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I have a dozen donuts and 84 friends. Since there is not enough to go around and no one wants to share, I will put all 84 names in a hat (a hat that perfectly randomizes) and draw 12 names. What are the chances of getting a donut? 12 out of 84 or ~14.3%.

Let's calculate the chances of getting a donut another way by calculating the chances as we draw names out of the hat.
1st draw 1.19% 1 out of 84
2nd draw 1.20% 1 out of 83
3rd draw 1.22% 1 out of 82
4th draw 1.23% 1 out of 81
5th draw 1.25% 1 out of 80
6th draw 1.27% 1 out of 79
7th draw 1.28% 1 out of 78
8th draw 1.30% 1 out of 77
9th draw 1.32% 1 out of 76
10th draw 1.33% 1 out of 75
11th draw 1.35% 1 out of 74
12th draw 1.37% 1 out of 73
Sum equals = 15.32%

The sum total of these is ~15.32% Why isn't it equal to 14.3%?

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    $\begingroup$ The probability of getting the donut on the third draw is the probability of not getting it on the first, and (conditioned on that) not getting it on the second, and (conditioned on both those) getting it on the third, which is $\frac{83}{84}\times\frac{82}{83}\times\frac{1}{82}=\frac{1}{84}$. Similar calculations apply for all the others, making the overall probability of getting a donut $\frac{12}{84}$ as you might expect intuitively. $\endgroup$
    – Henry
    Nov 7, 2023 at 11:03

3 Answers 3

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They're different because the numbers you're adding up shouldn't be added.

A smaller example

I have three donuts and four friends. Since there is not enough to go around and no one wants to share, I will put all four names in a hat (a hat that perfectly randomizes) and draw three names. What are the chances of getting a donut? Three out of four or 75%.

Let's calculate the chances of getting a donut another way by calculating the chances as we draw names out of the hat.

Draw Probability Reasoning
1st draw 25% 1 out of 4
2nd draw 33.3% 1 out of 3
3rd draw 50% 1 out of 2

Sum equals = 108.3%. That's not good.

What happened?

Each outcome depends on the one before it: if my name came up in the second draw, it can't come up again in the third. These probabilities are conditional probabilities, so you can't add them directly. The 50% is the chance of getting drawn on turn 2 given that you weren't picked on turn 1.

Let's introduce some notation. Let $W_i$ be the draw on turn $i$. The outcomes $w_i$ and $\neg w_i$ mean that my name did or didn't come up on turn $i$. The values in our Probability column, then, are $p(w_1)$ then $p(w_2 \mid \neg w_1)$, then $p(w_3 \mid \neg w_1, \neg w_2)$.

How do we make the results match?

Well, I know my name comes up if I draw it on the first, or I draw it on the second, or I draw it on the third.

\begin{align} p(\text{donut}) =& p(w_1) \\ & + p(\neg w_1) \times p(w_2 \mid \neg w_1) \\ & + p(\neg w_1) \times p(\neg w_2 \mid \neg w_1) \times p(w_3 \mid \neg w_1, \neg w_2) \\ =& 0.25 \\ &+ 0.75 \times 0.\overline{3} \\ &+ 0.75 \times 0.\overline{6} \times 0.5 \\ =& 0.75 \text{ which is the same as } \frac{3}{4}\text{.} \end{align}

Using the conditional probabilities and the chain rule of probability, we get to the same answer as the first method.

You can extend this reasoning to your example with bigger numbers.

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    $\begingroup$ given that you weren't picked on turn 1 ahhh and the light bulb goes on! thanks $\endgroup$ Nov 7, 2023 at 4:02
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    $\begingroup$ A small note as convenience for those who don't want to read through the rather complex notation on the linked wikipedia page: You can more easily calculate probabilities like this by using the fact that it's one (100%) minus the probability of never getting picked. Which is 1 - 3/4 * 2/3 * 1/2 = 1 - 1/4 = 3/4. $\endgroup$
    – Syndic
    Nov 9, 2023 at 13:44
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Because probabilities don't add like that - if you'd have 54 donuts surely the chance of winning one is not $\sum_{d=0}^{53}1/(84-d)\approx 102\%$?

You have to keep in mind that the chance of winning the $n$th donut is conditional on not winning any of the $1, ..., n-1$ donuts. The probability of something not happening is one minus the probability of it happening, or not winning the first donut has probability $1-\frac{1}{84} \approx 98.81\%$. The chance of then not winning the second one is multipicative with that, or $(1-\frac{1}{84})(1-\frac{1}{83})\approx97.62\%$. In general the probability of not winning any of the first $n$ donuts is $\prod_{d=0}^{n-1}1-\frac{1}{84-d}$.

The probability of winning any of the first $n$ donuts is then again one minus the above. Back to your example with 12 donuts, where the chance of winning any of those is $1-\prod_{d=0}^{11}1-\frac{1}{84-d}\approx 14.29\%$ which is $\frac{12}{84}$. You can check this for any other number of donuts and they will match.

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Let's calculate the chances of getting a donut another way by calculating the chances as we draw names out of the hat.<br>
1st draw    1.19%   1 out of 84<br>
2nd draw    1.20%   1 out of 83<br>
3rd draw    1.22%   1 out of 82<br>
4th draw    1.23%   1 out of 81<br>
5th draw    1.25%   1 out of 80<br>
6th draw    1.27%   1 out of 79<br>
7th draw    1.28%   1 out of 78<br>
8th draw    1.30%   1 out of 77<br>
9th draw    1.32%   1 out of 76<br>
10th draw   1.33%   1 out of 75<br>
11th draw   1.35%   1 out of 74<br>
12th draw   1.37%   1 out of 73<br>
Sum equals   =              15.32%<br><br>

This table shows an increasing chance of getting the donut, but the chances of getting a donut on a particular time/draw should be continuous function.

For example: For a friend of you, whether they get a donut in the 1st draw or in the 12th draw, that should be equal, as your hat should make it for a person as much likely to be drawn 1st as 12th. It is not more likely to get the donut on the 1st draw as on the 12th draw. You have 1/84 probability for both cases.

What your table shows instead are conditional probabilities. The probability to get the donut in the k-th draw, conditional on not having had it during the previous (k-1) draws. The comment by Henry explains well how you can create a sum with these conditional probabilities such that you still end up with 12/84.

$$\text{P(donut on k-th turn) = P(donut no donut before k-th turn) $\times$ P(donut on k-th turn conditional on no donut before k-th turn)} $$

You computed the terms $\text{P(donut on k-th turn conditional on no donut before k-th turn)}$ on the right-hand side and used it as $\text{P(donut on k-th turn)}$ on the left-hand side.

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