# Help me calculate the oddsSeptember 27, 2020 9:54 AM   Subscribe

I need some MeFi experts to help me calculate the odds of something happening.

A friend just had a gender reveal for her child, where she had an array of 16 black balloons arranged in a 4 x 4 grid, and had friends take turns throwing a dart at them. The balloons were filled with multi-colored glitter, except for the gender reveal balloon that contained pink glitter. Now it would be not as interesting if someone broke the pink glitter balloon on the first try, since you want suspense for this event, but as it turned out, all the other balloons were popped first, and the pink was the last one remaining. So it worked out well, with lots of audience reaction after each pop, but I wondered how unusual it was for this to happen. The odds of hitting the pink balloon on the first throw are 1 in 16, but it becomes more likely to hit as the others are popped. So tell me the odds, i.e., the number of times one would have to recreate this event before you could duplicate leaving the one special balloon in 16 to the very last, given random dart throws.
posted by weapons-grade pandemonium to Grab Bag (20 answers total) 1 user marked this as a favorite

Still 1/16.

Another way to calculate it is to calculate the probability of NOT popping the pink balloon: 15/16, then 14/15, then 13/14, all the way down to 1/2. Multiply all those together to get the probability of doing it 15 times in a row, and everything cancels, leaving 1/16.
posted by alexei at 10:00 AM on September 27, 2020 [7 favorites]

There are 16! (factorial) possible sequences of balloon popping here in the sequence you created - there were 16 choices for the first balloon, 15 choices for the second, etc. However, that's not a complete answer - there are many such sequences that result in the same outcome. As an example, say you picked 1 - 2 - 3 - ... - 16 where 16 is the reveal balloon. You could have equivalently picked 2 - 1 - 3 ... - 16.

For the first choice, there were 15 balloons that you could have picked that would have the exactly same outcome. For the second choice, there were 14, and so on. We have to remove all the sequences that are simply rearrangements of the first 15 balloon popping - we're only looking for the sequence that ends in one particular balloon.

In combinatorics terms, this is a combination of "16 choose 15". As alexei correctly indicates, "16 choose 15" is 1/16.
posted by saeculorum at 10:05 AM on September 27, 2020

Here's another way to think about it: imagine you choose an order to pop the balloons in at random. What are the odds the last balloon is the unique one? Since you have 16 balloons, it's 1/16.
posted by LSK at 10:06 AM on September 27, 2020 [2 favorites]

Here are all the odds:

Probability of hitting the balloon on the 1st throw: 1/16
Probability of hitting the balloon on the 2nd throw: 1/16
Probability of hitting the balloon on the 3rd throw: 1/16
Probability of hitting the balloon on the 4th throw: 1/16
Probability of hitting the balloon on the 5th throw: 1/16
Probability of hitting the balloon on the 6th throw: 1/16
Probability of hitting the balloon on the 7th throw: 1/16
Probability of hitting the balloon on the 8th throw: 1/16
Probability of hitting the balloon on the 9th throw: 1/16
Probability of hitting the balloon on the 10th throw: 1/16
Probability of hitting the balloon on the 11th throw: 1/16
Probability of hitting the balloon on the 12th throw: 1/16
Probability of hitting the balloon on the 13th throw: 1/16
Probability of hitting the balloon on the 14th throw: 1/16
Probability of hitting the balloon on the 15th throw: 1/16
Probability of hitting the balloon on the 16th throw: 1/16

The last line is the answer to your question: 1/16. Note that the probabilities sum up to 1.

The procedure you described generates a random permutation, and the odds of any given balloon occurring on any throw are equal.

> it becomes more likely to hit as the others are popped.

Here you're saying that the conditional probability of the balloon occurring on the last throw becomes higher once you have information that it did not occur on the first few throws. Given that the balloon does not occur on the first throw, the probability of it occurring on the last throw has risen to 1/15. And given that it does not occur on the first two throws, now the probability of it occurring on the last throw rises to 1/14. And so on (until after 15 failed throws you become certain that it must be the last balloon). But prior to the event, when there is no information about what has happened, the probability is 1/16 that the balloon occurs on any given throw.
posted by Syllepsis at 10:16 AM on September 27, 2020 [1 favorite]

So tell me the odds, i.e., the number of times one would have to recreate this event before you could duplicate leaving the one special balloon in 16 to the very last, given random dart throws.

