Math Problem

[kermit911]

If your odds are 1 in 146,107,962 with one ticket. What are my odds if I have 100 tickets?

Thanx

[Loosehead]

100 in 146,107,962 or 1 in 1,461,079.62

[gen2]

100 in 146,107,962 or 1 in 1,461,079.62

[gen2]

snap!

[shammydodger]

Note that if there are 146,107,962 possible outcomes and you have 1 ticket containing 1 possible outcome, than the odds against you winning are 146,107,961 to 1.

This highlights the significant difference between odds and chances.

It might be easier to consider on a smaller scale. If there were 3 blue socks and 1 red sock in a drawer and you were blindfolded, you would have a 1 in 4 chance of pulling out the red sock with your first attempt. However, the odds against you doing so are 3/1 (not 4/1.)

[kermit911]

How did you get that answer? What was the equation u used?

Thanx

[BigMac]

Loosehead and gen2 are right if the ticket is for a raffle, say, with a finite number of tickets.

In the UK National Lottery the odds of winning with one ticket are approx. 14m to one, but they remain at 14m:1 even if you have two tickets

[shammydodger]

Surely that can't be right Mac?

There is a finite number of combinations in the National Lottery. If you had 2 tickets and if they had different combinations, then your chances of hitting the 6 numbers drawn would be twice the chance than if you only had 1 ticket.

Or am I not eating enough brain food?

[Loosehead]

No big mac if you have 2 (different) tickets for the lottery you have 2 chances in 14m thus 1 in 7m. Dunno what you are thinking of.

[mibn2cweus]

Your odds of breaking even in the long run are even less since a large part of the money collected is used to operate, advertise, promotion, etc. and to fund programs.

Even if you buy all the lottery tickets you loose, d�oh!

[THECORBYLOON]

There are 13,983,816 possible winning combinations in the Lottery. If you buy one ticket, there are 13,983,815 OTHER combinations that may appear so the odds are 13,983,615 to 1. If you buy two different tickets, you reduce the remaining possible combinations by two so the new odds are 13,983,614 to 1.

If Loosehead's theory of halfing the odds with each extra ticket were true, you need buy only 25 to be guaranteed a win which can not be correct.

[baileys]

what was the odds of Loosehead's and gen2's answers coming in like that!!! WOW!!!!!!!!!! The mind boggles -------------------------------boggle ------boggle -----------------------------

[Loosehead]

I think people we are confusing odds with probablity.

[JudgeJ]

Loosehead did not suggest that each extra ticket purchased halves the odds. As he says, we must be careful not to confuse odds with probability. Certainly buying the second ticket doubles your chances of winning from one in (approx) 14m to 2 in 14m (or 1 in 7m). Buying the third does not double your chances again but increases them to 3 in 14m (about 1 in 4.67m). To be absolutely certain of winning you must buy 13,983,816 tickets.

The odds against any one ticket winning obviously remain the same throughout. Your chances of winning increase as you buy more tickets (each with different number combinations, of course).

[Loosehead]

and it's an inverse geometric progression. if you by half of all the tickets you have a 50% chance of winning. That chance increses to 100% the closer you get to buying all the tickets.

[jimmer]

the easiest way of thinking about this problem, so you don't confuse yourself, is to convert the probability into a percentage: Buying 1 ticket gives you a chance of .00000714 percent chance of winning the jackpot; Buying 2 tickets gives .000014 percent chance etc. To have a 50 percent chance of winning ( a coin toss ) you need to spend 7 million quid. The equation used is: no of tickets bought / total available combinations * 100. This is much easier than trying to work out odds which are pretty meaningless at this scale of things.

jim

[Reader]

Have a look at http://www.urbanrim.org.uk/Lottery%201100.htm (and http://www.urbanrim.org.uk/drop%20chart.htm)