Say you have a deck of twelve cards, Ace --> 9 and Jack, Queen, King.
You need to predict what four cards will be dealt.
For example:
9, 2, J, 3
What are the odds on getting that right?
Also, how many possible permutations are there for the four cards dealt?
A little tricky, because 4,4,4,4 is very different from 4,5,6,J
Specifying exactly which card makes this an easier problem. Then it’s 1/52 * 1/51 * 1/50 * 1/49
Other than this useless insight, I’m too lazy right now to get down into the figures.
There are 12 * 11 * 10 * 9 different ways for the cards to be dealt; that’s 11,880 different 4-card combinations, assuming that the order IS important (so dealing 5-6-7-8 is different than 8-7-6-5.)
Of those 11,880 outcomes, only 1 is “favorable” - the one you’ve predicted.
So the odds of predicting the exact 4 cards, in order, are 1:11799.
(If the order is NOT important, then there are 495 possible combinations of 4 cards, and so the odds are 1:494.)
I hate probability and am bad at it, so I’m sure if I’m incorrect someone will come along soon and point it out (thankfully).
nm
The OP specified a 12-card deck with 12 different cards, not a 52-card deck.
deleted post…I need to check something
Of course, 11880 - 1 = 11879, so the odds are 1:11879, not what I said above.
I misstated the setup in the OP.
My apologies…I am trying to understand a “lottery” someone is running to give something away and it wasn’t clear.
Here is the deal.
They have removed three jacks, three queens and three kings from a deck of cards. Aces are in there still (four of them) and one each of jack, queen and king.
He then deals four cards.
What are the odds you will be able to guess correctly?
Also, suit does not matter for this. Just the number/letter.
So the deck contains 43 cards (4 each of 1-10, and 1 each of 11-13)? And four cards are dealt from this? Do you have to guess the order correctly, or just guess which 4 cars are dealt?
So same idea - but calculate the odds of a hand with no face cards, with one face card, two, three; those are all possible hands when totalled.
Then you can figure out the number of possible combinations of 4-card hands with 0,1,2,or 3 face cards if order does not count.
I believe it’s 4/12 x 3/11 x 2/10 x 1/9. In other words, it’s the odds that the next card dealt will be one of the four you picked, four times in a row. It works out to just over two tenths of a percent or approximately 500-1.
In your revised example, it depends on what guess you make, since some hands are more likely than others. Assuming that the payoffs are the same, your best guess would be four different number cards. But I suspect that the payoffs aren’t the same, with a higher payoff for things close to poker hands, which are incidentally also less likely.
In the revised example, assuming the payoffs are the same, does it matter what you guess? There is a 64% chance of getting 4 different number cards, but any specific hand is as likely to be drawn as 4 aces.
Sorry: if you are allowed to pick sets of hands, e.g. A-2-3-4 corresponds to 256 different hands, while A-A-A-A to only one.
For some reason I want to say 4/43 x 4/42 x 1/41 x 4/40 to draw 9, 2, J, 3 in exact order …
For any order: (13/43 x 9/42 x 5/41 x 1/40) + (13/43 x 9/42 x 5/41 x 4/40) + (13/43 x 9/42 x 8/41 x 4/40) + (13/43 x 12/42 x 8/41 x 4/40) … each of these four cases depend on when we draw the single jack in the deck …
There’s a 51/86 chance I am completely wrong …
Does the order matter? I was assuming not, like poker hands.
In which case, if you pick a hand like A-2-3-4 (but not A-2-3-J !) you will have the roughly 480-to-1 odds calculated above…
I’m frustrated… I tried to calculate the total number of combinations, counting the aces (etc.) as non-unique. My first effort was 43 * 42 * 41 *40 / 24 – to cover the combinations of A234, AA23, AAA2, and AAAA – but I quickly bumped into the case of AA22 and I don’t know how to divide out those cases. Argh.
It may be easier to think of each card as unique, so that with 43 cards you have 123410 possible hands. Then AA22 corresponds to 36 different hands, so it is straightforward to calculate the odds of any guessed hand winning. (36/123410 in the case of AA22.)
Yes. Order matters.
I appreciate the attempts to solve this so far. It was beyond me but makes me feel better to see it is a difficult problem and I am not quite as dumb as I feared. 