The solution of 1/16 for the probability of the event occurring only maps to a probability of repeating that event after a number of trials. Which I have no clue about how to calculate. Think of it as flipping a 16 sided die. There's no number of throws guaranteed to result in the magic 16. You could roll forever and never actually hit the 16. The best you can get is some sort of probability along the lines of after N trials there's a 99.999999% chance that you'll have hit that 16. But there is no number of trials that makes it certain. Hail Eris!
posted by zengargoyle at 10:24 AM on September 27, 2020

And I don't see how audience information has any bearing. These aren't Schrodinger's balloons. They're not even Monty Hall's balloons. The dart throws are random; you could set this up mechanically.
posted by weapons-grade pandemonium at 10:42 AM on September 27, 2020

A simpler way of thinking about it without worrying about all the intermediate throws: There are 16 balloons that could occupy last place in the popping order. One on them is the pink one. So there’s a 1/16 chance that the pink balloon will occupy that last place.

If you wanted to calculate the odds of the pink balloon being last and all the other balloons being popped in a specific order, that would be different. And I guess if some were easier to pop (eg, if they’re arranged in rows with some in front) that would be different again.

But if you just want to know the odds of one of those 16 balloons occupying last place in the popping order when popped truly at random, it’s precisely that - one in 16.
posted by penguin pie at 10:54 AM on September 27, 2020 [2 favorites]

Response by poster: Yes, that is correct for the first throw. For the second throw the odds are one in 15, and so on. On the 15th throw, the odds are one in two, because there are only two balloons left. It is the cumulative sequence that puzzles me. Syllepsis: all your 1 in 16 probabilities are correct only when you have 16 balloons. That's only the first throw.
posted by weapons-grade pandemonium at 11:05 AM on September 27, 2020

It is the cumulative sequence that puzzles me.
You can work it out the long way to check…

There are 16 balloons: 15 regular and 1 special. I pop one randomly. My odds of popping a regular balloon are 15/16. Suppose the balloon that pops is a regular one.

Now there are 15 balloons: 14 regular and 1 special. I pop another one randomly. My odds of popping a regular balloon are 14/15. Again, I pop a regular one.

Now there are 14 balloons: 13 regular and 1 special. Et cetera, et cetera. Repeat this process until you get to the base case:

There are 2 balloons: 1 regular and 1 special. My odds of popping the regular one on my second-to last throw are 1/2, and I succeed in popping the regular one.

What was the likelihood of all these chances succeeding in sequence, leaving the special balloon for last? It's

(15/16) × (14/15) × (13/14) × (12/13) × (11/12) × (10/11) × (9/10) × (8/9) × (7/8) × (6/7) × (5/6) × (4/5) × (3/4) × (2/3) × (1/2)

…which, as demonstrated earlier in the thread, is 1/16 or 0.0625: see here.
posted by letourneau at 11:15 AM on September 27, 2020 [7 favorites]

Think of it this way:

Imagine you're picking balloons one at a time out of a basket, and you hang them on the wall and when you're done, you will break them all from right to left until you get to the special one. What are the odds that the first one you pick will be the special balloon? 1 in 16, and nothing you do after that makes a difference to whether the first balloon was or was not the right one.

Now break them from left to right instead. What are the odds that the last balloon you break is the special one?

It was still a 1/16 chance that it would be in that one position -- changing the order in which you broke them didn't change that.
posted by jacquilynne at 11:20 AM on September 27, 2020 [2 favorites]

the number of times one would have to recreate this event before you could duplicate leaving the one special balloon in 16 to the very last, given random dart throws

This is an application of the geometric distribution, which is the distribution of the number of binomial trials needed to get to the first success. In this case, one binomial trial is the sequence of dart throws beginning with 16 balloons, and success is defined as the sequence of balloon pops described in the original question. As others have described above, the probability (p) of any trial ending as in the original question is 1/16. The expected value of the geometric distribution is 1/p, so 16 trials would be expected before observing the specified series of dart throws and balloon pops. (This is a random variable so there is uncertainty around this estimate.)

Slight digression and PSA:
Note that this is not the odds of this event. Odds are defined as the ratio of the probability of the thing occurring to the probability of the thing not occurring, or p/(1-p), so 1/15 in this case. Odds are not intuitive. For example, at first glance you'd think an event with odds of 1 in 2 would be based on equal probability of the thing happening or not happening, but for p=1/2 , the odds are 1 to 1. Odds of 1/2 means p=1/3. For small probabilities, odds and probability will be close because 1-p is basically 1.
posted by esoterrica at 12:13 PM on September 27, 2020 [2 favorites]

Yes, that is correct for the first throw. For the second throw the odds are one in 15, and so on. On the 15th throw, the odds are one in two, because there are only two balloons left. It is the cumulative sequence that puzzles me. Syllepsis: all your 1 in 16 probabilities are correct only when you have 16 balloons. That's only the first throw.

