**#1**Jan 22nd, 2007

1 Ticket

2 Tickets

3 Tickets

4 Tickets

Anybody know the answer?

In a raffle where a total of 3,000 tickets are issued and 10 prizes to be won (one prize with each draw), what are the odds of someone winning when he/she buys;

1 Ticket

2 Tickets

3 Tickets

4 Tickets

Anybody know the answer?

1 Ticket

2 Tickets

3 Tickets

4 Tickets

Anybody know the answer?

as long as it doesn't qualify me as a "wizard"...

Assuming the winning tickets are not returned to the pool the odds are 10, 20, 30 and 40 in 29955 respectively.

Assuming the winning tickets are not returned to the pool the odds are 10, 20, 30 and 40 in 29955 respectively.

In a raffle where a total of 3,000 tickets are issued and 10 prizes to be won (one prize with each draw), what are the odds of someone winning when he/she buys;

1 Ticket

2 Tickets

3 Tickets

4 Tickets

Anybody know the answer?

10 prizes out of 3000 tickets gives you a 1 in 300 chance of winning. So with 1 ticket, 1 in 300. 2 tickets, 1 in 150. 3 tickets 1 in 100. 4 tickets 1 in 75. Not bad odds.

There after each ticket would have a 1:3000 chance. Can a gambler chime in here to figure the odds on the other quantities?

If 1 ticket is taken from 3000, that would be a 1:3000 ratio. Seems like 1 in 3000 chance on one ticky.

There after each ticket would have a 1:3000 chance. Can a gambler chime in here to figure the odds on the other quantities?

10 prizes out of 3000 tickets would mean a 1 in 300 chance of winning something, yes no?

*lol* never mind, more coffee needed karrie.....

Thanks for your replies.

The way I am understanding it;

3000 tickets / 4 tickets / 10 prizes = 1 in 75 to win

3000 tickets / 3 tickets / 10 prizes = 1 in 100 to win

3000 tickets / 2 tickets / 10 prizes = 1 in 150 to win

3000 tickets / 1 ticket / 10 prizes = 1 in 300 to win

I suck at math, but that's what my brain is telling is correct.

Suggestions?

The way I am understanding it;

3000 tickets / 4 tickets / 10 prizes = 1 in 75 to win

3000 tickets / 3 tickets / 10 prizes = 1 in 100 to win

3000 tickets / 2 tickets / 10 prizes = 1 in 150 to win

3000 tickets / 1 ticket / 10 prizes = 1 in 300 to win

I suck at math, but that's what my brain is telling is correct.

Suggestions?

Are the tickets returned to the raffle after they're drawn, can you win more than one prize?

Thanks for your replies.

The way I am understanding it;

3000 tickets / 4 tickets / 10 prizes = 1 in 75 to win

3000 tickets / 3 tickets / 10 prizes = 1 in 100 to win

3000 tickets / 2 tickets / 10 prizes = 1 in 150 to win

3000 tickets / 1 ticket / 10 prizes = 1 in 300 to win

I suck at math, but that's what my brain is telling is correct.

Suggestions?

i think it has to do with the functions nPr and nCr.... maybe not though

The way I looked at it came out with similar odds to Bit, but I'm off by a factor of ten...

If the tickets aren't returned it's something like (1/3000)+(1/2999)...(1/2991)=

Like I said it's not the same answer as Bit, but I think I'm in the general area.

If the tickets aren't returned it's something like (1/3000)+(1/2999)...(1/2991)=

Like I said it's not the same answer as Bit, but I think I'm in the general area.

The way I looked at it came out with similar odds to Bit, but I'm off by a factor of ten...

If the tickets aren't returned it's something like (1/3000)+(1/2999)...(1/2991)=

Like I said it's not the same answer as Bit, but I think I'm in the general area.

the factor of 10 is accounted for by remembering that each shot at a draw bumps the chances up by the number of tickets you hold.

1 in 3000 plus 1 in 2999 is 2 in 5999 and so on.

So if you have 2 tickets the chances of winning are 2 in 5999?

With one ticket the answer is 1 - (2999/3000 x 2998/2999 x 2997/2998....x 2990/2991) = 1 in 300

With two tickets....... 1- (2998/3000 x 2997/2999 x 2996/2998.....x 2989/2991) = 1 in 150.2 (but there's a chance of winning twice)

etc...

In other words, it's 1 minus the chance of NOT winning.

I think that's right...but it's been a while since I took math.

With two tickets....... 1- (2998/3000 x 2997/2999 x 2996/2998.....x 2989/2991) = 1 in 150.2 (but there's a chance of winning twice)

etc...

In other words, it's 1 minus the chance of NOT winning.

I think that's right...but it's been a while since I took math.

An interesting little problem in compound probabilities. Been a very long time since I had to think about anything like this though.

