Interpretation of a rare event - lucky draw

1. Feb 2, 2012

musicgold

Hi,

I was at a party where there was a lucky draw. There were about 200 people and each of them had put a chit with their name on it, in a transparent bowl. There were 10 prizes. The announcer chose a person at random to pick a chit for each prize. For one of the prizes, the picker picked his own chit ( there was no hanky-panky). Some one said what are the chances of that happening.

Now I am trying to calculate the chances of such a thing happening. Each chit had a 0.5% chance of being picked. If we also assume that the announcer selected the picker randomly (0.5% chance of getting selected to pick) . Then the chance of a picker picking his own chit = 0.5% * 0.5% = 0.0025%

2. My interpretation of how likely is such an event: As likely as any random event out there. I think we can’t use the 0.0025% probability figure here because we know that this experiment will not be carried out hundreds of times. The calculated figure is meaningful only if we repeat the experiment many times. Is that a correct interpretation?

3. If the announcer had pre-determined the pickers, would the chance of such an incident go to 0.5%?

Thanks.

Last edited: Feb 2, 2012
2. Feb 2, 2012

SW VandeCarr

I'm assuming 10 chits are drawn without replacement and pickers can only be selected once and are then no longer eligible. If you are asking for the probability that a random person will pick their own chit out of 200, I would do this in the following way. For each selection of a picker, the average denominator is 200-4.5= 195.5. 4.5 is average of being between the 0th picker and the 9th picker inclusive. I use this assignment because the first picker will have the full 200 chits to choose from. The denominator is adjusted accordingly so my answer would be P= (1/195.5)^2 = .0.000026. Since there are ten opportunities we multiply this by ten to get 0.00026 This would be a reasonable approximation vs a more computationally demanding exact alternative.

EDIT: A more nearly exact way would be to first calculate the probability of not being selected: (199/200)(198/199).......(190/191), subtract from 1 and multiply the result by 0.005115 since each picker has just one opportunity to draw (adjusting again for the unreplaced chits)

Last edited: Feb 2, 2012
3. Feb 2, 2012

SW VandeCarr

I made an error in failing to account for the fact that everyone has 10 opportunities to be a picker. I've corrected this in the above post.

Last edited: Feb 2, 2012
4. Feb 2, 2012

awkward

Here is another approach which yields an approximate result.

The probability that any single picker will not pick his own chit is 199/200. The pickers' outcomes are not independent-- if the first picker picks his own chit then this alters the second picker's chances slightly-- but they are nearly so. If we assume independence, then the probability that none of the ten pickers gets his own chit is (199/200)^10, so the probability that at least one picker gets his own chit is 1 - (199/200)^10, which is about 0.049.

Last edited: Feb 2, 2012
5. Feb 2, 2012

SW VandeCarr

I don't follow this. There are two steps. Your probability is the probability of being chosen as a picker given 10 chances. However the picker gets one chance to draw where the probability of getting their own chit is about .005. So the approximate probability of a random member being selected as a picker and drawing their own chit is 0,05*0.005 = 0.00025.

EDIT: I reread the problem and it's a bit confusing. Apparently one picker is chosen who then draws ten chits. The probability of being a picker is then .005. The probability of the picker getting their own chit in 10 tries is .05. So we still have .05*.005=.00025 unless I misunderstood something.

Last edited: Feb 2, 2012
6. Feb 3, 2012

musicgold

No. One picker picks only one chit. After that the announcer selects another picker.

Guys, what is your take on #2 & 3 ?

7. Feb 3, 2012

SW VandeCarr

For a random audience member, P=0.05 to be selected given ten opportunities, and P=0.005 to pick their own chit given they are selected. .
Therefore the probability that a random audience member selects their own chit is 0.00025, not 0.0025 These are approximate since the denominators have not been adjusted.

2. My interpretation of how likely is such an event: As likely as any random event out there. I think we can’t use the 0.0025% probability figure here because we know that this experiment will not be carried out hundreds of times. The calculated figure is meaningful only if we repeat the experiment many times. Is that a correct interpretation?

Not really. These are the prior probabilities that result from following assumptions: 1)each person and each chit are equally likely to selected. 2) all selections are random.

3. If the announcer had pre-determined the pickers, would the chance of such an incident go to 0.5?

If the pickers are predetermined, then the only random step is picking chits. If each picker picks 1 chit, then the approximate probability is P=0.005 that the picker selects their own chit. However, since there are ten pickers, the approximate probability that at least one picks their own chit is .05. This is a different situation than starting with a random audience member.

Last edited: Feb 3, 2012
8. Feb 3, 2012

awkward

My interpretation of the original question is that we are asked how likely it is that at least one picker will pick his own chit. That event has a probability of about 0.049. That's much more likely than the probability that Fred will be both chosen as a picker and also pick his own chit, where Fred is some member of the audience.

I am assuming that there are 10 distinct pickers, which I gather is what the OP intended.

9. Feb 3, 2012

SW VandeCarr

Yes. I understood the question as the probability of a random member of the audience picking their own chit. So to be the first picker and draw your own chit, the probability is (.005)(.005)=.000025. However that same person is eligible to be the second pick (as is everyone else who wasn't chosen in the first pick). For the second pick the probability is (1/199)^2, for the third (1/198)^2 and so on for a given point in the sequence. However, for those not chosen the probability that they will be chosen at some point in the sequence of 10 would be the sum of the remaining probabilities at any given point in the sequence. So the probability of picking your own chit after 9 pickers have been chosen and 9 chits have been drawn is (1/191)^2 = 0.000027

I'm satisfied to say that before any picker is selected or any chit is drawn, the probability of any random person in the audience of being selected as picker, and picking their own chit in one attempt is approximately (.05)(.005)= 0.00025. In fact it will be a slightly higher. I estimated 0.00026 in a previous post..

I agree that the probability that at least one picker will pick their own chit is (.05) given they are already chosen as picker. Also the probability that at least one member of the audience will pick their own chit before the contest starts is (0.05)(0.05) = 0.0025. I think this is a different question than the probability for any given random member of the audience to select their own chit.

Last edited: Feb 4, 2012
10. Feb 4, 2012

awkward

I see we disagree on the relevant probability. To me, the answer to "what are the chances of that happening" is the probability of at least one picker picking his own chit.

The OP went on to ask about the probability that a designated member of the audience will be chosen as a picker and pick his on chit. But that's the wrong probability if we want to assess how likely we are to see one of the pickers pick his own chit.

11. Feb 4, 2012

SW VandeCarr

Well, the OP asked three questions. In any case I worked out three answers, one of which agrees with yours.

Probability of at least one picker picking their own chit, about 0.05.

Probability of at least one member of the audience picking their own chit: about (.05)(.05)=.0.0025 if pickers are selected randomly.

Probability that any given random audience member will select their own chit on one draw: about (.05)(.005)= 0.00025 if pickers are selected randomly.

I think its useful to show these, irrespective of the OP's specific questions. I think the second answer was what the OP was looking for, given his/her calculation.

Last edited: Feb 4, 2012