This can't possibly be correct, since then the probability of popping the special balloon at some point would be the sum of all these probabilities. But 1/16 + 1/15 + 1/14 + ... + 1 ≈ 3.308, and the overall probability can't be greater than 1.
posted by Johnny Assay at 12:38 PM on September 27, 2020

Product, not sum. Look at that answer closely.
posted by j_curiouser at 1:29 PM on September 27, 2020

Response by poster: When I witnessed the event I thought it must be a more rare occurrence, partly because the suspense built up perfectly for a reveal, but the easiest way for me to accept the 1 in 16 probability is to realize that for every sequence one of the 16 balloons must be last. So I think everyone here has the correct answer. Solved. Thanks.
posted by weapons-grade pandemonium at 1:44 PM on September 27, 2020

Syllepsis: all your 1 in 16 probabilities are correct only when you have 16 balloons. That's only the first throw.

No, you're not understanding Syllepsis' claim. The point is that for any number n you pick in the set {1, 2, ..., 16}, the probability that n will be the number of the special balloon is 1/16. A couple of people have given the calculation, but take a look again at the calculation for the probability that the third balloon broken is the special one:

If the third balloon broken is the special one, then what has happened in the game? Well, we broke two non-special balloons and then the special one. The probability of that happening is (15 / 16) * (14 / 15) * (1 / 14). That is, 15 non-special out of 16 to start, and then 14 non-special out of 15 remaining after one non-special broken, and then 1 special out of 14 remaining after two non-special broken. The 15s and 14s cancel, leaving 1 / 16.

So, as Syllepsis said: The probability that the nth balloon broken is the special one is 1 / 16, for every n in the range of the problem.

You need to be careful about the difference between conditional and unconditional probability. The unconditional probability that the special balloon is, say, the fourth one is 1 / 16. But the conditional probability that the special one is the fourth broken, given that the first three broken are not special is 1 / 13. To get the unconditional probability, you have to multiply by the probability of the condition (that the first three broken are not special).
posted by Jonathan Livengood at 1:50 PM on September 27, 2020 [1 favorite]

There are 16 balloons. One of them has to be last. The chances of any one of the balloons being last is 1 out of 16.

There's nothing else to think about here.
posted by Winnie the Proust at 7:53 PM on September 27, 2020

There are 16 balloons. One of them has to be last. The chances of any one of the balloons being last is 1 out of 16.

...except that the game stops as soon as the gender-reveal balloon is popped. So the gender-reveal balloon will always be the last balloon popped, and that's what makes the last-balloon probability calculation not immediately obvious.

The 1/16 result is correct, but it's not immediately obvious. The kind of reasoning laid out by Johnathan Livengood for the case where the game stops after three throws does need to be thought through. Because there are lots of problems in this general class where the numerators and denominators in successive probabilities don't all tidily cancel out.
posted by flabdablet at 11:39 PM on September 27, 2020

i.e., the number of times one would have to recreate this event before you could duplicate leaving the one special balloon in 16 to the very last, given random dart throws.

My going to sleep thoughts went like this.

Each trial has a 15/16 chance of failure to reproduce the last balloon being the reveal balloon. Two trials have (15/16)^2 chance of failure. Three trials have (15/16)^3 chance of failure.

