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Probability and Statistics
Lady Eri
Registered User regular
Registered User regular
I am trying to find the probability of that any three cards I draw out of a deck will add up to 21 or more (this is not for a school project or gambling theory or anything like that, real world application I promise), but I am having trouble coming up with the equation. I don't fancy typing out each possible combination out in excel either, and am starting to suspect it would take a very long time regardless.
Anyone know of a site where I can put the probability of a number being selected (like 4/13 for the number ten, 1/13 for the number six, ect) then set a threshold where it will give me a probability. Or alternatively, if someone knows an easier equation then writing out each possible combination of cards? I did well in statistics, but it was a looong time ago.
Anyone know of a site where I can put the probability of a number being selected (like 4/13 for the number ten, 1/13 for the number six, ect) then set a threshold where it will give me a probability. Or alternatively, if someone knows an easier equation then writing out each possible combination of cards? I did well in statistics, but it was a looong time ago.
Lady Eri on
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But in all seriousness...first of all, what are the value of face cards in this hypothetical? And what is an Ace worth? 1 or 11 or something else? Or are face cards omitted from this deck? Or what?
edit: The first problem (for us, since you probably already know exactly what you mean) is defining "deck," especially when you are talking about point values. Face cards can be worth anything (including zero) depending on the game. And some cards, depending on the game, can be worth different amounts depending on the context. So, yeah, we need some more info first.
SO (assuming you're counting all face cards as 10s and aces are only 1):
1 + 10 + 10
2 + 9 + 10
3 + 8 + 10
3 + 9 + 9
4 + 7 + 10
4 + 8 + 9
5 + 6 + 10
5 + 7 + 9
5 + 8 + 8
6 + 6 + 9
6 + 7 + 8
7 + 7 + 7
Now, there are 16 cards that count 10 and 4 cards each for 1-9. so the number of ways to make each of the aforementioned sets are:
1 + 10 + 10: 4 * 16C2 = 4 * 120 = 480 +
2 + 9 + 10: 4 * 4 * 16 = 256 +
3 + 8 + 10: 4 * 4 * 16 = 256 +
3 + 9 + 9: 4 * 4C2 = 4 * 6 = 24 +
4 + 7 + 10: 4 * 4 * 16 = 256 +
4 + 8 + 9: 4 * 4 * 4 = 64 +
5 + 6 + 10: 4 * 4 * 16 = 256 +
5 + 7 + 9: 4 * 4 * 4 = 64 +
5 + 8 + 8: 4 * 4C2 = 24 +
6 + 6 + 9: = 24 +
6 + 7 + 8: = 64 +
7 + 7 + 7: 4C3 = 4 +
= 1772
then the probability of getting 21 would just be the sum of all that divided by 52C3 or 1772 / 22100 or ~0.0802 or just over 8%.
Yay maths!
Thats awesome, is there a way for me to run with that and get all outcomes over 21 included in the result?
I think you'd need to calculate the total probability for each possible score and add them to each other.
So you'd have to work out the possibility of getting a 21, a 22, a 23, a 24, a 25, a 26, a 27, a 28, a 29, and a 30 and then smoosh them all together into an uber-probability.
I got a ~45.85% probability of getting 21 or greater without replacement, and a ~8.02% probability of getting exactly 21 without replacement (which matches the other guy's result!).
Here's the program (it's in Python):
threshold = 21 cards = [] for suit in range(0, 4): for value in [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]: cards.append((suit, value)) total_equal = 0 total_over = 0 total_cards = 0 for card1 in cards: for card2 in cards: if card2[0] == card1[0] and card2[1] == card1[1]: continue for card3 in cards: if card3[0] == card1[0] and card3[1] == card1[1]: continue if card2[0] == card3[0] and card2[1] == card3[1]: continue total_cards += 1 sum = min(10, card1[1]) + min(10, card2[1]) + min(10, card3[1]) if (sum == threshold): total_equal += 1 if (sum >= threshold): total_over += 1 print "total over %d: %d, percent: %f" % (threshold, total_over, 100 * total_over / float(total_cards)) print "total equal to %d: %d, percent: %f" % (threshold, total_equal, 100 * total_equal / float(total_cards))Edit: I did it by hand, so I fully expect to have made an error somewhere.
