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Who's Good at Odds and Probabilities ?

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posted on Feb, 28 2016 @ 10:11 AM
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Lets say you buy 1 ticket for a raffle with 1 million tickets sold.
This raffle has only one prize, no 2nd or 3rd, etc.
Your odds of winning are 1 in 1 million.

What are your odds if:
You buy 1 ticket for 100 identical raffles.
Is it truly 1 in 100,000 to win any prize ?



posted on Feb, 28 2016 @ 10:12 AM
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a reply to: samkent

How did you come up with 1 in 100,000?



posted on Feb, 28 2016 @ 10:24 AM
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a reply to: samkent

well assuming you got 1:100000 from the amount of players (the variable) you tell me?

or do you want me to ask?



posted on Feb, 28 2016 @ 10:25 AM
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Oppps! my typo.
I meant 1 in 10,000.



posted on Feb, 28 2016 @ 10:25 AM
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originally posted by: samkent
Your odds of winning are 1 in 1 million.

What are your odds if:
You buy 1 ticket for 100 identical raffles.
Is it truly 1 in 100,000 to win any prize ?


1/1,000,000 x 100 = 1/10,000, if all of the drawings are completely independent and you can't re-use the ticket for another drawing. Real drawings aren't always that way, sometimes you can re-use the ticket to enter a second drawing.

Non-winning tickets can be used again.



posted on Feb, 28 2016 @ 10:31 AM
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a reply to: samkent

Odds are set: it's a 1 in a million chance you'll win. There is only one outcome.

Probabilities change: 1 outcome with an increased probability of winning based on increasing your number of raffle entrants. Expected outcome (payout) is based on number tickets purchased - 10 tickets purchased would make your expected outcome/payout -$0.99999 as opposed to -$0.999999. If you bought all 1 million raffle tickets your expected payout would be $1.

You can increase your probability of winning but never change the odds.

ETA: If, as mentioned above, tickets are reusable for a 2nd drawing nothing changes, given all tickets are reusable.



edit on 28-2-2016 by BeefNoMeat because: included caveat based on above comment



posted on Feb, 28 2016 @ 11:16 AM
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Probabilities work in multiples sometimes.
Try rolling a die and getting a three. Your chance is one in six or 16.7%
Try doing it twice in a row, one in 36 or 2.8%
Three in a row, one in two hundred sixteen or 0.5%

Your scenario is a bit different,

1 in a million for the first. .0001%
2 in a million for the second. .0002%
3 in a million for the third. .0003%
Etc etc etc.

100 in a million for the last. .001%



posted on Feb, 28 2016 @ 11:20 AM
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a reply to: samkent

Think of it like roulette.

You can make the same inside bet every time, your chances are still 1 in 38.



posted on Feb, 28 2016 @ 12:36 PM
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a reply to: samkent

your odds are 100 in 100million

Look up "the gamblers fallacy" for an explanation



posted on Feb, 28 2016 @ 12:38 PM
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originally posted by: samkent
Oppps! my typo.
I meant 1 in 10,000.


in that case your odds with 100 tickets in 100 identical lotteries with each having a 1 in 10,000 chance is 100 in 1,000,000. Or, put more plainly: 1 in 10,000



posted on Feb, 28 2016 @ 07:41 PM
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Odds are your real dad is down at the county jail doing favors for meth because his son didn't pay attention in 5th grade math. It is a 7:1 Meth:Math ratio.



posted on Feb, 28 2016 @ 07:48 PM
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originally posted by: samkent
Lets say you buy 1 ticket for a raffle with 1 million tickets sold.
This raffle has only one prize, no 2nd or 3rd, etc.
Your odds of winning are 1 in 1 million.

What are your odds if:
You buy 1 ticket for 100 identical raffles.
Is it truly 1 in 100,000 to win any prize ?


Not quite but close.

This is a fairly typical problem I've used to interview candidates for an analytic job.

If p is the probability of winning one event, the exact probability of winning 1 or more times out of N independent tries is

1 - (1-p)^N, or the converse of the probability of losing all attempts.

Step 2: expand in one and then two terms in O(p) for p close to 0.


edit on 28-2-2016 by mbkennel because: (no reason given)

edit on 28-2-2016 by mbkennel because: (no reason given)



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