The chance of success after N trials is: 1-(15/16)^N
```
\$ raku -e '
for 1 .. 50 -> \$n {
say "trial: \$n prob: {1 - (15/16)**\$n }"
}'
trial: 1 prob: 0.0625
trial: 2 prob: 0.121094
trial: 3 prob: 0.176025
trial: 4 prob: 0.227524
trial: 5 prob: 0.27580357
trial: 6 prob: 0.321065843
trial: 7 prob: 0.3634992279
trial: 8 prob: 0.40328052617
trial: 9 prob: 0.440575493281
trial: 10 prob: 0.47553952495127
trial: 11 prob: 0.508318304641818
trial: 12 prob: 0.5390484106017048
trial: 13 prob: 0.56785788493909828
trial: 14 prob: 0.594866767130404636
trial: 15 prob: 0.6201875941847543464
trial: 16 prob: 0.6439258695482072
trial: 17 prob: 0.6661805027014442
trial: 18 prob: 0.687044221282604
trial: 19 prob: 0.7066039574524412
trial: 20 prob: 0.7249412101116637
trial: 21 prob: 0.7421323844796847
trial: 22 prob: 0.7582491104497044
trial: 23 prob: 0.7733585410465978
trial: 24 prob: 0.7875236322311855
trial: 25 prob: 0.8008034052167364
trial: 26 prob: 0.8132531923906904
trial: 27 prob: 0.8249248678662722
trial: 28 prob: 0.8358670636246301
trial: 29 prob: 0.8461253721480908
trial: 30 prob: 0.8557425363888351
trial: 31 prob: 0.864758627864533
trial: 32 prob: 0.8732112136229997
trial: 33 prob: 0.8811355127715622
trial: 34 prob: 0.8885645432233396
trial: 35 prob: 0.8955292592718808
trial: 36 prob: 0.9020586805673882
trial: 37 prob: 0.9081800130319265
trial: 38 prob: 0.9139187622174311
trial: 39 prob: 0.9192988395788416
trial: 40 prob: 0.924342662105164
trial: 41 prob: 0.9290712457235912
trial: 42 prob: 0.9335042928658668
trial: 43 prob: 0.9376602745617502
trial: 44 prob: 0.9415565074016408
trial: 45 prob: 0.9452092256890382
trial: 46 prob: 0.9486336490834734
trial: 47 prob: 0.9518440460157562
trial: 48 prob: 0.9548537931397715
trial: 49 prob: 0.9576754310685358
trial: 50 prob: 0.9603207166267522
```
If you do it once, you have 6.25% of having been successful, if you've done it 50 times you have a 96.0% chance of having been successful. Doing 16 trials only gives you a 64.4% chance, doing 11 gives you a respectable 51%.

...except that the game stops as soon as the gender-reveal balloon is popped. So the gender-reveal balloon will always be the last balloon popped

And a lost item is always in the last place you look for it.
posted by zengargoyle at 6:15 AM on September 28, 2020

the number of times one would have to recreate this event before you could duplicate leaving the one special balloon in 16 to the very last, given random dart throws

The word "could" is doing a lot of heavy lifting here. I mean, you could duplicate leaving the one special balloon in 16 to the very last in one re-creation of this event, with a bit of luck.

All that a probability calculation is ever going to be able to tell you is how many 16-balloon runs you'd expect to see in any arbitrarily large number of re-creations.

To put a probability of 1/16 into some kind of intuitive context, it's exactly as probable as flipping four coins together and having them all turn up heads. And if you want to get a feel for just how unlikely this is, flipping four coins is a process that requires a lot less setup and cleanup than preparing and demolishing endless glitter bombs.
posted by flabdablet at 6:53 AM on September 28, 2020

