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I am trying to figure out the odds of a certain situation in the card game French Tarot, specifically what are the odds of the taker being out of the same suit of the $R$ that a defender has played on round 1. This is a bad situation (for the defenders) since the taker would win at least $6$ points on the first round.

As a start to this, I want to calculate the odds of any player who takes the dog ending up out of a suit of cards (i.e. having 0 cards of that suit). The dog is a set of 6 cards that the attacker can pick from and must put back 6 cards before starting play.

To calculate this, I want to see the probability that the person has at most 6 of one suit after adding the 6 cars from the dog to their 18 cards (assuming a 4 player game). With 6 or less cards for one suite they can all be placed in the dog, ending up with none of that suit for the final 18 cards before the playing begins.

I wrote the following formula to capture this: $$C(78-14,24-6)×C(78-(24-6),6)/C(78,24)$$

The first term $C(78-14,24-6)$ represents choosing $18$ cards that are not from that suit, and the second term $C(78-(24-6),6)$ represents picking 6 more cars (of any suit) from the remaining cards, which is 78 total minus the first 18. The last term in the division $C(78,24)$ represents the total number of hands of $24$.

However, when I plug this into Wolfram Alpha I get $\approx2280$ which means my equation is fundamentally wrong since it should be less than $1$.

If someone can tell me where I am going wrong I'd appreciate it.

Update: I realize that in a real game of French Tarot the attacker who wins the bet is likely to have a better hand than the others, but I am not factoring that in to my calculations at this point, and am assuming an even distribution. Mainly I want to see what is wrong with my equation and then I can build off that.

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  • $\begingroup$ Is 'R' for "roi"? $\endgroup$
    – joriki
    Commented Sep 17, 2015 at 18:37
  • $\begingroup$ Yes, it's for 'or' or King. Rules for the game here:en.wikipedia.org/wiki/French_tarot $\endgroup$
    – Locksleyu
    Commented Sep 17, 2015 at 18:41
  • $\begingroup$ Yes, I found those rules, but was asking since I didn't find an 'R' there. What's 'or' in your comment? $\endgroup$
    – joriki
    Commented Sep 17, 2015 at 18:42
  • $\begingroup$ Sorry I just realized that typo and tried to edit it but couldn't. It should be 'roi' like you said. $\endgroup$
    – Locksleyu
    Commented Sep 17, 2015 at 18:52

1 Answer 1

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I'll assume that by "at most $6$ of one suit" you mean "at most $6$ of one particular suit" (namely the one whose 'R' the defender plays).

I'll also ignore, like you did, the fact that a defender is known to have played the 'R', so this card is known not to be in the attacker's hand.

Your calculation overcounts hands that have more than $18$ cards not from the given suit (equivalently, hands with less than $6$ cards from that suit). For instance, a hand with $19$ cards not from the given suit is being counted $\binom{19}1=19$ times, since any of the $19$ cards can be taken to be the one in the group of $6$. What you need is

$$ \sum_{k=0}^6\binom{78-14}{24-k}\binom{14}k\;. $$

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  • $\begingroup$ Thanks. This looks good though I'll wait until I actually get a number to validate before I accept your answer. I need to figure out how to easily input this into wolfram or some other calculation engine. $\endgroup$
    – Locksleyu
    Commented Sep 17, 2015 at 19:42
  • $\begingroup$ @Locksleyu: Interestingly, Wolfram|Alpha accepts $\TeX$ code. It has a weird bug with juxtaposed binomials, which it tends to nest instead of multiplying them. It works if you copy the $\TeX$ code from above (Right/Ctrl-click on the formula, select "Show Math As ... TeX Commands", select all and copy) and insert a '*' between the two binomials before feeding it to Wolfram|Alpha. $\endgroup$
    – joriki
    Commented Sep 17, 2015 at 19:45
  • $\begingroup$ Thanks. I tried that but getting some crazy number. Can you please confirm if this is what you mean: \sum_{k=0}^6\binom{78-14}{24-k}*\binom{14}k\;. $\endgroup$
    – Locksleyu
    Commented Sep 17, 2015 at 19:54
  • $\begingroup$ I think the problem is that I need to add the denominator, but not sure how to add that to the TEX form. $\endgroup$
    – Locksleyu
    Commented Sep 17, 2015 at 19:56
  • $\begingroup$ @Locksleyu: you don't need the trailing \;.. And you can mix styles, so you can add the denominator in $\TeX$ form, or in standard form like this: \sum_{k=0}^6\binom{78-14}{24-k}*\binom{14}k / binom (78,24). $\endgroup$
    – joriki
    Commented Sep 17, 2015 at 19:57

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