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Title: Question Post by: the-oneeye on January 07, 2011, 01:47:55 AM Sorry to appear dumb but could someone please give me the odds of being dealt Aspades 7s
Title: Re: Question Post by: Robert HM on January 07, 2011, 01:50:54 AM I would guess 2,652/1 but I am always wrong on these odd thingies.
Title: Re: Question Post by: the-oneeye on January 07, 2011, 01:53:07 AM i was thing around 640/1 and i am also always wrong
Title: Re: Question Post by: gatso on January 07, 2011, 01:54:12 AM 1325 to 1
1/26 chance of getting either card first then a 1/51 chance of getting the second card=1 in 1326 robert, you've done the chances of getting them in that exact order Title: Re: Question Post by: the-oneeye on January 07, 2011, 01:58:55 AM many thanks im doing a lucky hand bonus at my live games and we are doing world series winning hands just to explain the question
Title: Re: Question Post by: JK on January 07, 2011, 02:10:05 AM 50/50 innit. You either get it, or you don't.
Seriously though, Gatso's right (on the basis he did the maths) Title: Re: Question Post by: Robert HM on January 07, 2011, 02:51:59 AM 1325 to 1 1/26 chance of getting either card first then a 1/51 chance of getting the second card=1 in 1326 robert, you've done the chances of getting them in that exact order Doh, I was wrong again. Title: Re: Question Post by: ggeseriesgg on January 07, 2011, 11:03:53 AM The chances are 50:50. You either get the hand, or you don't ;)
(just kidding by the way, it's not 50:50) Title: Re: Question Post by: cambridgealex on January 07, 2011, 01:28:49 PM The chances are 50:50. You either get the hand, or you don't ;) (just kidding by the way, it's not 50:50) nice fresh line. Title: Re: Question Post by: cambridgealex on January 07, 2011, 01:29:16 PM is it different for Ahrt 7h?
Title: Re: Question Post by: boldie on January 07, 2011, 01:30:11 PM is it different for Ahrt 7h? Depends if you're in a Labour council or Tory council. Title: Re: Question Post by: MC on January 07, 2011, 01:36:44 PM The chances are 50:50. You either get the hand, or you don't ;) (just kidding by the way, it's not 50:50) Never get tired of hearing this line [ ] Title: Re: Question Post by: boldie on January 07, 2011, 01:38:53 PM The chances are 50:50. You either get the hand, or you don't ;) (just kidding by the way, it's not 50:50) Never get tired of hearing this line [ ] [ ] never worked out where the box should go either :) Title: Re: Question Post by: gatso on January 07, 2011, 01:43:37 PM The chances are 50:50. You either get the hand, or you don't ;) (just kidding by the way, it's not 50:50) Never get tired of hearing this line [ ] [ ] never worked out where the box should go either :) [ ] whereas you sure don't seem to have figured out how to combine boxes and negatives Title: Re: Question Post by: boldie on January 07, 2011, 01:44:58 PM The chances are 50:50. You either get the hand, or you don't ;) (just kidding by the way, it's not 50:50) Never get tired of hearing this line [ ] [ ] never worked out where the box should go either :) [ ] whereas you sure don't seem to have figured out how to combine boxes and negatives DOH! WP. Title: Re: Question Post by: Cf on January 07, 2011, 01:45:49 PM is it different for Ahrt 7h? Prob about the same. Ad 7d is more common though. And i'd say you get Ac 7c the least. Title: Re: Question Post by: GreekStein on January 07, 2011, 01:46:47 PM The chances are 50:50. You either get the hand, or you don't ;) (just kidding by the way, it's not 50:50) Never get tired of hearing this line [ ] this!! Title: Re: Question Post by: MC on January 07, 2011, 01:50:30 PM lol, I think you'll find I'm an expert at using checkboxes, just thought I'd be a little creative this time!
Title: Re: Question Post by: bhoywonder on January 07, 2011, 03:40:53 PM 52 times 51...divided by 2...maybe?
Title: Re: Question Post by: Boba Fett on January 07, 2011, 04:16:49 PM While we are on the subject, whats the math for the odds of winning the Mint Glasgow jackpot (See thread in Live poker for full details) once you have won the tourney outright?
