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Mokele

Raffle strategy

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Ok, this should be interesting, and will hopefully settle a debate between my wife and I.

 

Let's say there's a raffle, with 10 prizes, and you can put your tickets in individual boxes for each prize (thus there will be 10 separate, non-overlapping drawings). 9 other people have already bought 10 tickets each, and as a result, each cup now has 9 tickets in it.

 

If I buy ten tickets myself (and I'm the last person to enter), would I have a better chance of winning if I put one ticket in each cup, or if I put all 10 in one cup?

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Ok, this should be interesting, and will hopefully settle a debate between my wife and I.

 

Let's say there's a raffle, with 10 prizes, and you can put your tickets in individual boxes for each prize (thus there will be 10 separate, non-overlapping drawings). 9 other people have already bought 10 tickets each, and as a result, each cup now has 9 tickets in it.

 

If I buy ten tickets myself (and I'm the last person to enter), would I have a better chance of winning if I put one ticket in each cup, or if I put all 10 in one cup?

 

If you put all your tickets in one cup, your chances of winning would be:

 

[math]\frac{10}{19} = 52.6\%[/math]

 

If you put one ticket in each cup, your chances of winning would be:

 

[math]\frac{1}{10} \cdot 10 = 100\%[/math]?

 

yeah... that's not right... The first one is though. >.< What have I done wrong?

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I think that's right because:

 

[math]P(XuY)= P(X)+P(Y)-P(XnY)[/math]

 

[math]P(XuY)=\frac{1}{10}\bullet10=\frac{10}{10}[/math]

 

I cannot think of any intersect so I would have to agree with you.

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Ahah! For the second case I have to use a binomial distribution! I forgot about that, I took Statistics two years ago >.<.

 

Anyway... The probability of winning given ten drawings, each with a probability of 1/10 each is:

 

Binomial distribution given 10 trials, 1 success, and p = 0.10:

 

[math]_{10}C_1 \cdot (0.10)^1 \cdot(1-0.10)^{(10-1)} = 38.7\%[/math]

 

In conclusion, you would have better chances dumping all your tickets into that prize you really, really want. :cool:


Merged post follows:

Consecutive posts merged
I think that's right because:

 

[math]P(XuY)= P(X)+P(Y)-P(XnY)[/math]

 

[math]P(XuY)=\frac{1}{10}\bullet10=\frac{10}{10}[/math]

 

I cannot think of any intersect so I would have to agree with you.

 

No, that could not have been right. It is obviously not a sure thing to win in that circumstance. I don't remember why that approach does not work, however, I just remember that one must use binomial distributions in this case.

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Sometimes it's easier to figure out the probability that you won't win. Remember, there is exactly one possibility, 10 losses, on one side, and every other combination of wins and losses on the other. Focus on the 10 losses.

 

When you distribute the tickets evenly, the chance that you'll lose for any given one is 9/10. The chance that you'll lose ten times in a row, then, is 9^10/10^10 = 0.34867844 = ~35%. Therefore, the chance that you'll win at least once is about 65%, better than 10/19, which is about 52.6%.

 

Put 1 ticket in each pot.

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You don't just use the binomial distribution for 1 success. Because winning twice or three times or more should still be a success in the winners eyes.

 

Your first calculation was correct. Each trial, you have a 1 in 10 chance of winning. And, when you do 10 chances -- on average when you do 10 chances, you will get 1 winner. Just because the average is 1 (or 100%), does not mean it will happen. But, it is the average. If the same raffle (with 10 mini-raffles in each where the chance of winning a prize was 10% in each mini) were held 10,000 times, the total number of prizes at the end would be near 10,000 -- some raffles you would win 0 prizes, some raffles you would win 5 prizes, but in the end the average is 1 prize per set of 10 mini raffles that are ran.

 

As a related example. On average, if you roll a fair 6 sided die 6 times, how many time will the number 1 appear? Once. Same for all the numbers, because on average each number appears once every 6 rolls. The key phrase is on average, because how any individual trial plays out is unknown. All we can do is calculate averages (and variances and other broad numbers), not specifics for any individual case.

