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Given a 52 card deck of playing cards, I am trying to calculate the probability of the following in a 4 person game.

I am the dealer, the person to my left (player 1) is to be dealt the first card and this goes around the circle.

We are dealing 6 cards in total to each player, one by one around the group.

What is the probability of the person to my left getting 9(suit), 10(suit), J(suit), Q(suit), K(suit), A(suit), with me then declaring that the suit "of choice" (a rule in the game) is the suit that they were dealt (let's assume 1/4)

I started off with:

$${\binom{6}{52}\binom{5}{48}\binom{4}{44}\binom{3}{40}\binom{2}{36}\binom{1}{32}\binom{1}{4}}$$

Is that all I have to do? Or am I missing some other calculation? Do I need to calculate that the other players aren't dealt one of those cards also as they come before player 1 is dealt their next card?

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  • $\begingroup$ It looks as if the coefficients are upside down here. Check please. $\endgroup$ – drhab Aug 28 at 12:55
  • $\begingroup$ Do you mean that the cards 1st player gets have the exact order 9,10,J,Q,K,A, (the last card of the suit you say)? Do the cards player1 gets have the same suit? $\endgroup$ – Lada Dudnikova Aug 28 at 12:56
  • $\begingroup$ As an aside., if you are talking about calculating a probability then your answer should be between $0$ and $1$. Your answer seems to be counting the number of outcomes, not a probability yet. $\endgroup$ – JMoravitz Aug 28 at 12:59
  • $\begingroup$ Next, it is far more convenient to work in a sample space where the only information recorded is about the cards held by the player on your left. Equivalently, you could instead describe the scenario as though rather than taking turns dealing cards to each person one at a time, we just give the first six cards to the player on your left (and then continue if we like giving cards to the remaining players). The specific pattern taken when delivering the cards does not in any way affect the probabilities involved. This is just like how an ace is just as likely to be a second card as first $\endgroup$ – JMoravitz Aug 28 at 13:02
  • $\begingroup$ Similarly, it is more convenient for calculations to have declared the "suit of choice" before having started dealing cards. You see then the final probability is very simply $\dfrac{1}{\binom{52}{6}}$ $\endgroup$ – JMoravitz Aug 28 at 13:04
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There are four ways to deal the six highest cards of any suit to the player to your left. The probability of the player getting dealt these cards thus equals:

$$\frac{4}{52 \choose 6} = \frac{4}{20358520} \approx 1.965 \cdot 10^{-7}$$

Assuming the choice for the suit is $\frac{1}{4}$ (independent of the cards the dealer gets), we arrive at:

$$\frac{1.965 \cdot 10^{-7}}{4} \approx 4.912 \cdot 10^{-8}$$

Alternatively, look at it this way. If the dealer always chooses a suit with probability $\frac{1}{4}$, we can choose the suit first. Then, there is only one combination of cards the player to your left should get, or again:

$$\frac{1}{52 \choose 6} \approx 4.912 \cdot 10^{-8}$$

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  • $\begingroup$ Thanks! This makes sense, apologies for the confusing terminology / incorrect coefficients! $\endgroup$ – Jamie Pollard Aug 28 at 13:20

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