Question: If we have cards numbered 1-9 how do the odds change when we have four each of cards 1-9? (Ace = 1)
After that we need to figure in the face cards but one step at a time.
Are there 10s in the deck?
Ok - pick 4 cards. 40 cards (A-9) plus 3 face cards. Suite does not matter. Order does matter.
For no face cards - 40x39x38x37 =2,024,640 hands.
For 1 face card - 40x39x38 = 59,280 <== Three non-face cards.
Now pick one of 3 face cards; then that could be first, second, third or fourth. (FNNN, NFNN, NNFN, NNNF)
59,280x3x4 = 711,360
For two face cards - 40x39 = 1560 for the two number cards.
Pick 2 of 3 face cards (i.e. don’t pick 1 so 3 choices) You can pick A then B or B then A.
Hands: ABNN, ANBN, ANNB, NABN, NANB, NNAB; and swap A,B -BANN, BNAN, etc. Total of 12.
1560x3x12 = 56,160
Total possible hand choices - sum of the above - 2,736,000
So odds of picking the right hand - 1:2736,000
Then comes the fun part - eliminate hands where suite doesn’t matter … good grief…
(I.e. a had with 2C 2H is same as 2H 2C)
I’m thinking its something along the lines of -
Number of hands where two number are the same divided by 2, and 2 pairs divided by 16, number of hands where 3 numbers same divided by 6, and there are 10 hands of 4 that can be picked 160 different ways…
Not a trivial problem.
Still not sure what cards are in the deck, but for the sake of argument let’s use the 43-card deck described in Post #8.
In that case, nothing is substantially different from the previous calculation. There are 43×42×41×40 ways to deal 4 cards. A, A, A, A corresponds to 24 of them, so you would not want to guess that. A, 2, 3, 4 has a 1-in-11570 chance of winning.
ETA the worst guess is something like A, J, Q, K, of course.
md2000, you are just making things unnecessarily complicated 
And missing a bunch of permutations; there are nearly 3 million ways of drawing 4 cards, not 2,7
Actually there are no tens (since he wants the guess to be a four digit guess) but I felt bad changing the parameters again. I figured the math would be the same and easy to recalculate without the tens if we ever figured the math out here.
To be clear the cards are:
-
1-9, four of each number (Aces are “1”)
-
Jack, Queen, King (one each)
You can see him do it here: https://youtu.be/xdsakEafFRU?t=46
Well, with only 39 cards, there are 1974024 different results, so A,2,3,4 will match approximately 1 time in 7711.
Hijack: obviously A = 1, but who started to print the letter A on the card? I have a French-style Tarot deck (78 cards) where it just says “1”.
Correction/clarification: way above I wrote, “There is a 64% chance…” Just to be clear, that is the chance of an arbitrary hand containing no pair or above; that was for the 43-card deck though.
Aha! The little light bulb comes on. I don’t want to divide by 24; I want to subtract 23! (And also for all possible combinations like AAA2, AA23, AA22, and so on.) That’s how to get the total number of distinct possible deals.