LOL. Very funny. The probability of completion before STARTING is ZERO! Now, would you like to know the probability of going through the entire deck? ;-)
Prob/Stat problems
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sorry about the wording. but yeah thats what i ment to say
is it in a random order? (the cards)
the deck is suffled
hmmm
1/2 x 1/3 x 1/4 x ... x 1/52? = the answer your looking for :P
it is much lager than that. remeber a regular 52 card deck has 13 numbers and 4 suits.
HINT getting though the first card is 12/13
May 2, 2011
0
#9
I've become intrigued by this problem. First, I think the probability of getting through the deck is much less than previous estimates. In fact, I think the simple (12/13)^52 might be the answer. That works out to: .01557, or just over 1%. By the way, I actually did 50 sample runs, and failed even once to get through the deck without a match (although, on one try, I matched on the 50th card). A thought occurred to me: what would be the average number of tries to match? My intuition told me, very simple, 13, since there is a 1/13 chance of matching on any particular turn. I recorded the number of the card that matched in the 50 tries, and the average was 13.254. I guess I might be on to something.
Jun 1, 2011
0
#10
Further runs seem to confirm my formula above. After 150 runs, I got through the deck one time w/o a match. The average first card matched was 11.24, a little lower that the expected 13, but well within range. I also continued the last 100 tries through to the end, whether a match occured or not, to find the average number of total matches per complete run. My prediction wa 4, based on the simple 1/13 over 52 tries. The actual average was 4.17. Unless anyone has better, I think it's time to declare this one solved.
Kumann wrote:
Is it 12/13 x 47/51 x 23/25 x 45/49....x 9/10
Kumann, I love your thinking. But, when you go to make the second draw, you assumed that a 2 wasn't already drawn on the first draw. If that's the case, then your second number, 47/51 is correct, but if a two was drawn, then your probability of not drawing one on the second draw is 48/51. So, the thing to do is take the probability of a two being drawn in the first place, and multiply it by 48/51, and add that product to the product of the probability of a two not being drawn times 47/51, and carry on from there.
same here. I gave it a few tries earlier this summer, then I got lazy. I'll ask my Calculus teacher if she knows of any easier methods.
Jul 13, 2013
0
#13
Let's try [(48/52)(47/51)(46/50)(45/49)]^13, which equals 0.01365865315, using an online calculator, or about 0.01366. Thus, we have a 1.366% probability of getting through the entire deck without a match. This is a little smaller than mrd55's suggested probability.
My reasoning goes like this: We want the probability that the cards numbers 1, 14, 27, and 40 are not aces, and that cards numbers 2, 15, 28, and 41 are not twos, and that cards numbers 3, 16, 29, and 42 are not threes, and so on. This is equivalent to asking the probability that none of the first four cards is an ace, that none of the next four cards is a two, that none of the next four cards is a three, and so on. What's the probability that none of the first four cards is an ace? Well, the probability that the first card isn't an ace is 48/52; and that the second card isn't an ace given that the first card isn't is 47/51; and that the third card isn't an ace given that neither of the first two is an ace is 46/50; and that the fourth card isn't an ace given that none of the first three is an ace is 45/49. But then for the next set of four cards--the ones that can't be twos--we start over. It doesn't matter whether any of the first four cards were twos or not--all we care about is that none of the second set of four cards is a two. So, once again we get (48/52)(47/51)(46/50)(45/49). And so on, through the thirteen denominations. So, we get [(48/52)(47/51)(46/50)(45/49)]^13 (which is the same as [(4/13)(47/17)(23/5)(9/49)]^13, which is the same as (38,916/54,145)^13, which is roughly 0.718737^13, which is roughly 0.01366).
I'm not sure how to work out the average time to a first match.
Pretend you have a regular 52 card deck. you flip a card if it isn't an ace go on. flip another card if it isn't a 2 then keep going. you keep doing this A-K 4 times. What is the probability of completion before starting? i'll give a great student trophy for the first to give me the exact answer.