# Hypergeometric Probability - Average chances after multiple trials

Hopefully someone can help me out here. I'm stuck on a few hypergeometric probability calculations I'm working on for an assignment. I'm stuck on finding the average chances from Trial 3 and onwards.

You have a normal shuffled deck of 52 cards (13 ranks * 4, no jokers). Specific suits do not matter, only that there are 4 successes of each rank. Each trial involves selecting a rank and then drawing 10 cards without replacement except for the final trial where only 1 card is drawn from 2 remaining in the deck. I need to find the average chances in each trial of drawing atleast one card of the selected rank. IMPORTANT: The rank selected for each trial will always be one that has the (equal) most successes left in the deck.

I simulated the trials 10 million times with code and got each trials estimated %, but I need to know how to get to those numbers at the core of it all. Below is a table of the results. A fail in a trial meant all the cards were reshuffled and began from Trial 1 again.

The results of the simulations are here.

I need to work out how to get the average chances for a success in each trial (column highlighted in bold text) from Trial 3 and onwards. No matter what spread of cards are drawn in Trial 1, there will always be a rank to select in Trial 2 that still has 4 cards remaining in the deck. However where I'm stuck is after Trial 2, the number of cards drawn (20) is greater than the number of ranks (13), therefore in Trial 3 the rank to select with the most successes remaining could have 3 and other times 4 remaining.

So for Trial 1, the chances atleast one card drawn of the selected rank is: $$1 − \frac{48}{52} \frac{47}{51} \frac{46}{50} \frac{45}{49} \frac{44}{48} \frac{43}{47} \frac{42}{46} \frac{41}{45} \frac{40}{44} \frac{39}{43} \approx 0.586555 = \mathbf{58.6555\%}$$

For Trial 2 with 42 cards remaining in the deck you now select another rank. It must be one with the best chances, therefore it will be one that hasn't been drawn yet. The chances of the selected card being drawn atleast once would be: $$1 − \frac{38}{42} \frac{37}{41} \frac{36}{40} \frac{35}{39} \frac{34}{38} \frac{33}{37} \frac{32}{36} \frac{31}{35} \frac{30}{34} \frac{29}{33} \approx 0.678728 = \mathbf{67.8728\%}$$

Now Trial 3 and onwards is where I need help. Trial 3 has a population size 32 and sample size 10. Chances if selected rank has:

3 successes remaining = 68.9516%

4 successes remaining = 79.6580%

How to work out the average chances there will be 3 remaining, and other times 4, to get the overall chances?

Further, how would you work out the average chances for the next trials where the best rank to select could have 2, 3 or 4 cards remaining depending on the average spread of the previously drawn cards?

For Trial 6, 50 cards have now been drawn with only 2 remaining in the deck. Only 1 card will be drawn this time. What would the chances be of selecting the rank of the drawn card? In the unlikely possibility the last 2 remaining cards are the same rank, your selection would be 100% correct. So to find the average chances would it be this?:

($$\frac{3}{51}*100%$$) + ($$\frac{48}{51}*50%$$) = 52.9412%

Since there is a 3/51 chance the 51st card is the same rank as the 52nd? Is there some other variable to consider because that seems too far off from my simulated results but maybe they aren't accurate enough.

If someone could please help and provide the proper notation too that would be very much appreciated :-)

• Welcome to MSE. Please edit and use MathJax to properly format numbers and math expressions. Also, I personally think that you should define the "game" more clearly before the detailing the trials. – Lee David Chung Lin Apr 14 at 4:31
• Sorry, new here. I have updated best I can. I'm not sure how else to better define each trial – user663934 Apr 14 at 4:50
• Using MathJax is for better readability. It serves a practical purpose and not just some "rule" one needs to conform to. I just did a partial edit as an example to show you how the post might be improved. As for defining/describing the trials, maybe it's just me so don't worry about it. – Lee David Chung Lin Apr 14 at 5:02