General Betting
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Hi all,
I have a maths odds question.
If I was to buy 100 UK national Lottery tickets, and assuming that getting 3 numbers up is 56-1, what are the odds of me getting 3 numbers up 3 seperate times in those 100 tickets.
Basically I had an argument with a friend who thinks its around even money, and I personally think its much bigger.
I couldnt quite work out the probability exactly, but did come up with my own solution. Which gets me a figure of 5 (ie 4-1)
But my method is probably incorrect. If anyone can help me with the methodology of working out this problem, it would be much appreciated. Specially if you can show how to work out this problem.
(this whole thing started after me and my mate bought a 100 tix on the nationaly lottery re launch last wknd in hope of winning one of the 20k prizes, we ended up with just 2 lines of 3 numbers!)
I have a maths odds question.
If I was to buy 100 UK national Lottery tickets, and assuming that getting 3 numbers up is 56-1, what are the odds of me getting 3 numbers up 3 seperate times in those 100 tickets.
Basically I had an argument with a friend who thinks its around even money, and I personally think its much bigger.
I couldnt quite work out the probability exactly, but did come up with my own solution. Which gets me a figure of 5 (ie 4-1)
But my method is probably incorrect. If anyone can help me with the methodology of working out this problem, it would be much appreciated. Specially if you can show how to work out this problem.
(this whole thing started after me and my mate bought a 100 tix on the nationaly lottery re launch last wknd in hope of winning one of the 20k prizes, we ended up with just 2 lines of 3 numbers!)
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I think this is going to be
56!/(3!x53!) X (1/57)^3 X (56/57)^53 Can work it out on my phone. Also, try excel binomial dist N=56 P=1/57 R=3 |
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well if you can post the calculation answer tht would be awesome also!
But we are working out the odds of getting 3 in 100 goes. I say that as didnt see 100 in the calculation anywhere. Another way of thinking about this would be to get 3 56-1 winners in 100 bets on only 56-1 shots! Thanks in advance :) |
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ok, so 100 ticket and exactly 3 with 3 numbers (assuming 1/57 probability of 3 numbers) = 15.6%
100 ticket and at least 3 with 3 numbers = 25.7% |
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Hey, thanks for that, so its a 3-1 shot!
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It'd be different if it's "getting 3 numbers up 3 seperate times" or you mean "getting 3 numbers up AT LEAST 3 seperate times" as JLivermore suggests.
I quite believe "100 ticket and at least 3 with 3 numbers = 25.7%" might be right but exactly 3 numbers being 15.6%, that sounds instinctively a bit high? and .. just worked it out, JLivermore is correct, instincts were wrong. Read this to understand: http://en.wikipedia.org/wiki/Binomial_distribution and you can plug the values into pre-made calculators. (remember probabiity of 0.15 is probably what you'd express as 15%) |
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Actually that wiki page is a bit difficult. But that't theequation, as stated in shorthand by JL on his first post.
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yes he was correct!
I also found an online resource a binomial calculator, which will definitely come in handy! Strange thing is buying 28 tickets doesnt make us even money, 57/28 aprox .5, but according to binomial calc its around 39 % ! Once again thnx :) |
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http://stattrek.com/online-calculator/binomial.aspx
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maybe think about it with a dice instead.
You're trying to roll 6's. If you roll 3 times are you 50% to get at least one 6? Easiest way to see it is say what is the probability of rolling NO 6's (fail, fail, fail = 5/6x5/6x5/6) = 57.8% So chance of rolling at least 1 6 is 1-57.8% = 42.2% This is the same result you found with your 28 tickets. To try to explain it: If you were pulling balls out of a sack (and not putting them back in) (say 6 balls, numbered 1-6) you would indeed have a 50% chance of picking the 6 ball after 3 attempts But here the dice/tickets are independent (failing once does not improve subsequent chances of success), so going on an extended losing/failing run is easier. |
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you're welcome!
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isn't there a bit more to this problem? I don't think that the trials are independent as losing on one ticket increases the chances of winning on another ticket (assuming that the 100 tickets contain 2000 different combos of 3 -- is that possible anyway?)
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Yeah, my answer assumed random numbers on tickets.
You could improve your chances of getting a small number of winning tickets at the expense of the chance of getting a large number |