Introducing Scooby Snacks!

I've run through a few different calculations (one thing is for sure - I can't add on my own!) but first this is what I've settled on as a base:

To calculate probability of getting one number on either of 2 20-sided die:

19 of the 20 sides are wrong.

19/20 * 19/20 = 90%
100% - 90% = 10% probability

If you want a certain number(s) on both die:

19/20 * 19/20 = 90%
100% - 90% * 2 = 20% probability

Sticky part, I now have 3 die but I only need 2 to show my number(s).

19/20 * 19/20 * 19/20 = 85.7%
100% - 85.7% * 2 = 28.6% probability that 2, ONLY 2, of the 3 will show my number(s)

This is where it gets downright murky. I *think* the following is right.

19/20 * 19/20 * 19/20 = 85.7%
100% - 85.7% * 3 = 42.9% probability that 2 OR 3, of the 3 will show my number(s)

So we have 13 cards in a suit, 4 suits, and 2 cards we picked. The 2 cards can be in any order but their positions will take up 5 of the 13 cards.

Using the die examples, we could do..
5 good, 8 bad positions of 13, 4 sets
8/13 * 8/13 * 8/13 * 8/13 = 14.3%
100% - 14.3% = 85.7%

2 cards, 13 possible, 11 are bad
11/13 * 11/13 * 11/13 * 11/13 = 43.3%
100% - 43.3% / 5 = 91.4%

Or combine the possibilities of the 2 positions (next to & one away) - each would take up 3 of 13 cards:

Number 1: 10/13 * 10/13 * 10/13 * 10/13 = 35%
Number 2: 10/13 * 10/13 * 10/13 * 10/13 = 35%
So each, taken by itself, has a 65% chance of showing up in a given spot.
35% * 35% = 12.2%
100% - 12.2% = 87.8% combined probability

Or combine the possibility of EITHER combination showing up
5 positions, and 3 positions (one away, or next to)
8/13 * 8/13 * 8/13 * 8/13 = 14.3%
10/13 * 10/13 * 10/13 * 10/13 = 35%
35% * 14% = 5%
100% - 5% = 95% of either combination showing up

I would like to know what the ratio is for each combination when it does show up. My son got fascinated with this and actually took my deck of cards to Homecoming with him and I never got them back. I'll try to wrestle them away from him tomorrow and test it out.

Still waiting for te coffee but something does not seem right. using your example, the probability of having 4 die with your number would be 40%. Bring that out to 10 die and we have a winner every time (100%). Again, no coffee yet, but this does not seem right. Could be why they only use 2 die for craps.

edit: Be sure to see my next comment where I disprove my assumption here.
I do not understand where the 5 comes from. I think you are referring to BxAxx and xxAxB. I do not think this is correct and I would assume 3 positions are used for AxB and BxA. This is a revolving position in the circle and at any time would not require more than 3 positions to satisfy the required conditions.

Ahh, coffee is ready

Nope, you don't remove the numbers. That same number can show up on the other die and there is an equal chance of it doing so on each die - plus, I never stated the number I wanted from each was the same. It could be the same or different, it doesn't really matter, as long as it is just the 2nd number.

The reason I used 13 is because there are 4 sets of the same 13 numbers. There can be duplicates (another reason you don't remove the numbers from the other die). You could end up with 1, 2, 1, 2 for your roll.

My coffee is not ready, yet

You are correct that you could get 1,2,1,2, but they are distinct 1's and 2's which can never be repeated. Splitting it into 13 sets up 4 piles of suits. You may get a die with two 2's and a die with no 2's making you 1,2,1,2 impossible. Die numbers are not shuffled in the same manner as a deck of cards. You have to assume each position could be any of the remaining cards.

If you shuffle your deck of cards and place them on the table. Turn over the first card and place it in the center. There are now 51 cards left in the deck. The chances of your second number being on that second card (assuming it was not in the first) is 4/51. OR 47/51 incorrect. Place this card on the left of the first card.

Then chances of the next card being your number is now 4/50 since two cards have been removed. OR 46/50 incorrect. Place this one on the left of the two cards on the table.

Working to the right you get 4/49 and 4/48 OR 45/49 and 44/48 incorrect.

Incorrect chances would therefore be 47/51 * 46/50 * 45/49 * 44/48
Also shown as 47*46*45*44 / 51*50*49*48
(See how I so casually prove myself wrong without making note of the epiphany)
4280760 /5997600 = .7137 or 71.4% chance of getting them all wrong.
That would leave a 28.6% chance of getting it right.

We can't use the number remaining, as we don't know, and don't know if any of our cards have shown up yet.

We need either a 2-number combo or a 3-number combo in a 5 number block in a big circle of 52.

132600 3-number combinations
2652 2-number combinations

There are 2598960 combinations of 5-block numbers.

If we're only going to calculate out the 4, how do we know if we've gone through 1 or 20 cards so far?