| | | # Calculating Raffle Odds
Asked Aug 12, 2009, 02:34 PM
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** 13 Answers** | I have been trying to figure this out, have done many, many searches, read hundreds of pages, and still cannot understand the calculation. (Most questions have to do with lotto odds... )
The specifics are, for example:
100,000 entries
5 prizes of equal value
50 entries.
I thank anyone for assistance, and would learn quite a bit in the process. I have been fine understanding the process until the 5 equal prizes come into play, and how they affect the numbers. |
13 Answers
| Ultra Member | |
Aug 12, 2009, 02:41 PM
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First draw would be 50:100000 right? Second would be 50:99999 since presumably the first winning number is not put back.
Basically you add the odds for each separate event together and that would equal the odds of winning once. | | | Ultra Member | |
Aug 12, 2009, 02:46 PM
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I think the confusion lies in the fact that 5 prizes equals 5 drawings. | | | New Member | |
Aug 12, 2009, 02:47 PM
| | Originally Posted by **stevetcg** First draw would be 50:100000 right? Second would be 50:99999 since presumably the first winning number is not put back.
Basically you add the odds for each separate event together and that would equal the odds of winning once. I don't understand. So then the initial would be 50:99,950?
Where do the 5 chances come into play and how?
I assumed the odds with 1 entry would be 1:20,000 (since 5 equal prizes in 100,000)
I am unsure of how the number of entries apply, and then how the number of prizes apply.
Are the number of total entries, and then played entries calculated first? or are the 5 prizes calculated first and then the 50 entries applied?
I really do thank anyone for any assistance. I can't understand why I am having such a problem with this.. | | | New Member | |
Aug 12, 2009, 02:48 PM
| | Originally Posted by **stevetcg** I think the confusion lies in the fact that 5 prizes equals 5 drawings. The 5 is certainly not helping me. :confused::) | | | Ultra Member | |
Aug 12, 2009, 02:55 PM
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Ok... think of it like this.
5 prizes. That means there are 5 drawings, one for each prize.
For each draw you hold 50 tickets of a possible 100000.
So to win the first prize you have 50 chances in 100000. For the second prize you have 50 chances in 99,999 since one possibility has already been won by (presumably) someone else.
For a raffle, each prize is given by a separate draw, so it is a separate event.
Of course my math is simplistic because it does not account for the possibility of winning more than once... but it gives you an idea of the direction of thinking.
One of the math geeks will likely be along to give a real answer... I'm just trying to point you in the direction of solving it yourself. | | | New Member | |
Aug 12, 2009, 03:06 PM
| | Originally Posted by **stevetcg** Ok... think of it like this.
5 prizes. That means there are **5 drawings, one for each prize**. **for each draw **you hold **50 tickets of a possible 100000**.
So to win the first prize you have **50 chances in 100000**. For the second prize you have **50 chances in 99,999 **since one possibility has already been won by (presumably) someone else.
For a raffle, each prize is given **by a seperate draw, so it is a seperate event**.
Of course my math **is simplistic **because it does not account for the possibility of winning more than once... but it gives you an idea of the direction of thinking.
One of the math geeks will likely be along to give a real answer... im just trying to point you in the direction of solving it yourself. All above were very helpful, and did give me more of an idea, but still not able to arrive at how to calculate it all through... ( *unless essntially, it means in the end, odds of any one prize are 1:99,950 *(using ballpark accounting for 1:99,999, 1:99,998, and so on... ) *This is* a real life scenario I am currently playing, *but not being able to calculate it myself, just out of curiosity has been driving me absolutely mad*.:D
My initial thinking 1:100,000/5 prizes/1:20,000 chance of any one prize, but then again the 50 different entries and how to apply confused me. (The I thought 1:100,000, 50 entries, 5 prizes with a lower number that I knew could not be correct... )
Simplistic is fine. I'm just trying to understand the process of calculating it... | | | New Member | |
Aug 12, 2009, 03:18 PM
| | Originally Posted by **stevetcg** Ok... think of it like this.
5 prizes. That means there are 5 drawings, one for each prize.
for each draw you hold 50 tickets of a possible 100000.
So to win the first prize you have 50 chances in 100000. For the second prize you have 50 chances in 99,999 since one possibility has already been won by (presumably) someone else.
For a raffle, each prize is given by a seperate draw, so it is a seperate event.
Of course my math is simplistic because it does not account for the possibility of winning more than once... but it gives you an idea of the direction of thinking.
One of the math geeks will likely be along to give a real answer... im just trying to point you in the direction of solving it yourself. Unless the starting point would be:
1:100,000
1:99,999
1:99,998
1:99,997
1:99,996
For the 5 prizes, but then how would the 50 entries be applied?
Would 50:100,000 be translated into 1:2,000 for simplicity? Making it 1:2,000,1:1,999, etc?
I thank anyone for their help... it's very important to me personally to be able to figure this out... | | | Ultra Member | |
Aug 12, 2009, 03:19 PM
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Simplistically, 50 in 100000 5 times (generously rounded) | | | Ultra Member | |
Aug 12, 2009, 03:22 PM
| | Originally Posted by **ImJustCurious** Unless the starting point would be:
1:100,000
1:99,999
1:99,998
1:99,997
1:99,996
for the 5 prizes, but then how would the 50 entries be applied?
would 50:100,000 be translated into 1:2,000 for simplicity? making it 1:2,000,1:1,999, etc?
I thank anyone for their help......it's very important to me personally to be able to figure this out..... Except 1:1999 is not the same as 50:99999. Its closer to 1999.98. | | | New Member | |
Aug 12, 2009, 03:31 PM
| | Originally Posted by **stevetcg** Except 1:1999 is not the same as 50:99999. Its closer to 1999.98. Of course. But then understanding how the 5 come into play, unless it would be *5 chances, each chance 1:2,000, then 2nd prize 1: 1,999, then 3rd prize 1:1,998, and so on*... (I know would be slightly different... trying to understand the overview , and when do will run numbers myself for exact calculations... )
But do those all remain separate? or are they combined into an overall final number?
I know, I know... no Fields Medal for me :p | | | Ultra Member | |
Aug 12, 2009, 03:42 PM
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Your first paragraph nailed it. You can then calculate your chances of winning any of the prizes by adding the odds together. | | | New Member | |
Aug 12, 2009, 03:46 PM
| | Originally Posted by **stevetcg** Your first paragraph nailed it. **You can then **calculate your chances of winning any of the prizes **by adding the odds together**. What was confusing me was thinking: 1:2,000 , *then 5 prizes, (5/2000=400*... meaning end figure *of 1:400*... (Which I assumed *could not *be right... )
The highlighted above then confused me. :)
Thank you, sincerely for assisatance... I'm unable to rest until I understand this. | | | Uber Member | |
Aug 13, 2009, 10:21 AM
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What is your question exactly?
Here, I saw that steve has given you the odds, but these are in the situation when none of the numbers got out.
Let's see it to a smaller scale.
100 entries.
5 winning prizes
10 is the number of entries you bought.
The probability that you win, on the first draw is 10/100.
If you lost, then the probability that you win on the second draw is 10/99.
But if you won, the probability that you win on the second draw is 9/99 since one of your entries was drawn (except if they draw some number twice, and that winning number gets 2 prizes, which I doubt)
Basically, winning changes the whole probability... | | Question Tools | Search this Question | | | ### Add your answer here. ##
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