Probability VP Questions

[COLTON]

With all those math wizards out there, I would love to know the answers to the following questions:
 
1.  I play a lot of 10 play VP.  If you are dealt 3 of a kind, what are the odds you will connect for:  (a) 1 four of a kind;
                       (b) 2 four of a kinds;
                       (c) 3 four of a kinds;
                       (d) 4 four of a kinds; and
                       (e) zero four of a kinds.
(I remember one memorable play getting dealt 3 aces on 10 play and receiving Quad aces four times.)
 
2.  on 9-6 DDB UX (10 Play), does anyone know the frequency of receiving:
                      (a) No multiplier on any given hand;
                      (b) Probabilitiy of getting a 2 multiplier;                                                   
                      (c) Probability of each: 3x, 4x, 8x, 10x and 12x.
(I've hit 14 RF on this game over the past couple of years.  4 times I received a 2X multiplier.  The other 10 times I had no multiplier at all.  I'm sure that's just plain bad luck---when it comes to multipliers---but I would like to know just how bad that is!)
 
Thanks, in advance, for your sabermetrics.

[craig_ga]

I THINK question #1 is a binomial distribution problem:

Given the fact that holding 3ok results in 47 cards left in the deck, of which one is needed to complete the 4ok, chances of getting a 4ok on the draw in single play is 1/47 = 0.21276596.

From this we can apply the binomial probability formula (link) to determine the chance of obtaining 1 success out of 10 trials, 2 successes out of 10 trials, etc:

Chance of n 4OKs out of 10 trials:     
1     17.5324%
2     1.7151%
3     0.0994%
4     0.0038%
5     0.0001%
6     1.78761E-08
7     2.22063E-10
8     1.8103E-12
9     8.74539E-15
10     1.90117E-17
Cumulatively, you have a 19.35086% chance of getting 1 or more 4ok. Getting all 10 4ok on the draw is so unlikely it isn't worth thinking about.

I have no idea how to attack question #2

[COLTON]

Thanks for the response, craig ga.  But, don't we have two bites at the apple for each draw to 4OAK?  Since you are holding three cards, you have two opportunities on each hand to draw the fourth matching card, correct.  I would assume that would adjust the probability significantly. 

[onemoretry]

I would assume that would adjust the probability significantly. 
For sure.

The probability of making quads when drawing two cards to trips works out to exactly twice that, i.e., 1081 (the total number of possible outcomes) divided by 46 (the number of combinations containing the desired card), which calculates to 1/23.5.

[BobDancer]

9/6 DDB UX Ten Play uses a strategy very close to regular 10/7 DB. Therefore, the distribution of resulting hands at 10/7 DB is as a first approximation reasonably close to what that distribution would be on 9/6 DDB UX.Using Video Poker for Winners, the information is easy to come by. Now I've provided the "road map" I'll let somebody else assemble the data requested.Bob