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I also posted this on 2p2 but it is time to just answer it myself.
We need to know the probability of fouling under Scenario A vs Scenario B.
We are drawing two more cards and if at least one of the two cards is not 6, 7, 8, 9, T or J then we foul under both scenarios. We'll call these cards our outs.
Scenario A:
We also foul if we draw the last 3 in the deck (3h) with any of our outs.
Scenario B:
We also foul if we draw 2 or 4 with our 7, 8, 9 and T outs.
If we draw our 7, 8, 9 or T out before drawing the 6 then we also foul. This is the only spot where the order matters.
Cards Exposed: 22
Cards Left for Drawing Pool: 30
Breakdown of cards left for drawing:
2: 2
4: 2
3: 1 (only the 3 of hearts is left)
6: 2
7: 2
8: 2
9: 3
T: 2
J: 4 (all four of the jacks are live)
Other: 10
Total: 30
This is the probability for missing our outs altogether such that we foul under both scenarios:
P(blank and blank) = (15/30)*(14/29) = 24%
The order is important for part of Scenario B so we'll use permutations for the scenarios instead of combinations.
The number of ways we can draw our last 2 cards in our sample space:
Order Matters ==> Permutations ==> P(n,k) = n!/(n-k)! = P(30,2) = 30!/28! = 30*29 = 870
Scenario A: 3h issue. If the case 3 comes with any of our 15 outs in either of 2 orders (3h, 7c or 7c, 3h) then we foul. The chances are 15*2/P(30,2) or 3.5%.
*This means the total chances of fouling under Scenario A are 27.5% (24 + 3.5).
Scenario B: Deuces and Fours Issue. There are 4 deuces and fours left (two of each). If these combine with 9 of our outs (7, 8, 9, T, J) in either of 2 orders then we foul. The chances are 4*9*2/P(30,2) or 8.3%.
Scenario B: Sixes Issue. There are 2 sixes left. These 2 sixes cause fouls only if they come AFTER 9 of our outs (7, 8, 9, T). The chances are 2*9/P(30,2) or 2%.
*This means the total chances of fouling under Scenario B are 34.2% (24 + 8.3 + 2).
Let me know if these probabilities look correct. We'll need to do more of this type of thing if we are to build a foul calculator.
Notes to Self:
We could have used combinations instead of permutations for the parts where order doesn't matter:
Order Doesn't Matter ==> Combinations ==> C(n,k) = n!/k!(n-k)! = C(30,2) = 30!/(28!*2!) = (30*29)/2 = 435
3h Issue. If this case 3 comes with any of our 15 outs in any order then we foul. The chances are 15/C(30,2) or 3.5%
Deuce Four Issue. There are 4 deuces and fours left (two of each). If these combine with 9 of our outs (7, 8, 9, T) in any order then we foul. The chances are 4*9/C(30,2) or 8.3%
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