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Lottery spread betting

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shyted | 15:42 Mon 15th Dec 2003 | How it Works
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What would you make Saturdays bonus ball multiplied by Wednesdays bonus ball ? There are two schools of thought .. 49 squared , divided by 2 = 1200.5 or 49 divide 2 , squared = 600.25 ?? I believe someone else told me that 70% of the possible outcomes would fall under 470 or something like that. Can anyone out there in mathland shed some light on this ?
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If you know how to translate standard odds into spread-betting format - I can't! - here are the standard figures. Basically, you seem to be asking: "What are the odds re guessing beforehand the bonus-ball number correctly over two separate lotteries?"

On the first lottery occasion - before the selection even starts, that is - there are 49 balls in the machine, any one of which might be that night's bonus-ball. The odds against any given one - let's say you've guessed No 15 - of these coming up are, therefore, 48 to 1 against...ie only one is 'right' (No 15) and the other 48 are 'wrong'.

The same 48 to 1 against odds apply to the second lottery occasion, when you chose perhaps No 37. The odds against getting both 15 and 37 in that sequence over the two occasions are calculated by multiplying the odds for each occasion...ie 48 X 48, which = 2304 to 1 against. As I said, I don't know how to relate that to 'spread' betting, but I hope you can.

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