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davver | 10:12 Wed 05th Oct 2005 | Science
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Whould I be more likely to win a prize if I bought 3 tickets to one raffle or a ticket to each of three raffles? All raffles have the same number of tickets, say 100 for the sake of argument. Instinct tells me the probabilities are equal, but my calculations tell me that 3 tickets to one raffle edges it. But I don't trust either! What's the answer?
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3 tickets of the same raffle give a better probability to win one prize.
1 ticket of each raffle gives you a probability to win more than 1 prize.
So each method has its pros and cons, and in the end, their value is equal.

there is not enough information to work out the probabilities of winning a prize. For a start we would need to know how many prizes were in each raffle. But if we assume there is 1 prize of equal value in each raffle then each ticket has a 1 in 100 chance of winning regardless of the raffle. However, if you have three tickets for one raffle you actually have a 3 per cent chance of winning which is marginally better than the 1 percent chance you get by entering the three raffles.

 

jim

 

jim

assuming there is 1 prize in each raffle:

In the 1st case you have 3% chance to win 1 prize.

In the 2nd case you have 2.9701% chance to win at least 1 prize,
including :
2.9403% chance to win exacly 1 prize,
0.0297% chance to win exacly 2 prizes
and 0.0001% to win all 3 prizes.

Aren't there 100x100x100 different outcomes across all 3 raffles?
I'm sure other people know better but I'd have thought that winning all 3 raffles was one outcome from a possible 1,000,000 meaning that the probability of winning all 3 was
0.000001
Am I right or can someone explain it better?
Exactly the same chance of winning assuming there is only one prize in each raffle.
.000001 = .0001%

Space wrote:

2.9403% chance to win exactly 1 prize,
0.0297% chance to win exactly 2 prizes
and 0.0001% to win all 3 prizes.

stevie21 wrote:

Aren't there 100x100x100 different outcomes across all 3 raffles? [...] I'd have thought that winning all 3 raffles was one outcome from a possible 1,000,000 meaning that the probability of winning all 3 was
0.000001
Am I right or can someone explain it better?

That's what Space meant by 0.0001%
 

Surely if you buy 3 tickets for one raffle you have a 3 in 100 chance of winning.  If you buy 1 for each raffle you have a 3 in 300 chance of winning.  3% chance or 1% chance!  I know where my money would be.  But I'm probably wrong!
Sorry - I missed Space's percent sign.
Now.. onto the 2.9 etc : how do we get each of the other probabilities such as winning exactly 2 prizes?
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But a ticket in each of 3 raffles isn't 3/300, it's 1/100 + 1/100 + 1/100 = 3/100. However, this isn't quite right either.
to win one prize: 3(1/100 x 99/100 x 99/100)
to win two prizes: 3(1/100 x 1/100 x 99/100)
to win three prizes: 3(1/100 x 1/100 x 1/100)

Add all these up, and it doesn't quite make 3/100, which surprised me. I posted the question as I thought either my arithmetic or my methodology was wrong.

davver, what your instinct tells you is that both 2 methods offer the same value, which they do:
the average number of prizes you will win is exactly 0.03 for both methods.

I can't get my head round this one. What if there were 2 raffles and two tickets for each raffle. If you bought 2 tickets in one raffle, you would have a 100% chance of winning right? Easy one.

But if you bought one of each, you would have a 50% chance of winning each one. 50% + 50% = 100% so you would definately win one prize, but this is obviouly not true. Where have I gone wrong?

you don't just add probabilities of different events to find out the probability of one of them happening.
... I meant: independent events

that's right Space. Assuming there is only one prize in each raffle, only one ticket can win each event. buying three tickets for one raffle gives a .03 chance of winning. Buying 1 ticket per raffle gives a .01 chance of winning each raffle. After each raffle is drawn, the result of that raffle does not effect the outcome of the next raffle so you can not add them together to get a probability of .03. If you want to find out the probability of the man winning two raffles then you need to multiply the events. for two events this would give: (1/100)*(1/100) = 0.0001 or 1 in 10 000. but having won (or lost) the first event the chances of winning the second event would still be .01

 

jim

OK, but multiplying two individual probabilities wouldn't be right either would it? In my 2 raffles, 2 tickets in each example, buying one ticket in each gives you prob 1/2 of winning each raffle.

What is the probability of winning either raffle? Surely not 1/2 * 1/2 = 1/4 ??

But also obviously not 1/2 + 1/2 = 1.

How does this work?

(I should remember this - I've got a degree in maths lol)

Multiplying indicates the probability of both independent events happening. That is: winning both raffles. (1/4).

The probability of winning at least one is the sum of the probability of these cases :
winning the 1st and loosing the 2nd ( 1/2 * 1/2 = 1/4 ) ,
loosing the 1st and winning the 2nd ( 1/2 * 1/2 = 1/4 ),
winning both ( 1/2 * 1/2 = 1/4 )
The sum of those is ... 3/4

i don't think its 3/4. its bloody confusing i know, but i think it boils down to relative frequency. so, for every independent event you would have a 50 per cent chance of winning if you competed in 50 raffles and won 28 times you could use the relative frequency 28/50 = .56 or a 56 per cent chance of winning. as you competed in more raffles this would tend towards 50 per cent which reflects your actual chance in each raffle and confirms that the events are independent.

 

jim

 

http://www.stats.gla.ac.uk/steps/glossary/probability.ht ml#indepevents

A two ticket raffle same as flipping a coin, (provided it doesn't land on edge,etc. . .)

Six ticket raffle simular to tossing dice.

Give it a whirl.

jimmer, you just calculated the probability of winning in a raffle .
My last calculation (3/4) is the probability of winning at least 1 when entering 2 raffles.

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