Raffle ticket odds, does buying say 10 out of 28,000, give you 1/2800 chance on every ticket or 1/28000?

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Raffle ticket odds, does buying say 10 out of 28,000, give you 1/2800 chance on every ticket or 1/28000?

In: Mathematics
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Each ticket has a 1 in 28000 chance but you as a person have 1 in 2800 chance because you have 10 in 28000. I hope that makes sense.

Each ticket will have a 1/28000 chance of winning, but overall you have a 1/2800 chance of winning. You can think about it like this: for each individual ticket, the number of ways that ticket can win is obviously 1 (if that ticket is drawn), and the number of total outcomes is 28000 (there are 28000 tickets, any one of which could be drawn). Probability is really just (number of ways the desired outcome could happen)/(number of total outcomes).

The reason you have an overall chance of winning that’s 1/2800 is because the (number of ways the desired outcome could happen) is now 10. The desired outcome is winning, and the number of ways you win is if any one of your 10 tickets is drawn.

Each ticket has a 1/28000 chance to win on its own, or 0.0035714285714%.

If you attempt a 1/28000 chance 10 times, there is a 0.035708546% chance that at least 1 of the attempts will win, which is *not* simply 10x more.

I see a few people saying you’d have a 1/2800 chance to win but actually this is false. Because each ticket that is NOT the winning ticket reduces the overall pool count i.e.

ticket 1 is competing against 27999 tickets but fails to win

ticket 2 is competing against 27998 tickets but fails to win

ticket 3 is competing against 27997 tickets but fails to win etc.

So what you do is you multiply each tickets probability of not winning then subtract that total from 1 to get the probability that you will win.

In this case:

27999/28000 * 27998/28000 * … * 27990/28000 = 0.99803

Then subtract that value from 1.0 which gives you a probability of 0.00197 or 0.197% chance to win given that you bought 10 tickets in a 28000 ticket raffle.

You can use the expected value probability algorithm to calculate other values :).

Maybe this will help people get what u/berael and u/Medved are saying.

Obviously, if we have zero tickets in a lottery/raffle we have zero chance of winning.

After buying the first ticket the increase in the chance of winning is increased 100% because we started with zero and now have some chance.

What is the percent increase in the chance of winning when we buy the second ticket?

It’s so low as to be almost zero.

The same applies for subsequent tickets.

Each ticket increases the chance of winning mathematically but in a practical sense almost not at all.

I stayed away from the actual numbers to keep it simple (and because I just woke up from a nap) but I’m sure some big-brain out there will run the numbers for us.

Here’s another way to look at it.

The chance of winning a 6/49-type lottery is 1 in 13,983,816.

Source: [Wikipedia Lottery Mathematics]( https://en.wikipedia.org/wiki/Lottery_mathematics#%3A%7E%3Atext%3D7_External_links-%2CChoosing_6_from_49%2Chappening_is_1_in_13%2C983%2C816.?wprov=sfla1)

That’s approximately the same as the odds of being struck by lightning – twice – in the same place.

Using OP’s example of having a 1-in-10 chance means buying 10% of the tickets.

The odds of a favorable occurrence are dependent on how many possibilities there are and how many are favorable. Since we don’t care which ticket wins, we add the probabilities of any given ticket winning. If there are 28,000 tickets, and you have 10 of them, your odds are 1/28000 * 10 or 1/2800. That’s the odds of ANY of your tickets winning. The odds increase if there’s multiple drawings, like if there were 3 prizes, your odds of winning at least once are slightly more than tripled, since your odds are 1/28000 + 1/27999 + 1/27998 (since we have to account for previous winners) but your odds would be slightly greater than 3/28000 or roughly 1/9333.33.

Lots of confusion here. Some people are treating this like the typical dice rolling problem (if you roll 2 dice, your chance of getting at least one 6 is not double your chance with one dice).

I think the confusion lies in the question being a little ambiguous. When I think of a raffle, I think: 28,000 tickets are sold, 28,000 stubs get put in a hat, one is pulled out, and that is the winner. Done.

But did you mean that maybe fewer than 28,000 are sold, 28,000 stubs are put in a hat, and they keep pulling tickets out of a hat until they get one that was sold?

Or maybe they pull ten tickets stubs out and give a prize to each?

Also, what exactly do you mean by a 1/2800 chance on _every_ ticket? I don’t understand that.