What are e odds of winning the lotto with 6 numbers all up to 42. Workings please?
153297
There are 42!/((42-6)!6!) ways of pulling 6 balls out of a set of 42. That's (42)(41)(40)(39)(38)(37)/(6)(5)(4)(3)(2)(1)\nSimlifying, that's (7)(41)(13)(38)(37) = 5245786.\n\n1 combination is the winning combination, so 1 in 5245786.
1009585
If they are asking for money for full details then it is a con. Otherwise it is just another hairbrained theory.\n\nThe odds for every draw remain approx 14 miilion to 1 for each line you purchase.\n\nit is very easy to baffle people with numbers, particularly when dealing with probabilities because what frequently seems common sense is actually wrong. \n\nFor example how many people do you need at a gathering to have a 50/50 chance that two of them will share the same birthday. Common sense would state to many that you would need about 183. However you only actually need 23.\n\nIf someone is quoting odds at you and asking for money then leave quickly.
1009251
Ahh!  True, 6 consecutive flips of a coin in which each lands heads has a (1/2)^6 probability of happening.  But what are the odds that the seventh flip will be heads, too?
370301
The best way of covering all 15 numbers evenly is simply to start by selecting 1,2,3,4,5,6 then 7,8,9,10,11,12 then 13,14,15,1,2,3 etc. That way, you're sure to hit each number an equal amount of times in the long run. If you're saying "but these combinations are totally unlikely to be selected!" then yes, you are right. But they are precisely just as likely as any other of the combinations. So, in other words, don't play the lottery.
430637
First choose which number it is.  There are 6 numbers, so there are C(6,1) = 6 choices.\nThe probability of that number is 1/6\nSince there are three\n(1/6)(1/6)(1/6) = 1/216\nNow multiply this times 6\n6/216 = 1/36
251181
at least a gazillion....ps did you win with my winning lotto #'s????
731770
You have 5 dice, with 6 numbers each, for a total of 6^5 possible rolls on your first roll.\n\nThere are 6 ways you can get a Yahtzee:\n1-1-1-1-1\n2-2-2-2-2\n3-3-3-3-3\n4-4-4-4-4\n5-5-5-5-5\n6-6-6-6-6\n\nTherefore your chances of getting a Yahtzee on your first roll is 6/6^5 = 6/7776 = 1/1296 ≈ 0.00077\n\n~ ~ ~\n\nIf you already have a 5-5-5, you only roll two dice -- with two dice you have 6^2 = 36 possible outcomes.\n\nP(yahtzee on next roll) = P(5-5 on next roll) = 1/36\n\nP(full house on next roll) = P(1-1 or 2-2 or 3-3 or 4-4 or 6-6 on next roll) = 5/36\n\n{ notice I left out 5-5, but technically 5-5-5-5-5 is a full house too, in addition to a yahtzee, so depending on how your teacher interprets "full house" the answer to this might be 6/36 rather than 5/36 }\n\nP(4 of kind) = P(5-1 or 5-2 or 5-3 or 5-4 or 5-6) = 5/36\n\n~ ~ ~\n\nYou have 2,3,4,5 and are rolling one die -- you have 6 possible outcomes for rolling that one die\n\nP(large straight) = P(1 or 6) = 2/6 = 1/3\nP(small straight) = 100% since you already have a small straight with your 2-3-4-5\n\n~ ~ ~\n\nYou have 3-4-5 and are rolling 2 dice -- you have 36 possible rolls with those 2 dice\n\nP(full house) = 0, since a full house is 3 of one number and 2 of another, and you already have three different numbers showing!\n\nP(3 of a kind) = P(3-3 or 4-4 or 5-5) = 3/36 = 1/12\n\nP(large straight) = P(1-2 or 2-6) = 2/36 = 1/18\n\n~ ~ ~ \n\nI hope you understood how I approached and solved these, so that you can do them on your own when it's time to take your test on this material.  Good luck!\n\n~ ♥ ~
514377
There are 20! possible orderings for picking the numbers from the hat. Only one has them in order, so 1/20!.\n\nAnother way to look at it is that when there are 20 numbers in the hat, your chances are 1/20 of picking the 1. Then they are 1/19 of picking the 2, then 1/18 of picking the 3 etc.\n\nSo the chances are (1/20)(1/19)(1/18)...(1/2)(1/1)\n\nIf you threw the numbers back in each time, the probablity of pulling them out in order would be 1/(20^20).
