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extropalopakettle
No offense, but....

 Posted: Sat Apr 25, 2009 1:49 pm    Post subject: 1 Sorry, this is not a puzzle. I just happen to know that the folks who can answer it are here. I toss N fair coins in the air. I repeat this 2^M times. What are the probabilities that: 1) "All heads" never occurs. 2) "All heads" occurs exactly once. 3) "All heads" occurs more than once. If it's a lot simpler, I'm more interested in the cases where M = N-1, N and N+1
Zag
Tired of his old title

 Posted: Sat Apr 25, 2009 2:56 pm    Post subject: 2 Hi extra. You should take a look at the probability problem that I just posted in "Some Original Riddles." This is similar, as least for some of your questions. If you toss N fair coins, the chances of "all heads" in one toss is 1/2^N. Call this p. So the chances of it NOT happening for one toss is 1-p So the chances of it NOT happening for M tosses is (1-p)^M So the chances of it happening at least once in M tosses is 1 - (1-p)^M You can plug in 2*M for M, or 2*(N-1) for M, etc. I'm less sure how to answer #2 and #3.
Amb
Amb the Hitched.

 Posted: Sun Apr 26, 2009 5:26 am    Post subject: 3 Ive been meaning to post this. Imagine a four player card game, where a whole deck of cards (no jokers) has been dealt out (13 cards each): 1. What are the odds you will be short suited at least one suit. 2. When short suited, what are the chances of one of your opponents being a) Short Suited of any suit, b) Short suited of the same suit you are short suited in.
Thok
Oh, foe, the cursed teeth!

 Posted: Sun Apr 26, 2009 5:47 pm    Post subject: 4 For extropal's question For one flip, the probability of getting all heads of N coins is 1/2^N. If you do this 2^M times, you get 1) All heads never appear, (1-1/2^N)^(2^M) times. For N,M large and close you can probably approximate this by some power of e (taking advantage of the fact that (1-1/n)^n approaches 1/e as n gets large.) 2) All heads occurs exactly once: there are 2^M places for the all heads to appear, and you get 2^M*1/2^N*(1-1/2^N)^([2^M]-1) as the desired probability. Again, you can approximate this by some power of e times a power of 2 in the case you care about that. 3) The sum of the probabilities in the three cases is 1, so the probability in this case is 1-(1-1/2^N)^(2^M)-2^M*1/2^N*(1-1/2^N)^([2^M]-1) @Amb: does short suited=void in that suit?
Amb
Amb the Hitched.

 Posted: Sun Apr 26, 2009 7:29 pm    Post subject: 5 If you are short-suited, you are missing a suit eg No Hearts.
Busted!

 Posted: Thu Apr 30, 2009 1:36 am    Post subject: 6 For Amb, question 1: There are 52 C 13 possible hands. There are 39 C 13 hands that are short suited at least one suit (all possible hands of 13 if you remove one suit from the deck). About 1.28% 2. b. (39 C 13)/(52 C 13) * (26 C 13)/(39 C 13) (Answer from question 1) * (Possible hands made from the remaining cards not of the short handed suit) / ( The possible hands that can be made with the remaining cards after "you" are dealt your cards) I don't see why the order in which you deal the cards will have an effect. =(26 C 13)/(52 C 13) About 0.00164% a. It makes sense to me that it would just be the answer from part one squared but I'm not 100% sure on that._________________"Heads you're good, tails you fail" says my fortune cookie
Zag
Tired of his old title

 Posted: Thu Apr 30, 2009 1:55 am    Post subject: 7 ~smacks head~ I knew there was an easy way to answer that question. I'm annoyed with myself that I didn't think of it. Well done, SS.
Busted!

 Posted: Thu Apr 30, 2009 10:33 am    Post subject: 8 I didn't actually compute the value for 2a. It turns out to be 10 times more likely than 2b. That's quite nice but probably just coincidental._________________"Heads you're good, tails you fail" says my fortune cookie
ChickenMarengo
Daedalian Member

Posted: Thu Apr 30, 2009 8:38 pm    Post subject: 9

 ShadowSword wrote: For Amb, question 1: There are 52 C 13 possible hands. There are 39 C 13 hands that are short suited at least one suit (all possible hands of 13 if you remove one suit from the deck). About 1.28%

I've got to disagree with that. That's the probability that you're short-suited in hearts (or in any given suit).

There are also 39 C 13 hands that are short in each of the other suits too, so you need to multiply by 4. But that double-counts hands that are short in 2 suits, and so on.

Using inclusion-exclusion, there are 4 * (39 C 13) - 6 * (26 C 13) + 4 = 32427298180 hands short at least one suit, so the probability is about 0.0511.

For question 2 you only looked at one of the opponents, and Amb wants the probability that any of them is short-suited. It's also complicated by the fact that you can be short 2 suits, and then (presumably) either of them counts towards part b. I'll have to think some more about that one.
Busted!

 Posted: Thu Apr 30, 2009 8:57 pm    Post subject: 10 Thanks CM. I was starting to think I wasn't as bad at probability as I thought I was. Order has been restored back to the universe._________________"Heads you're good, tails you fail" says my fortune cookie
ChickenMarengo
Daedalian Member

 Posted: Sat May 02, 2009 6:42 pm    Post subject: 11 As I said, ShadowSword's answer for 2b is out by a factor of 12 (4 suits * 3 opponents). But that double-counts various deals and I can't get my head round all the odd cases. It should be very close though. I'm going to have to settle for a simulation. Here are my results: Trials: 10000000 Me not short-suited: 9489360 Me short-suited: 512607 Me and opponent short-suited: 101712 Me and opponent short-suited in same suit: 1967 (26 C 13)/(52 C 13) * 12 * 10000000 = 1965, so that's pretty close. Also I noticed Amb wants the probability of an opponent being short-suited given that I am. 2a P(short in any suit) ~= 101712/512607 = 0.198 2b P(short in a suit I'm short in) ~= 1967/512607 = 0.00384 Here's one of my own: Look through a shuffled deck and note down a) the order in which you find the first card of each suit b) the order in which you find the last card of each suit. What is the probability that 1) the two orders are the same, 2) b is a reversed. I have an answer for 2, but not 1.
lostdummy
Daedalian Member

 Posted: Mon Jun 15, 2009 10:02 am    Post subject: 12 Unless I misunderstood what "order in which you find the first card of each suit " means, wouldn't both a and b be very close to: pa~pb~ 4!^-2
ChickenMarengo
Daedalian Member

 Posted: Sun Jun 21, 2009 9:29 pm    Post subject: 13 Here's an example deck with 4 cards of each suit: DSCCHHCDHCSDSHDS The first card of each suit is highlighted in bold, and the last card in bold italics, so the order of the first cards is DSCH and the order is of the last cards is CHDS. Since there are 4! possible orders for each, you'd expect the probability that the two are the same to be approximately 1/4!, but since they aren't independent, "approximately" turns out to mean "not within 10% of". (I do have answers for both parts now.)
ralphmerridew
Daedalian Member

 Posted: Mon Jun 22, 2009 4:44 pm    Post subject: 14 If my programs are correct, I get the answers of: a: 512410035775437975643/10745521445375105130000 b: 288/8075
ChickenMarengo
Daedalian Member

 Posted: Mon Jun 22, 2009 7:14 pm    Post subject: 15 Yes, that's what I get.
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