0
$\begingroup$

I regularly play 7-card brag with 4 other people at my local pub.

The rules of 7-card brag:

  • Each player is dealt 7 cards.
  • Each player chooses 2 hands of 3 cards and discards the remaining card.
  • The player to the left of the dealer starts by playing their best hand.
  • The other players then play their best hands.
  • The player with the winning hand of that round then plays their second hand.
  • The other players then play their other hands.
  • To win the 'game' (both rounds), a player must win both the first and the second round.
  • If no one has won, then the pot roles over to the next game.

The hands, in order, are:

  • Three of a kind (7s are the best, then aces, then kings, then queens etc.).
  • Running Flush (three consecutive cards in the same suit, again ace, king, queen is the highest and so on).
  • Run (three consecutive cards - higher the better).
  • Flush (three cards of the same suit - the higher the highest card is the better).
  • Pair (2 of the same card).
  • High card (age is the best, then kings, then queens etc.).

My question (or the one that I was asked) is: "What is the probability of getting the `unbeatable hand'"?

As you might be able to work out, there are technically 2 unbeatable hands. The first is the most obvious: Three 7s and three aces (case A). The other is slightly less obvious: Three 7s and three kings (case B). Case B is also unbeatable because the second hand (three kings) can only be beaten by three aces (I have the three 7s), but the three aces have to be played in the first round since it must be their best hand - which would be beaten by my 7s.

My Thinking: For simplicity I am going to assume I am the person to the left of the dealer. Also, I think it is probably better to take the 2 unbeatable hands as separate cases. The probability of the two cases should be the same, so I will look at case A (7s and aces). The probability of being dealt either a 7 or an ace as the first card is $\frac{2\times4}{52}=\frac{8}{52}=\frac{2}{13}$. I think we can assume that I was dealt either a 7 or an ace. The probability of my second card being a 7 or an ace is $\frac{7}{52-5}=\frac{7}{47}$. The third is $\frac{6}{42}=\frac{3}{21}$, fourth is $\frac{5}{37}$, fifth is $\frac{4}{32}=\frac{1}{8}$, sixth is $\frac{3}{27}=\frac{1}{9}$, and seventh is $\frac{2}{22}=\frac{1}{11}$. Multiplying these probabilities together gives $$\Pr(\text{unbeatable})=\frac{2\times3\times4\times5\times6\times7\times8}{22\times27\times32\times37\times42\times47\times52}=\frac{40320}{72191927808 }=\frac{210}{375999624 }\approx5.85\times10^{-7}.$$ So the probability of getting either case 1 or case 2 should be $\approx1.117\times10^{-6}$.

My problem: Honestly I think I have probably over-simplified this somewhere (if not everywhere). Does anybody know if this is correct. If it isn't, how do I do it?

$\endgroup$
  • $\begingroup$ Use the multi-hypergeometric pmf to help: $\displaystyle \frac {\displaystyle \binom {4} {3} \binom {4} {3} \binom {52 - 4 - 4} {1}} {\displaystyle \binom {52} {7}}$ $\endgroup$ – BGM Jul 9 '17 at 16:42
  • $\begingroup$ @BGM could you explain a little please? $\endgroup$ – JSharpee Jul 9 '17 at 21:28
0
$\begingroup$

We can compute the chance of three sevens and three aces, then double it to account for the hands with kings. The number of hands like this comes from four choices of the sevens, four choices of the aces, and fourty-four choices of the odd card. There are $52 \choose 7$ total hands, so the probability including the kings is $$\frac {2\cdot 4\cdot 4 \cdot 44}{{52 \choose 7}}\approx 1.05 \cdot 10^{-5}$$

$\endgroup$
  • $\begingroup$ I've had a think about this. Don't we need to use conditional probability (so complicated that computation should probably be used) because we are assuming that after I've been dealt my first card, no one else is dealt one of the cards I want? $\endgroup$ – JSharpee Jul 10 '17 at 13:31
  • $\begingroup$ You don't have to worry about the cards anybody else is dealt. You might as well just be drawing the top seven cards of the deck, then they get theirs after you. $\endgroup$ – Ross Millikan Jul 10 '17 at 13:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.