Let's just run a simulation...
```
\$ raku -e '
my \$N = 100_000;
my %h;
for 1..\$N -> \$i {
for 1..* -> \$n {
if (1..16).pick(*).[*-1] == 16 {
%h{\$n}++;
last
}
}
};
for %h.pairs.sort(-*.value) -> \$p {
state \$sum=0;
my \$r = \$p.value/\$N;
\$sum+=\$r;
once say "throws\tcount\tprob\tcumulative";
say "{\$p.key}\t{\$p.value}\t{\$r.fmt("%f")}\t{\$sum.fmt("%f")}"
}
'
```
100,000 shall we shuffle 1-16 and count when the magic 16 is the last one in the sequence.
```
throws  count   prob    cumulative
1       6260    0.062600        0.062600
2       5870    0.058700        0.121300
3       5476    0.054760        0.176060
4       5154    0.051540        0.227600
5       4810    0.048100        0.275700
6       4515    0.045150        0.320850
7       4271    0.042710        0.363560
8       3964    0.039640        0.403200
9       3768    0.037680        0.440880
10      3466    0.034660        0.475540
11      3212    0.032120        0.507660
12      3113    0.031130        0.538790
13      2915    0.029150        0.567940
14      2692    0.026920        0.594860
15      2532    0.025320        0.620180
16      2392    0.023920        0.644100
17      2247    0.022470        0.666570
19      1964    0.019640        0.686210
18      1909    0.019090        0.705300
20      1902    0.019020        0.724320
21      1753    0.017530        0.741850
22      1656    0.016560        0.758410
23      1483    0.014830        0.773240
24      1407    0.014070        0.787310
25      1301    0.013010        0.800320
26      1246    0.012460        0.812780
27      1119    0.011190        0.823970
28      1111    0.011110        0.835080
30      1004    0.010040        0.845120
29      969     0.009690        0.854810
31      923     0.009230        0.864040
32      801     0.008010        0.872050
33      787     0.007870        0.879920
34      766     0.007660        0.887580
35      726     0.007260        0.894840
38      626     0.006260        0.901100
36      609     0.006090        0.907190
37      576     0.005760        0.912950
39      534     0.005340        0.918290
40      519     0.005190        0.923480
41      463     0.004630        0.928110
44      446     0.004460        0.932570
42      445     0.004450        0.937020
43      409     0.004090        0.941110
45      381     0.003810        0.944920
46      362     0.003620        0.948540
49      313     0.003130        0.951670
47      308     0.003080        0.954750
48      294     0.002940        0.957690
50      276     0.002760        0.960450
51      240     0.002400        0.962850
53      231     0.002310        0.965160
52      214     0.002140        0.967300
56      196     0.001960        0.969260
54      192     0.001920        0.971180
57      189     0.001890        0.973070
55      159     0.001590        0.974660
58      157     0.001570        0.976230
59      155     0.001550        0.977780
60      143     0.001430        0.979210
61      134     0.001340        0.980550
62      123     0.001230        0.981780
64      109     0.001090        0.982870
63      103     0.001030        0.983900
65      100     0.001000        0.984900
66      93      0.000930        0.985830
67      92      0.000920        0.986750
72      78      0.000780        0.987530
70      78      0.000780        0.988310
68      72      0.000720        0.989030
74      65      0.000650        0.989680
69      63      0.000630        0.990310
75      59      0.000590        0.990900
78      59      0.000590        0.991490
71      57      0.000570        0.992060
73      54      0.000540        0.992600
77      54      0.000540        0.993140
79      54      0.000540        0.993680
80      45      0.000450        0.994130
76      45      0.000450        0.994580
82      41      0.000410        0.994990
81      38      0.000380        0.995370
84      31      0.000310        0.995680
83      28      0.000280        0.995960
86      27      0.000270        0.996230
89      24      0.000240        0.996470
87      23      0.000230        0.996700
88      22      0.000220        0.996920
90      20      0.000200        0.997120
85      17      0.000170        0.997290
100     15      0.000150        0.997440
91      14      0.000140        0.997580
92      13      0.000130        0.997710
96      13      0.000130        0.997840
102     13      0.000130        0.997970
99      12      0.000120        0.998090
95      11      0.000110        0.998200
101     11      0.000110        0.998310
94      11      0.000110        0.998420
108     10      0.000100        0.998520
97      10      0.000100        0.998620
103     9       0.000090        0.998710
93      9       0.000090        0.998800
117     8       0.000080        0.998880
104     8       0.000080        0.998960
105     8       0.000080        0.999040
98      7       0.000070        0.999110
115     6       0.000060        0.999170
114     5       0.000050        0.999220
113     5       0.000050        0.999270
119     5       0.000050        0.999320
111     4       0.000040        0.999360
123     4       0.000040        0.999400
127     4       0.000040        0.999440
107     4       0.000040        0.999480
109     4       0.000040        0.999520
110     3       0.000030        0.999550
133     3       0.000030        0.999580
132     3       0.000030        0.999610
112     3       0.000030        0.999640
136     3       0.000030        0.999670
120     3       0.000030        0.999700
147     2       0.000020        0.999720
126     2       0.000020        0.999740
116     2       0.000020        0.999760
145     2       0.000020        0.999780
138     2       0.000020        0.999800
122     2       0.000020        0.999820
125     2       0.000020        0.999840
135     1       0.000010        0.999850
121     1       0.000010        0.999860
169     1       0.000010        0.999870
153     1       0.000010        0.999880
124     1       0.000010        0.999890
144     1       0.000010        0.999900
134     1       0.000010        0.999910
130     1       0.000010        0.999920
171     1       0.000010        0.999930
160     1       0.000010        0.999940
139     1       0.000010        0.999950
106     1       0.000010        0.999960
146     1       0.000010        0.999970
143     1       0.000010        0.999980
150     1       0.000010        0.999990
141     1       0.000010        1.000000
```
If I flipped the four coins looking for heads 100,000 times... one day... OMG there's a 171 in there. 171 tosses looking for that 1/16 chance.

It still comes out that around 11 tries gets you better than 50/50 and 16 tries is call it 2/3.

Seems a fun bar bet would be to have a 16 sided die and bet the rube that they can't roll a 16 in 9 rolls while assuring them that since 16 is 2x8 giving them 9 rolls puts the game in their favor. Slim margins, but house wins by 6%.
posted by zengargoyle at 9:22 AM on September 28, 2020

« Older Now that location doesn't matter, should we...   |   Am I Overreacting To My Boss' Discouraging Comment... Newer »