The probability of winning anything at all is the sum of the probabilities of winning any particular draw. Obviously the probability of winning the first draw is 1/3000 if you hold one ticket; I'm sure everybody sees that. The probability of winning the second draw, however, is not 1/2999. In order to win the second draw, you have to lose the first one (I'm assuming tickets aren't replaced), which you will with a probability of 2999/3000. You have to multiply those two fractions together. 2999/3000 is almost 1, but not quite, so the probability of winning on the second draw is a little less than 1/2999. In fact the 2999s cancel and you're left with 1/3000. That'll happen at every draw. The probability is thus as BitWhys stated, 1/3000 for every draw, so in 10 draws it's 10/3000, or 1/300. Makes sense: if there were 3000 draws you're guaranteed to win if winning tickets aren't put back in the draw, so the logic of that argument must add up to a probability of 1 after 3000 steps, and it does. Putting winning tickets back into the pool after each draw makes no practical difference, the probabilities differ only after the fifth decimal place.

Holding multiple tickets seems to complicate it a bit more than is obvious though. Strictly speaking it appears you can't just multiply the results for a single ticket by 2, 3, or 4. I was a little surprised when I found that result, so if anyone can point out a flaw in it I'd be grateful. This one's easier to approach from the other end: what's the probability that you'll lose all 10 draws in a row? If you have four tickets, the probability you'll lose the first draw is 2996/3000. Again assuming no replacement, the probability you'll lose the first and second draws is 2996/3000 times 2995/2999. For three draws, there's a third factor, 2994/2998. Just keep multiplying fractions that way and stop after the 10th one, which will be 2987/2991. There's not as much nice cancellation in that exercise as there is with only one ticket, and you actually do have to do a little multiplication of 4-digit numbers. The answer I get for that is 0.986726. That's the probability of losing all ten draws. The probability of winning anything is thus 1 minus that, which is 0.01327. If it were as simple as just multiplying the result for 1 ticket by four, that number would have been 0.01333, which is also, as somebody said, 1 in 75.

More specifically, the probability of winning with one ticket is a third of a percent, 0.003333 or 1/300, and the probability of winning with four tickets is very slightly less than four times that. I don't see why that should be so, frankly, so I reiterate, if anyone can show something's wrong with my argument... ? Has my creaky old brain blown a gasket somewhere?

The probability of winning anything at all is the sum of the probabilities of winning any particular draw. Obviously the probability of winning the first draw is 1/3000 if you hold one ticket; I'm sure everybody sees that. The probability of winning the second draw, however, is not 1/2999. In order to win the second draw, you have to lose the first one (I'm assuming tickets aren't replaced), which you will with a probability of 2999/3000. You have to multiply those two fractions together. 2999/3000 is almost 1, but not quite, so the probability of winning on the second draw is a little less than 1/2999. In fact the 2999s cancel and you're left with 1/3000. That'll happen at every draw. The probability is thus as BitWhys stated, 1/3000 for every draw, so in 10 draws it's 10/3000, or 1/300. Makes sense: if there were 3000 draws you're guaranteed to win if winning tickets aren't put back in the draw, so the logic of that argument must add up to a probability of 1 after 3000 steps, and it does. Putting winning tickets back into the pool after each draw makes no practical difference, the probabilities differ only after the fifth decimal place.

Holding multiple tickets seems to complicate it a bit more than is obvious though. Strictly speaking it appears you can't just multiply the results for a single ticket by 2, 3, or 4. I was a little surprised when I found that result, so if anyone can point out a flaw in it I'd be grateful. This one's easier to approach from the other end: what's the probability that you'll lose all 10 draws in a row? If you have four tickets, the probability you'll lose the first draw is 2996/3000. Again assuming no replacement, the probability you'll lose the first and second draws is 2996/3000 times 2995/2999. For three draws, there's a third factor, 2994/2998. Just keep multiplying fractions that way and stop after the 10th one, which will be 2987/2991. There's not as much nice cancellation in that exercise as there is with only one ticket, and you actually do have to do a little multiplication of 4-digit numbers. The answer I get for that is 0.986726. That's the probability of losing all ten draws. The probability of winning anything is thus 1 minus that, which is 0.01327. If it were as simple as just multiplying the result for 1 ticket by four, that number would have been 0.01333, which is also, as somebody said, 1 in 75.

More specifically, the probability of winning with one ticket is a third of a percent, 0.003333 or 1/300, and the probability of winning with four tickets is very slightly less than four times that. I don't see why that should be so, frankly, so I reiterate, if anyone can show something's wrong with my argument... ? Has my creaky old brain blown a gasket somewhere?

Chances of losing!!! That's what I had screwed up, I was thinking (2999-10)/2999 = 99.666556% then (2998-10)/2998 = 99.666444% ........ (2991-10)/2991 = 99.66566% .......... (2981-10)/2981= 99.664542% etc. but it seemed screwy and I was in the middle of figuring out what was screwy and I got a brain fart and then sidetracked by helping someone get unstuck.

More specifically, the probability of winning with one ticket is a third of a percent, 0.003333 or 1/300, and the probability of winning with four tickets is very slightly less than four times that. I don't see why that should be so, frankly, so I reiterate, if anyone can show something's wrong with my argument... ? Has my creaky old brain blown a gasket somewhere?

Probabbility of winning exactly once = 0.013213613

Probability of winning twice with 4 tickets = 0.0000597

winning three times = 0.000000107

four times = 6.23 x 10^-11

Probability of winning AT LEAST once = 0.01327342

0.013213613 + 2(0.0000597) + 3(0.000000107) + 4(6.23 x 10^-11) = 0.0133333333

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