Luck o' the Draw

[araya]

Minotaur's new puzzle Luck of the Draw..

Skinny has to know which pile is which suit. As long as he knows that, he knows that C beats H 5/9 of the time, S beats C 5/9 of the time, and H beats S 5/9 of the time.

code:

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Hearts:  2         7   9


Clubs:     3   5         10


Spades:      4   6   8

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Of course, this scheme really only works if you play the game a number of times, so even if Skinny doesn't know which pile is which suit, he will after the first draw.

I think I posted a problem very similar to this one a while ago, maybe in the old forum, involving 3 dice which you had to label with 3 numbers on each such that die A beat die B, B beat C, C beat A.

[mithrandir]

interesting side note:

code:

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9  2  7


4  6  8


5  10 3 

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is a magic square, each suit is a row...

[Wonko the Sane]

Agreed, if he doesn't know what piles are what suits, then it is, in fact, completely fair. If he knows which is which, his bets are best if he has you choose hearts, then spades. In that case, he wins 0/3 times if you pick the 2 of hearts, 2 out of three for the 7, and 3 out of three for the 9. This gives a relative probability of 5/9, as Araya already said. One of these days, I WILL post first. It is my quest...or something. Good call Araya.

[Ghost Post]

I think - i have devised a method by which the player going second can win 3/4 of the time.

If he simply doesn't pick the pile form which the first card was drawn, he will win 3/4 of the time.

If you go first and get the card shown in brackets, the cards that the second player will win with are shown next, (only out of the 6 from the untouched piles) then the propability that the 2nd player wins is shown.

(2H) - 3,4,5,6,8,10 so 1/6 * 6/6 = 6/36
(3C) - 4,6,7,8,9 so 1/6 * 5/6 = 5/36
(4S) - 5,7,9,10 so 1/6 * 4/6 = 4/36
(5C) - 6,7,8,9 so 1/6 * 4/6 = 4/36
(6S) - 7,9,10 so 1/6 * 3/6 = 3/36
(7H) - 8,10 so 1/6 * 2/6 = 2/36
(8S) - 9,10 so 1/6 * 2/6 = 2/36
(9H) - 10 so 1/6 * 1/6 = 1/36
(10C) - # so 1/6 * 0/6 = 0


Total Probability = 27/36 = 3/4

So pick from a different pile than the first card is selected, and you win 3/4 of the times.


Ashley England

[Quailman]

Ashley: I laid out the same table, but the probability of each first draw is 1/9, so the 27 winning positions are out of 54 possible outcomes, or 50-50. If you have a reason for using a 1/6 factor on each line, please explain. I used 1/9 there.

I just re-read the problem, though, and it appears that the cards are random within their piles, but you know which pile has which suit. Furthermore, so does Skinny, and he'll select which one you are to draw from next. That is why Skinny has a 5/9 advantage, no matter which suit you select. Each suit has a 5/9 advantage agaiinst one and a 4/9 disadvantage against the other, but Skinny tells you which one to draw from.

[Dragon Phoenix]

The solution is so straightforward and so simple that I have the sneaking suspicion we are missing something.... just don't know what. If this really is all to it, I am rather disappointed.

[Ghost Post]

Hmm, I dunno. This puzzle seems really easy if it's just all % of winning...There must be something more!

oh, btw. this has nothing related to this puzzle but, does any1 know any good sites on paradox's?

if you do plz tell! thanks!
PyroGx99@hotmail.com

[Amy]

One thing wasn't clear to me. The first player picks a card; does he then show that card to Skinny? Or does Skinny know nothing about the card except its suit? The puzzle, as posted, doesn't really say either way. I guess it doesn't really affect the playing strategy, though...funny, it seems like it should...

Pyro, try posting your question in the Off Topic forum, too. Some people frequent that forum that don't regularly visit this one.

[Ghost Post]

For any one who thought this puzzle was a mite easy, I've posted a more difficult variation, which you might want to consider, under "General Luck of the Draw" in the Visitor Submitted Puzzles forum.

[Throsby]

Query- there are other sets of cards from 2 to 10 in subsets of three cards of three suits which would give Skinny better odds (presuming he knew which pile was which)- eg 2H, 3S, 4S, 5C, 6C, 7C, 8H, 9H, 10S. Is there a set which gives even better odds?

[Ghost Post]

The catch to this game of chance is that Skinny will have a greater chance of drawing the higher card on the second draw then I will on the first. Skinny gets to determine which pile I choose from on the second draw, and he knows that his chances are greater that he will draw the higher card, however, in order for Skinny to score jack off of me, he would have to play many hands against me.

[Ghost Post]

In an earlier posting, Amy asked whether or not Skinny knows your card before he takes his. This is crucial in my mind as you can win if the card is kept secret. Assuming Skinny doesn’t know your card, then not only will he draw from the pile with best probability, but in the event of a tie he will be forced to choose the pile containing the highest card. Drawing a 2 will still force skinny to try for the 10. With the fields leveled in this way, the advantage sings to you because you get to go first. Running the probabilities reveals that picking Spades gives a slightly high advantage. Over 10000 games, Spades should average $600, Clubs $-300 and Hearts -$300.

[Ghost Post]

Sorry, my numbers above are wrong. A better definition would be to say that starting with Spades would yield a 20% profit, Clubs –10% and Hearts –10% (I hope)

[JVC]

The second card will beat the first card 5 out of 9 times, given that skinny chooses the most favorable pile for the second draw. I arrive at this by simply totalling three trials for each possible first draw, with three possible first draws for each pile, all equally likely. Of 9 possible pairs of first/second draws for each pile (not 18), five of these are favorable to skinny in all three cases. The game is fair only if the payout ratio is changed to 5:9.

[Calfaile]

On a completely unrelated topic... Minotaur, where did you find that deck of cards pictured in this puzzle? I've never seen anything like them before!