 

So, on average, it is better to play 1 ticket per cup (assuming that you can be reasonably sure that the cups will stay that way) -- the more interesting question is if everybody continues to distribute them evenly, is it still better to spread it around, or should you weigh more heavily on one side? Is there a certain number where a "break even" point is reached. I.e. 86 tickets per cup and you should spread around your 10, but if there are 87 you should put all 10 into 1 to maximize you chances of at least 1 win?

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You don't just use the binomial distribution for 1 success. Because winning twice or three times or more should still be a success in the winners eyes.

 

Alright, I did some recalculations: The cumulative probability of getting AT MOST ten wins in the given circumstances is: 1 (I see now, but then you have to...) subtract from that the binomial probability of zero successes: 34.9% = 65.1%. So, you should put a ticket in each pot.

 

Just playing around with numbers, now, here...

 

What if there were 100 pots (9 people, each with 100 tickets, 9 tickets in each)?

 

The chance of winning if you dump all 100 in one is much greater: 100/109 = 91.7%

 

But the chance if you put one in each is 1 - 0.0000266 ≈ 100%

 

It is still advantageous to put 1 ticket in each.

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-- the more interesting question is if everybody continues to distribute them evenly, is it still better to spread it around, or should you weigh more heavily on one side? Is there a certain number where a "break even" point is reached. I.e. 86 tickets per cup and you should spread around your 10, but if there are 87 you should put all 10 into 1 to maximize you chances of at least 1 win?

 

Interesting question. Shouldn't be too hard: where n is the number of other participants, the probability of winning by putting all 10 in one pot is:

 

10/(10+n)

 

The probability of winning by putting one ticket in each pot is:

 

1 - (n^10)/[(n+1)^10]

 

To find the n where the first becomes better (if there is one), set up the inequality we want...

 

10/(10+n) > 1 - (n^10)/[(n+1)^10]

 

...and algebra the hell out of it:

 

10/(10+n) + (n^10)/[(n+1)^10] - 1 > 0

 

10 + (10+n)(n^10)/[(n+1)^10] - (10+n) > 0

 

(10+n)(n^10)/[(n+1)^10] - n > 0

 

(10+n)(n^10) - n[(n+1)^10] > 0

 

10n^10 + n^11 - ..... something long that has to include n^11 and 10n^10, difference can't be greater than zero for any positive n. So, putting everything in one pot is never smarter. :)

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So, on average, it is better to play 1 ticket per cup (assuming that you can be reasonably sure that the cups will stay that way) -- the more interesting question is if everybody continues to distribute them evenly, is it still better to spread it around, or should you weigh more heavily on one side? Is there a certain number where a "break even" point is reached. I.e. 86 tickets per cup and you should spread around your 10, but if there are 87 you should put all 10 into 1 to maximize you chances of at least 1 win?

 

I realize that my calculations did not address this question. Let me try to tackle this one...

 

10 cups, 99 people, 10 tickets each. You are the hundredth. 99 tickets in each cup.

 

Probability of winning if you put all your tickets in one cup: 10/109 = 9.17%

 

Probability of winning if evenly spread: 1 - 0.904 = 9.60%

 

Hmm, the two seem to be converging. Time for some good ol' algebra...

 

[math]\frac{10}{n + 9} = 1 - (\frac{n-1}{n})^{10}[/math]

 

...Except I'm going to end up with like a tenth-order equation there. Oh well. That's the form of it, if anyone feels like solving tenth-order equations :doh:

 

And... Sisyphus beat me to it.

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Ahh, thanks!

 

 

So, to add to the math fun, what happens with fewer or more prizes? More or fewer tickets per pot? Is there ever a set of conditions where it's better to put all the tickets in one pot?

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Alright, I did some recalculations: The cumulative probability of getting AT MOST ten wins in the given circumstances is: 1 (I see now, but then you have to...) subtract from that the binomial probability of zero successes: 34.9% = 65.1%. So, you should put a ticket in each pot.

 

Just playing around with numbers, now, here...

