How to work out the odds video

[Frank]

Well I'm sure this will be nothing new to many members here, but what is interesting is the 'keep it simple' way of working out the odds without any fancy factorial notation. Also he works out the odds of NOT matching any numbers on your ticket, which many people would not know how to calculate. The text description along the bottom of the video needs to be followed as he goes through it. Its for 6/49 but the method can be adapted for any lottery.

http://www.youtube.com/watch?v=U7f8j3mVMbc&feature=channel&list=UL

I wouldn't advise it, but if you are tempted to read the viewer comments below the video, it illustrates how some people who don't understand maths prefer to deny the truth and accuse him of 'cheating' or faking the video.

 

[Icewynd]

Entertaining and informative video on the basics of probability as applied to the lotto. Well worth the time. I found the comments to be quite funny, in an appalling kind of way.

 

[Icewynd]

Here's another good description of how to calculate the odds of 2 sequential numbers. Its from a much longer and wider-ranging article that is worth a read. I especially enjoyed the story of how a computer programmer gamed Keno at a casino.

http://thelongrunblog.wordpress.com/2008/09/05/lottery-scams-can-you-game-these-games-of-chance/

Suppose the first number drawn is 12. We need a 11 or 13 as the second number drawn to create a two consecutive number result. The odds that the second number is going to be 11 or 13 is roughly 4% (2/48 and not 2/49 as there can be no repeating numbers, so there is now one less possible number). Let’s say the second number drawn is 4. That’s not 11 or 13 and that doesn’t much help us. But no worries, mate. We now pin our hopes on the third number, which can be 3, 5, 11, or 13. That’s roughly a 8.5% (4/47) chance. Suppose our third number comes up 40. If the fourth number is 3, 5, 11, 13, 39, or 41 it will satisfy our two consecutive number condition. That’s a 13% (6/46) chance. Our fifth number has 8 out of 45 numbers to satisfy the condition (about 18%). Our 6th and final number has 10 out of 44 possible numbers that will create a two consecutive number combination. That’s roughly a 23% chance. If I remember Stats 201 correctly, all those percentages get added to calculate the chance any position will satisfy the condition. So, pure chance assures roughly 66% of winning numbers will, over time, have two consecutive numbers. I’m not really taking into account that 2% of the results will have a 49. If 49 is drawn then only 48 satisfies the condition. But I don’t think this is going to radically alter the 66% figure.

 

[Frank]

This would be a great description if only it were correct. Had the author taken the trouble to analyse any 6/49 lottery with a significant number of draws (more than 1000) and counted the incidence of consecutives he would realise that this figure is considerably out. The correct figure is p=0.4951 or 49.51%.

I think the more accurate way to do this longhand is to calculate the probabilities of there NOT being a consecutive in the next drawing.

So to use the same assumptions and figures as described, after the first ball is drawn there are 48 left to chose from. Using the same logic, The probability of the next ball NOT being consecutive is 46/48. Probability =0.95833.
After ball 2 is drawn:- 47 left, 43 balls cannot be consecutive, Probability 43/47 =0.9149.
After ball 3 is drawn:- 46 left, 40 balls cannot be consecutive, Probability 40/46 =0.8695.
After ball 4 is drawn:- 45 left, 38 balls cannot be consecutive, Probability 38/45 =0.8444.
After ball 5 is drawn:- 44 left, 34 balls cannot be consecutive, Probability 34/44 =0.7727.

the overall probability of the balls NOT containing a consecutive is found by multiplying the above 5 ball by ball probabilities together. This comes to 0.4975.
So using this method the probability that the result will contain at least 1 consecutive is 1- 0.4975 = 0.5025 or 50.25%.

Note that this still suffers from the assumption that numbers 01 and 49 have 2 consecutive partners when in fact they each have only one. This would bring down that 50.25% to 49.51%.

I know that:

when drawing r balls without replacement from balls numbered 1, 2, . . ., n the chance of least one consecutive pair of numbers is 1 - C(n-r+1,r)/C(n,r).


For a 6/49 this equates to 1-C(44,6)/C(49,6) = 0.495198449 . or 49.519%



This agrees reasonably well with my figures for the UK lotto where in 1730 draws 47.91% of draws had at least 1 consecutive in the result.
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