The name of every player in the tourney goes into a bag/box/hat whatever and the winner gets 1 pick for every 5 players that entered the tourney (50 players=10 picks, 100 players=20 picks etc) and if they pick their own name they win the jackpot. So based on 50 players, it would be 50/1 then 49/1 then 48/1 ... 40/1 What is the total? How much does it change for 75 players with 15 picks and 100 players with 20 picks? TY Title: Re: Question Post by: kinboshi on January 07, 2011, 04:50:25 PM The chances are 50:50. You either get the hand, or you don't ;) (just kidding by the way, it's not 50:50) Never get tired of hearing this line [ ] [ ] never worked out where the box should go either :) [ ] whereas you sure don't seem to have figured out how to combine boxes and negatives I don't understand. Boldie correctly placed the box next to a sentence where he said he didn't know where to place boxes. So he was correct, wasn't he? Title: Re: Question Post by: Longines on January 07, 2011, 04:50:58 PM It's a fairly narrow range if I understand it correctly; best case you're 5/1 to win - you get to pick out exactly 20% of the players if there is a 'round' number of entrants i.e. 50, 55, 75, 90 etc.
Worst case you're 5.444/1 to win - you get to pick 9 from 49 players. Title: Re: Question Post by: boldie on January 07, 2011, 04:56:39 PM The chances are 50:50. You either get the hand, or you don't ;) (just kidding by the way, it's not 50:50) Never get tired of hearing this line [ ] [ ] never worked out where the box should go either :) [ ] whereas you sure don't seem to have figured out how to combine boxes and negatives I don't understand. Boldie correctly placed the box next to a sentence where he said he didn't know where to place boxes. So he was correct, wasn't he? YEAH!....erm....please don't confuse me anymore..my head hurts. Title: Re: Question Post by: gatso on January 07, 2011, 04:59:18 PM 50 player, 75 players, 100 players, all are 4/1 shots unless I'm completely misunderstanding what you've described
Title: Re: Question Post by: gatso on January 07, 2011, 05:04:37 PM I don't understand. Boldie correctly placed the box next to a sentence where he said he didn't know where to place boxes. So he was correct, wasn't he? no, he was suggesting that mc doesn't know where to place boxes which would be fine if he hadn't used a double negative therefore saying that he did know Title: Re: Question Post by: cambridgealex on January 07, 2011, 07:03:48 PM i'm the same. i don't often not understand check box usage.
[ ] Title: Re: Question Post by: Boba Fett on January 07, 2011, 07:07:31 PM 50 player, 75 players, 100 players, all are 4/1 shots unless I'm completely misunderstanding what you've described It was a discussion between 2 other people. 1 said its 4-5/1 with 1 pick for every 5 players but the other said it isnt as each pick is gonna be entrants/1, entrants-1/1 etc, 1st person counters that after all the picks it will work out around 5/1. I wasnt sure of the math so was interested in the exact math Title: Re: Question Post by: gatso on January 07, 2011, 07:28:05 PM It's a fairly narrow range if I understand it correctly; best case you're 5/1 to win - you get to pick out exactly 20% of the players if there is a 'round' number of entrants i.e. 50, 55, 75, 90 etc. Worst case you're 5.444/1 to win - you get to pick 9 from 49 players. 1 in 5 is not the same as 5/1. you're logic is correct, just your odds are wrong, you need to knock 1 off each of your numbers worst case (not counting cases with<5 entrants) is 8/1 when you're picking 1 from 9 but mostly your going to be in the range between 4/1 and 9/2 Title: Re: Question Post by: rex008 on January 07, 2011, 08:34:55 PM Does buying two lottery tickets (for the same draw) double your chance of winning the jackpot?
