Help me calculate the odds
September 27, 2020 9:54 AM
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.
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.
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
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
posted by LSK at 10:06 AM on September 27, 2020
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
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
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
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
Obviously I am not asking about certainty; I'm asking about probability.
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
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
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
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
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
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
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
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
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
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
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
posted by j_curiouser at 1:29 PM on September 27, 2020
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
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
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
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'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
...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
...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
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
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...
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
$ 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 1.000000If 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
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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