 

What if there were 100 pots (9 people, each with 100 tickets, 9 tickets in each)?

 

The chance of winning if you dump all 100 in one is much greater: 100/109 = 91.7%

 

But the chance if you put one in each is 1 - 0.0000266 ≈ 100%

 

It is still advantageous to put 1 ticket in each.

 

Ahh, thanks!

 

 

So, to add to the math fun, what happens with fewer or more prizes? More or fewer tickets per pot? Is there ever a set of conditions where it's better to put all the tickets in one pot?

 

So I think I covered more prizes, the probability that you would win with an even spread would approach 100% more quickly than the probability of winning with a boatload of tickets in one, so no good there.

 

What about fewer prizes?

 

3 pots, 2 people, 3 tickets each. 2 tickets in each pot:

 

All in one: 3/5 = 60.0%

 

Even spread: 1 - (2/3)^3 = 70.4%

 

So, with fewer prizes it is still more advantageous to go with an even spread.

 

This makes sense, too, because even though number of pots is variable, the way we've set it up is so that number of participants is equal to number of pots, so the math is identical to the above by Sisyphus.

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Alright, I did some recalculations: The cumulative probability of getting AT MOST ten wins in the given circumstances is: 1 (I see now, but then you have to...) subtract from that the binomial probability of zero successes: 34.9% = 65.1%. So, you should put a ticket in each pot.

 

This still isn't right. You are not taking into account the multiple wins in the expected value.

 

The expected value of wins is equal to (1/10 per mini raffle)*10 min raffles = 1.

 

The binomial distribution is exactly the same thing: The expected value = 0 * probability of 0 successes in 10 tries (denoted P(0,10))+ 1*P(1,10) + 2*P(2,10) + 3*P(3,10) + 4*P(4,10) + 5*P(5,10) + 6*P(6,10) + 7*P(7,10) + 8*P(8,10) + 9*P(9,10) + 10*P(10,10).

 

Calculate this number out and you get 1. This says that, on average, every time this raffle occurs (this raffle consisting of 10 mini raffles where you have a 1 in 10 chance of winning each mini), you will win 1 prize. Again, it says nothing specific about any individual raffle -- you may win 5 prizes once, 0 prizes if it is repeated, 2 prizes if repeated again, etc. But, in the long run, if the raffle is repeated time and time again, the average will be 1 prize. Not 0.651 prizes, 1 prize.

 

EDITED to add: this actually probably all comes down to the way Mokele asked the question. I am thinking that your answer answers the question better -- because his question was what is the probability of winning something, which would be 1-P(0,10). These questions are always confusing in being sure of answering exactly what the asker wants.

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The binomial distribution is exactly the same thing: The expected value...

 

This is the issue. The expected value and the probability according to the binomial distribution are different. The expected value is zero multiplied times the probability. The probability itself is:

 

[math]_{10}C_0 \cdot (0.10)^0 \cdot (1 - 0.10)^{10-0}[/math]

 

Since the first two terms are equal to 1, the probability is:

 

[math](0.90)^{10} = 34.9\%[/math]

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This is the issue. The expected value and the probability according to the binomial distribution are different. The expected value is zero multiplied times the probability. The probability itself is:

 

[math]_{10}C_0 \cdot (0.10)^0 \cdot (1 - 0.10)^{10-0}[/math]

 

Since the first two terms are equal to 1, the probability is:

 

[math](0.90)^{10} = 34.9\%[/math]

 

I put that in my edit -- the language of the question is important. If the question were "which strategy wins more prizes?", then you have to use the expected value.

 

-----------------------

 

The real question is whether Mokele or his wife won the debate ?!?

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I put that in my edit -- the language of the question is important. If the question were "which strategy wins more prizes?", then you have to use the expected value.

 

The question was never "which strategy wins more prizes", but rather "which strategy yields a greater chance of winning", meaning not-losing, meaning the inverse of the probability of getting zero prizes, which is what the math portrays.

 

EDIT: And here's my edit after reading your edit. I think we're on the same page.

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