{\small ♣}) \\[1ex] & - P(3 \,{\small ♠}\cap 3 \,\color{red}{♥}) - P(3 \,{\small ♠}\cap 3 \,\color{red}{♦}) - P(3 \,{\small ♠}\cap 3 \,{\small ♣}) - P(3 \,\color{red}{♥}\cap 3 \,\color{red}{♦}) - P(3 \,\color{red}{♥}\cap 3 \,{\small ♣}) - P(3 \,\color{red}{♦}\cap 3 \,{\small ♣}) \\[1ex] & + P(3 \,{\small ♠}\cap 3 \,\color{red}{♥}\cap 3 \,\color{red}{♦}) + P(3 \,{\small ♠}\cap 3 \,\color{red}{♥}\cap 3 \,{\small ♣}) + P(3 \,{\small ♠}\cap 3 \,\color{red}{♦}\cap 3 \,{\small ♣}) + P(3 \,{\small ♠}\cap 3 \,\color{red}{♥}\cap 3 \,{\small ♣})\\[1ex] & - P(3 \,{\small ♠}\cap 3 \,\color{red}{♥}\cap 3 \,\color{red}{♦}\cap 3 \,{\small ♣}) \end{align} $$

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However, note that just like all four events in the first line have equal probabilities, so do the six events in the second line and the four events in the third line; so we really only have to calculate three additional probabilities.

$$ \color{white}{text}\\ \begin{align} P(3 \,{\small ♠}\cap 3 \,\color{red}{♥}) & = P(3 \,{\small ♠}\cap 3 \,\color{red}{♦}) = P(3 \,{\small ♠}\cap 3 \,{\small ♣}) = P(3 \,\color{red}{♥}\cap 3 \,\color{red}{♦}) = P(3 \,\color{red}{♥}\cap 3 \,{\small ♣}) = P(3 \,\color{red}{♦}\cap 3 \,{\small ♣})\\[2ex] &= \frac{\binom{13}{3}\binom{13}{3}\binom{26}{7}}{\binom{52}{13}} \end{align} $$

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$$ \begin{align} P(3 \,{\small ♠}\cap 3 \,\color{red}{♥}\cap 3 \,\color{red}{♦}) &= P(3 \,{\small ♠}\cap 3 \,\color{red}{♥}\cap 3 \,{\small ♣}) = P(3 \,{\small ♠}\cap 3 \,\color{red}{♦}\cap 3 \,{\small ♣}) = P(3 \,{\small ♠}\cap 3 \,\color{red}{♥}\cap 3 \,{\small ♣})\\[2ex] &= \frac{\binom{13}{3}\binom{13}{3}\binom{13}{3}\binom{13}{4}}{\binom{52}{13}} \end{align} $$

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$$P(3 \,{\small ♠}\cap 3 \,\color{red}{♥}\cap 3 \,\color{red}{♦}\cap 3 \,{\small ♣})=0$$

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That last probability is zero as the hand would only have twelve cards. So the final answer would be

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$$4 \cdot \frac{\binom{13}{3}\binom{39}{10}}{\binom{52}{13}} \; - \; 6 \cdot \frac{\binom{13}{3}\binom{13}{3}\binom{26}{7}}{\binom{52}{13}} \; + \; 4 \cdot \frac{\binom{13}{3}\binom{13}{3}\binom{13}{3}\binom{13}{4}}{\binom{52}{13}} - 0$$

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I'll leave the final calculation to you!