754620
you can only get 6 - 36.you can only get 6 one way, and 36 one way.  you get 7 six ways, and 35 six ways.  figure out all the ways to get each number, including 19, and add them all up.  the odds of getting 19 is the number of ways to get 19 over the total number of outcomes
427726
There is only one other person who has a share of 220 million dollars with the punter you describe, but this other person, following a hunch, bought 219 identical tickets, with the same winning numbers on them, and scoops 219/220ths of the prize money as a result.\n\nThe mistake is in assuming there is one other winning ticket, when you are told there is one other winner. You choose what your 6 numbers from 49 will be, and there is nothing to prevent you choosing the same set of numbers more than once.
585757
The odds are based on the number of numbers you can select from.  For example, if the winning number is five numbers out of a total of 70, you have a chance of 1 in (70 x 69 x 68 x 67 x 66), since this is probability without replacement.  So the calculated odds are 1 in 1,452,361,680 in this example.\n\nAnd as for the answer that made fun of your use of commas, feel free to point out to him that his use of apostrophes was erroneous as well.  "comma's" was the word he used, and it really should have been "commas." :)
97799
i don't believe the person above me got the second part right.  he is right about the beginning - the first probability is 2/5:\n\npossible numbers to draw: 2 4 6 11 13 15\n\nso you can choose any of these six out of the fifteen given (6/15 = 2/5).  the next number you draw must be also a "winning" number.  BUT, you have to have an even number less than 7 AND an odd number greater than 10.  this gives us two scenarios:\n\nIF the first number picked was even and <7, then the possible numbers you can pick from are ONLY: 11 13 15, since picking another even number <7 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\n\nIF the first number picked was odd and >10, then the possible numbers you can pick from are ONLY: 2 4 6, since picking another odd number >10 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\nSO, finally, suppose you go with the first scenario.  the probability would be: (6/15)*(3/14) = 3/35  going with the second scenario will give you the same thing, 3/35.  thus adding these two, since you can have either scenario occur and still meet your requirements, gives you: 3/35+3/35 = 6/35.\n\nAnd that would be the odds: 6/35.
1133393
If the odds are 5-1, and you win, you get back $5 for every dollar bet,  plus your original dollar back, which is $6.\n\n$5 X 5 + 5 (original bet) = $30.00.\n\nIn the question as it stands, a 5-1 with a payoff of $14.40 is not a 5-1 odds, but 6.20 -1 odds.. A two dollar bet returning $14.40 is 6.20 x 1, and for a return on a two dollar bet would render 6.20  x 1 (odds) and the return of you original dollar, or $7.20 x 2 dollar bet would come to $14.40.\n\nSo, a $14.40 return on a $5 winning ticket would be $6.20 x 5 + your original $5 or $36.00.
894091
None. Two odds make an even
985314
its 1/36 chance of any two particular numbers coming up on each roll of the dice.  (1/6 is for one die)
745507
diidy, how could the odds get worse when you buy more tickets?\n\nI think there are actually two answers, depending on whether each ticket is independent of the others or not.\n\nIf each ticket is independent, and there's no guarantee as to how many tickets there are or how many winners there will be, then think of it this way.  The odds of any ticket NOT winning are 299:300.  Therefore the odds of NO ticket winning are 299^n:300^n, where n is the number of tickets.  The odds of at least one ticket winning are 1 - the odds of no ticket winning, or 1 - 299^n:300^n, or (300^n - 299^n):300^n.\n\nIf there are 300 tickets of which 1 is a winner, then the odds of winning are n:300.
239442
You're picking the wrong numbers. listen carefully, i'll tell you the winning numbers for last week 4-23-26-31-40-51.\nNo just joking, its a good idea to have three of the numbers to be even numbers and three to be odd numbers, these hit almost all the time, other than that it's all luck.