Title: Re: Question Post by: gatso on January 07, 2011, 08:57:57 PM Does buying two lottery tickets (for the same draw) double your chance of winning the jackpot? not if you pick the same numbers for both Title: Re: Question Post by: doubleup on January 07, 2011, 10:23:48 PM 50 player, 75 players, 100 players, all are 4/1 shots unless I'm completely misunderstanding what you've described It was a discussion between 2 other people. 1 said its 4-5/1 with 1 pick for every 5 players but the other said it isnt as each pick is gonna be entrants/1, entrants-1/1 etc, 1st person counters that after all the picks it will work out around 5/1. I wasnt sure of the math so was interested in the exact math one pick for every 6 entrants, so 5-1. The simple way to think of it is you are splitting the tickets into two pools one being 5 times bigger than the other. Presumably the debate was about the actual draw - ie the first draw is eg 59-1, the second 58-1 etc. However you have to take into account that the second draw only occurs when the first one fails. So the first success probability is .016667 (59/1), the second draw is .016949 (58/1) but this only occurs when the first draw fails .983333 x .016949 =.016667. So if you run through all those calculations for the 10 goes and add then together you still get 5-1. Title: Re: Question Post by: gatso on January 07, 2011, 10:34:13 PM one pick for every 6 entrants, so 5-1 have the rules changed? you're answering a different question to me by the looks of it Title: Re: Question Post by: doubleup on January 07, 2011, 10:47:06 PM one pick for every 6 entrants, so 5-1 have the rules changed? you're answering a different question to me by the looks of it from the relevant thread At the conclusion of the tournament, the outright winner will draw one name for every six “£10 Jackpot Add-ons” purchased. just clarifying what the question should have been..... Title: Re: Question Post by: rex008 on January 07, 2011, 11:51:41 PM Does buying two lottery tickets (for the same draw) double your chance of winning the jackpot? not if you pick the same numbers for both Sorry. It was a very tongue in cheek q. This thread reminded me of this absolute epic (http://www.avforums.com/forums/general-chat/1286973-winning-lottery-what-chances.html) on another forum I frequent. That epic then had 2 follow up threads that got nearly as big. Very scary indeed. Title: Re: Question Post by: gatso on January 08, 2011, 12:14:34 AM Does buying two lottery tickets (for the same draw) double your chance of winning the jackpot? not if you pick the same numbers for both Sorry. It was a very tongue in cheek q. This thread reminded me of this absolute epic (http://www.avforums.com/forums/general-chat/1286973-winning-lottery-what-chances.html) on another forum I frequent. That epic then had 2 follow up threads that got nearly as big. Very scary indeed. ldo, that's why it didn't get a serious answer. I've got to post 5 of the thread you linked to. it's a real jawdropper. may have to read the rest now Quote First ticket buys you are a 1 in 14,000,000 chance second ticket (assuming difference in ticket) buys you a 1 in 13,999,999 chance 1 in 13,999,999 a long long long long long way off 1 in seven million (next ticket buys you a a 1 in 13,999,998 chance) the difference between 14,000,00 and 13,999,999 is not that great at all... Title: Re: Question Post by: gatso on January 08, 2011, 12:30:04 AM wow
Quote Join Date: Apr 2007 Location: Wakefied Posts: 460 Thanks: Gave 40, Got 75 I've invented two new games. One is you are given a 100 sided die and need to roll a 1 to win. That's a 1:100 shot. In the other game you're given a 50 sided die and the object to win is to again throw a 1. That's a 1:50 shot You're twice as likely to win at the 1:50 shot, than the 1:100 shot. In the lottery, you have a 1:14,000,000 (aporx) chance of winning. To be twice as likely to win (double your chances), you'd need to lower the odds by half... In this case, that's lowing them by 7 mil (approx). To do this, you need to buy 7 million tickets, since 7 million is half of 14 million. All buying 2 tickets does is lowers your odds from 1:14,000,000 to 1 to 13,999,999. That in no way doubles your chances. Title: Re: Question Post by: gatso on January 08, 2011, 03:02:07 AM I can only read the thread a couple of pages at a time as it leaves me stunned. haven't yet dared to look at how long it goes on for
Quote But the problem of a 2nd ticket is it adds complication and the odds no longer seem so easy... 1 in 14000000 is easy enough but your odds of winning with 2 tickets do not equal 1 in 7000000, they are 2 in 14000000 Title: Re: Question Post by: rex008 on January 08, 2011, 11:18:51 AM I can only read the thread a couple of pages at a time as it leaves me stunned. haven't yet dared to look at how long it goes on for Quote But the problem of a 2nd ticket is it adds complication and the odds no longer seem so easy... 1 in 14000000 is easy enough but your odds of winning with 2 tickets do not equal 1 in 7000000, they are 2 in 14000000 Believe me, to anyone who even has the vaguest notion of odds and probability, it's incredibly scary. It got to 1000 posts, they closed it and opened another (improves forum performance, apparently), and that got to about 500. Then there was another 2 spin-off threads. There were still people seriously arguing that if you bought all but one possible combination, you still only had a 50/50 chance of winning the jackpot! ;gobsmacked; Title: Re: Question Post by: gatso on January 08, 2011, 11:47:23 AM There were still people seriously arguing that if you bought all but one possible combination, you still only had a 50/50 chance of winning the jackpot! I haven't yet got to anyone claiming that but was expecting it as it's the logical conclusion to the flawed logic that many are putting forward Title: Re: Question Post by: doubleup on January 08, 2011, 01:17:56 PM There were still people seriously arguing that if you bought all but one possible combination, you still only had a 50/50 chance of winning the jackpot! sadly ppl are only this emphatic in defending their ignorance on the internet, I have yet to meet someone irl who would allow me to draw two cards from a standard pack and pay me 11-1 if I drew an ace. Title: Re: Question Post by: Cf on January 08, 2011, 02:51:31 PM It continues to amaze me how many people don't understand basic probability. Even worse is people interested in gambling.
