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PROBLEM
The question is related to SET Card Game. I was thinking about that if I choose $12$ cards randomly from a deck then what will be the probability of no $3$-card-SET being there.

MY METHOD

  • It is easy to see that if you have chosen two cards, there is only one in the deck that could complete that as a set. So for the first two cards to choose there are $81$ and $80$ possibilities respectively.
  • Now there is a card in the deck which makes this two a set, we cannot choose that, there for the third card can be chosen from $78$ cards.
  • Now there is a SET-completing card for the third and first, and the third and second too. So there is two more 'forbidden' cards in the deck. Therefore we can only choose from $75$ cards, instead of $77$.
  • And just like this, in the $n^{th}$ step you can choose from $(81-n)$ cards (putting down the first card is step $0$ in my thinking). After you have your twelve cards you obviously have to divide it by $12!$, since the ordering of the cards doesn't matter.
  • Therefore I think the probability is:

$$P(\text{no SET in }12\text{ cards})=\cfrac{81 \cdot 80 \cdot 78 \cdot 75 \cdot 71 \cdot 66 \cdot 60 \cdot 53 \cdot 45 \cdot 36 \cdot 26 \cdot 15 \cdot \frac{1}{12!}}{81 \choose 12} \thickapprox 0.0105$$

This is different from the $30:1$ odds claimed here referenced from this page.

My question is where is the error in my calculations?

Rohit Singh
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math_inquiry
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  • Going from four cards where no set has occurred to a fifth card, you claim there are $81$ ways to have chosen the first card, $80$ ways to have chosen the second card. That's fine. You claim there are $78$ ways to have chosen the third card, that's fine. $75$ ways to have chosen the fourth. I could possibly believe that as well but am hesitant. You claim there are always only $71$ options to choose the fifth card, that $10$ of the options are now unavailable. Certainly, the four cards already chosen are not available... – JMoravitz May 15 '22 at 14:14
  • ...but you make the incorrect assumption that the third card for each of the pairs already chosen must be distinct for each of the pairs, but that need not be the case. For example, if the first two cards were both solid green circles (one circle and two circles) the card which completes that set would be solid green three circles. This could also have been the card to complete a different set, e.g. if the third and fourth cards were dotted green three circles, empty green three circles, the card which completes it would be again solid green three circles. – JMoravitz May 15 '22 at 14:17
  • Oh yeah, you are absolutely right, now i see how it went wrong. Thank you! – math_inquiry May 15 '22 at 14:20
  • The punchline is that multiplication principle as you have attempted will not easily work. As for what you can do to fix the calculation? I don't see any valid approach just yet beyond brute force computer assistance. I assume there must be some theory with finite projective fields or something which may help (like it does for dobble), but I haven't seen this problem before and would need more time to think – JMoravitz May 15 '22 at 14:20
  • Yeah with this newfound insight it really looks like its getting out of hand very rapidly with this method.
    I was also thinking about it's relations with dobble beforhand, but the transition from the game to a projective field is not that "trivial" in this case (for me at least) :)     – math_inquiry May 15 '22 at 14:25
  • I suggest following the wikipedia link for cited statistic: https://henrikwarne.com/2011/09/30/set-probabilities-revisited/ where it leads to a blog post of someone having investigated this problem and used computer simulation to approach it. – JMoravitz May 15 '22 at 14:27

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