Your solution happens to be simpler and more flexible than doing it by hand as I was. Assuming one has access to a microprocessor, that is. It's also probably more in line with what the OP was looking for.
like you could. If you wanted and kinda sucked at math. It's real easy and obvious. I'm fucking lazy of course so I'm not doing it.
Yeah - for a math class this would obviously not work, but in the real world, it would give close enough approximations of the probabilities for basically any purpose.
In this case, of course, there's likely a (relatively) quick analytical solution. It would just take longer to figure out and program.
There's no point in doing a random simulation when you can just deal out every single possible combination of cards and tally up the scores.
p(X = 1) = p(X = 2) = 0
p(X = 3) = .4589
p(X = 4) = .3582
p(X = 5) = .1435
p(X = 6) = .0338
p(X = 7) = .0052
p(X = 8) = .0005
p(X = 9) < .0001
p(X = 10) < .00001
p(X = 11) = negligible
Anyone want to check these numbers?
with 10,000,000 trials using
import random # we want to know how often we get a sum of at least 21 threshold = 21 trials = 10000000 deck = [] for suit in range(0, 4): for value in [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]: deck.append((suit, value)) results = {} for trial in range(0, trials): # a progress counter thing if uncommented #if (trial % (trials / 100) == 0): # print 100 * trial / float(trials) indices = {} sum = 0 while (sum < threshold): # select a random card index = random.randint(0, 51) # if we've already seen this index, try a new one while (index in indices): index = random.randint(0, 51) indices[index] = 1 sum += min(10, deck[index][1]) # the number of cards required is the number of indices we've # recorded cards = len(indices) if not cards in results: results[cards] = 0 results[cards] += 1 print "number of cards required to sum to 21 or over, as a percentage of cases" for (cards, number) in results.items(): print "cards: %d, percent: %.04f%%" % (cards, 100 * number / float(trials))which is pretty damn close to what you get.
using the following code:
import math # returns a list of all possible combinations of # num_cards cards, with no duplicates, where each # combination is listed in ascending order. def find_combos(num_cards): all_combos = {1: [[i] for i in range(1, 11)]} for i in range(2, num_cards + 1): prev_combos = all_combos[i - 1] combos = [] for combo in prev_combos: for j in range(1, 11): # ensure that we don't use a number more than # it is allowed (16 times for 10 value cards # and 4 times for all other values). times = (4 if j != 10 else 16) if j >= combo[-1] and combo.count(j) < times: combos.append(combo + [j]) all_combos[i] = combos return all_combos # returns total CHOOSE number def choose(total, number): return math.factorial(total) / (math.factorial(total - number) * math.factorial(number)) if __name__ == "__main__": total_cards = 12 all_combos = find_combos(total_cards) prev_prob = 0 for cards in range(3, total_cards + 1): total_over = 0 total_not_over = 0 for combo in all_combos[cards]: seen = {} sum = 0 for card in combo: if not card in seen: seen[card] = 0 seen[card] += 1 sum += card ways = 1 for (card, num) in seen.items(): # there are 4 cards of each value, except # there are 16 cards valued at 10 each choices = (4 if card != 10 else 16) ways *= choose(choices, num) if sum >= 21: total_over += ways else: total_not_over += ways prob = total_over / float(total_over + total_not_over) - prev_prob prev_prob += prob print "cards: %d, percent: %.06f%%, total probability: %.06f%%" % (cards, 100 * prob, 100 * prev_prob)It produces the right answer for 3, so I'm assuming it's correct.
EDIT: whoops, found my error. Had variable declarations outside the appropriate loop. *facepalm*
OLD : Although it's weird that the 4 card case is about 2% under the simulated result. I wonder why that is?