Title: Re: Question Post by: Cf on January 08, 2011, 02:52:44 PM Although I will add that buying a second lottery ticket doesn't necessarily double your chances of winning :)
Title: Re: Question Post by: gatso on January 08, 2011, 02:56:47 PM I'm gonna do the lottery tonight but I'm not gonna check the numbers for a couple of months. once all the other tickets have been checked mine is much more likely to be a winnah
Title: Re: Question Post by: Cf on January 08, 2011, 03:14:12 PM I'm gonna do the lottery tonight but I'm not gonna check the numbers for a couple of months. once all the other tickets have been checked mine is much more likely to be a winnah Make sure you use any numbers that are important to you such as birthdays etc. You're more likely to win with those. Then buy a second ticket to improve the odds from 1:1400000 to 1:1399999. Title: Re: Question Post by: Cf on January 08, 2011, 03:19:07 PM I've made the mistake of opening that thread up... quite scary stuff.
Title: Re: Question Post by: gatso on January 08, 2011, 03:23:46 PM I've made it up to page 21. it takes a while, you have to take breaks to take in what you've read. just seen there's 13 more pages to go
Title: Re: Question Post by: Cf on January 08, 2011, 03:29:03 PM Buying 2 tickets doesn't double your chances of winning.
I can't see how people believe it does... This is ridiculous. Title: Re: Question Post by: Cf on January 08, 2011, 03:30:09 PM lol same guy quoting someone else:
Quote:Originally Posted by zmoosa1 Voted YES.. one ticket = one chance in 14 million two tickets = two chances in 14 million. two is double of one, so the answer is yes. It's not that simple Perhaps you wish it was, but it really isnt. Title: Re: Question Post by: Cf on January 08, 2011, 03:34:58 PM In fact cos i've got nothing better to do i'm gonna quote every post by the idiot who doesn't get it.
If you have 1 ticket you have 1 covered 1 possible combination of 14 million combinations. If you pick another possibble combination by buying another ticket you will be taking combination 1 out of 13999999. You can't half the 14000000 because having one more ticket does not cover the 7000000 combinations you are removing by balancing it. Title: Re: Question Post by: Cf on January 08, 2011, 03:36:06 PM Quote:Originally Posted by imightbewrong
Iccz - Your first ticket has a chance of 1/14000000. Your second ticket also has a chance of 1/14000000 - as I think GD said, imagine these two tickets were bought by different people (and are different). The chance of one of these two people winning is 1/14M + 1/14M = 2/14M = 1/7M. Now if one person had bought these two tickets, their chance of a win would be 1/7M. Now if four people had bought tickets, the chance of one of them winning is double that - 2/7M. You can keep on doubling this - 7M people would have a 1/2 chance of winning between them. Your 2nd ticket has a chance of 1 in 14000000 on it's own, but as you have already picked one comination that doesn't need to be factored in to the 2nd ticket (unless you're stupid enough to go for duplicate tickets), so ticket two assuming you dont want to duplicate numbers will be 1 in 13999999. The chances of winning are based on combinations not on probability. There are 14000000 combinations you pick 2 of them you have a 2 from 14000000 chance of winning, it is not a ratio - you can't balance because you can't just scrap 7000000 combinations. It is not a ratio. I'm going to ignore the example because it's easy enough to talk about the tickets and I'll be repeating myself to talk about it. Title: Re: Question Post by: Cf on January 08, 2011, 03:36:28 PM lol @ saying this has nothing to do with probability!!
Title: Re: Question Post by: Cf on January 08, 2011, 03:37:01 PM Quote:Originally Posted by zmoosa1
You're dividing it the wrong way. 1 ticket is 1/14million which is 0.0000071 2 tickets is 2/14000000 which is 0.0000143 Your argument suggests the 2 ticket calculation is 14000000/2 which is 7000000. It is not a ratio. You have a 2 in 14000000 chance, you can't reduce the number of possible ways to win the lottery as it is in it's lowest form. What you do is you take your 1 in 14000000 chance and your 1 in 13999999 chance. You add them together: 2 in 27999999. Your odds are 1 in 13999999.5 Title: Re: Question Post by: Cf on January 08, 2011, 03:38:12 PM I'll stop now as I think we get the picture. I do like the conclusion we can draw from this though. I just wait until 1399999 tickets are sold before I buy my ticket. That ticket will have a 1 in 1 chance of winning! I'm gonna be rich!
Title: Re: Question Post by: gatso on January 08, 2011, 03:41:12 PM I like the argument that if you have a 10 sided die, choose 2 numbers and then roll it once you don't have a 1 in 5 chance of one of your numbers coming up because things don't act like that in the real worls
Title: Re: Question Post by: Cf on January 08, 2011, 03:45:35 PM He has to be winding everyone up. Surely?
Quote:Originally Posted by FruitBat Two chances at 14 million combinations is 1 in 7 million. That's how probability works. It is not, you are dealing with ratios, you can't use a ratio for this Title: Re: Question Post by: Cf on January 08, 2011, 03:49:42 PM Yup. Has to be a wind up.
The same way tossing a coin 10 times will not give you heads 50% of the time and tails 50% of the time. Title: Re: Question Post by: Cf on January 08, 2011, 03:52:17 PM This is just getting too much now. All of these posts are by the same guy...
But the problem of a 2nd ticket is it adds complication and the odds no longer seem so easy... 1 in 14000000 is easy enough but your odds of winning with 2 tickets do not equal 1 in 7000000, they are 2 in 14000000. Title: Re: Question Post by: Cf on January 08, 2011, 03:56:01 PM Enough quoting now. Question for gatso.
I take a random coin from my pocket and flip it. It lands on heads. I flip it again. Which side if any is it more likely to land on? Title: Re: Question Post by: boldie on January 08, 2011, 04:14:17 PM depends on the coin and whether you heated it or not
Title: Re: Question Post by: gatso on January 08, 2011, 04:16:01 PM heads innit. all I know about it is it's got at least one head, maybe it's got 2, maybe there's a tail on the other side, I've got no way of knowing so have to plump for the only bit of knowledge I have
I'm off to win the lotteryz now with my new skills Title: Re: Question Post by: Cf on January 08, 2011, 04:19:17 PM lol, good answer. but no. i can assure you the other side is a tails :)
Title: Re: Question Post by: Cf on January 08, 2011, 04:51:49 PM oh ffs. i got to the end of page 34 and found it continues in another 17 page thread.
Title: Re: Question Post by: Cf on January 08, 2011, 04:55:32 PM Not reading anymore of it but I did find this absolute gem. They emailed camelot asking about it. This was the response...
Originally Posted by Camelot Thank you for your email dated 15 July 2010. As stated the odds of matching all six numbers are 1 in 13,983,816. These odds apply for each individual ticket purchased. Should you purchase two tickets with different sets of numbers, this does not halve your chances of winning the lottery, it reduces the probability by 1. The number of tickets you purchase does not act as a denominator against the total odds, it reduces the total probability by that amount. For example, if you purchased 100 tickets, the odds of one of those tickets winning would not become 1 in 139,838 it becomes 1:13,983,716. With a dice role, any particular number can be rolled again. With the Lotto, there is a maximum number of combinations of 13,983,816 and once one combination is removed by buying a ticket, that combination cannot be produced again therefore it is minus 1. Therefore odds of one ticket = 1:13,983,816 " odds of two tickets= 1:13,983,815 By purchasing two tickets you do of course have two chances of winning although i hope you can see the odds are not halved. Title: Re: Question Post by: CHIPPYMAN on January 09, 2011, 12:53:40 PM rotflmfao
Title: Re: Question Post by: dagarbisa on January 10, 2011, 12:06:15 PM I would guess 2,652/1 but I am always wrong on these odd thingies. yes, You have mistaken but by the time you will learn it. This is pure mathematics(Probability) it would be 52C2 x 51C1 as in 1326. 1/26 of getting 1 or 1/51 of getting second card in 1326. Title: Re: Question Post by: kinboshi on January 10, 2011, 12:37:51 PM Just caught up on this thread.
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