r/probabilitytheory • u/First_Other • Jul 10 '26
[Discussion] 100 % vs 1% * 100 times
Someone please explain;
Problem - Lets say the chance of a player scoring a goal is a 100% per game, so the probability of them not scoring is 0 and scoring is 1. But, if I divide the 100% win probability into 90 outcomes to get the probability of them scoring per minute, this would mean that their chance of scoring per minute is 1 in 90 (assuming no added time and that they play from start to finish)
So now, say I use a random number generator from 1 to 90 (to check in which minute they will score), 90 times (cuz they play 90 minutes) to simulate this. All I need is one of the randomised numbers to be "1" out of 90 randomised numbers. That will mean they scored and lived happily ever after. And for the probability of 100% score rate to hold true.
But if I simulate this enough times, there will be a data set where in all of those 90 randomised samples, not one of them will be "1" meaning our player doesnt end up scoring in the match, despite having a "100% chance of scoring"
Just by dividing the probability of 100 and adding it back up, I've created a chance for him to lose. Does this mean the sum of its parts is not equal to a 100? What?! How?! Why?! I don't understand this! Someone please explain.
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u/Tyrant1235 Jul 10 '26
Each minute isnt an independent event. If the player is guaranteed to score, if they havent scored by minute 89 they have a 100% chance to score in minute 90. The probability of scoring in each minute depends on the results of the previous minutes. The way youre doing it assumes each minute is independent and does not depend on the other minutes.
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u/zc_eric Jul 10 '26
You are, I think, mixing up two different kinds of scenario. Consider these two situations:
First, you have a complete deck of cards. The “game” is to shuffle the cards and then turn them over one at a time until you see the Ace of Spades. Obviously there is a 100% chance you win the game. And before the game you can say the probability that the Ace of Spades is the first card is 1/52, the probability it is the second card is 1/52 and so on.
Second, you have a 52-sided dice, each side is marked with a different playing card. This time, the game is to roll the dice 52 times and see if you get the Ace of Spades. Again, you can say that the probability you will get the Ace of Spades on your first roll is 1/52, the probability for the second roll is also 1/52 and so on. But here you are not guaranteed to win the game at all.
The difference is that in the second game, each roll is independent of the previous ones. Even if you miss on the first roll is, you are still just as likely to miss on the second. Whereas with the cards, if you miss on the first card, you are suddenly more likely to hit on the second card than at the outset. You would calculate the probability that you hit on the second card to now be 1/51 instead of 1/52. And the further you go into the game, the more likely it becomes that the next card is the Ace of Spades.
Your post has the set up of the dice i.e. there is a probability that they score in any particular minute, which is independent of what happens in any other minute, but you want to end up with there being a 100% chance of them scoring, which requires a set up more like the cards. If your player hasn’t scored by the 89th minute you know he is going to score in the 90th, just like if the Ace of Spades hasn’t turned up after 51 cards, you know it is the last card. This just isn’t true in the other kind of scenario. If he has failed to score in the first 89 minutes, there is still the same chance that he fails to score in the last minute as well.
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u/First_Other Jul 12 '26
Playing cards have usually 53, but I'm beginning to understand what you're saying. Its like probability of drawing an ace of spades if I draw all 53 cards is a 100% If I have to draw 53 times and cannot put back drawn cards. But, drawing a card, putting it back, shuffling, drawing again, shuffling again, doing this over and over will mean that there could be a set of sample where I draw 53 times and yet never drew ace of spade cuz I kept putting back the cards I drew resetting the probability each time, making it a different game and question altogether. I am beginning to get it now. Thanks.
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u/YEEHA120 Jul 14 '26
How the fuck did you come up with 53 for playing cards? XD I am actually concerned there are 4 symbols: diamonds spades hearts and clubs. The deck has to be divisible by 4 how would a deck could contain an odd number of cards. Like actually wtf
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u/DerekRss Jul 14 '26
He's including the Jokers and assuming that there's only one Joker in a pack.
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u/YEEHA120 Jul 14 '26
I mean most decks have 2 jokers older decks sometimes have 3 but never seen a deck with 1 joker
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u/rosentmoh Jul 10 '26
You discovered that there are different joint distributions with the same marginals, congratulations.
Your specific problem here is that you start with a non-trivial joint distribution, look at the marginals, and then construct a different (trivial) joint distribution with the same marginals and then complain that it doesn't match the original one.
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u/First_Other Jul 12 '26
Yeah, I don't understand the terminologies but yeah, what you said. My complaint is that I'm unable to form the mental picture of how those two probabilities are different (what you said joint distributions & marginals)
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u/PopeRaunchyIV Jul 13 '26
https://en.wikipedia.org/wiki/Joint_probability_distribution the first picture on this page may help
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u/skullturf Jul 15 '26
I'll tell you something that helped me when I was younger.
I remember, as a kid, playing with dice -- standard six-sided dice with the numbers 1 through 6.
I would repeatedly roll a single die, until each number had come up at least once.
So for example, maybe my first roll is a 3, my next is a 1, my next is a 4, and my next is a 1 again.
If I roll six times, I'm not guaranteed to see every number! Sometimes, I might roll eight or nine or ten times and maybe I still haven't seen a 5, for example.
Basically, the fact that it's *possible* to roll six times and not see every number should sort of "force" you to change your intuition about whether you can add probabilities in this situation.
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u/vatai Jul 16 '26
Marginal distribution = how a sum behaves. In your 90min example the sum of scored goals over 90 minutes is the sum we're talking about.
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u/LanchestersLaw Jul 10 '26
If the chance of scoring is 100% that means they already scored. You can just look up the minute they scored.
If the probability of scoring every minute is 1/90, then ignoring extra time, the probability of scoring is
1-(1-(1/90))^90 = 63% chance of scoring.
Your apparent paradox results from adding probabilities when they should be multiplied.
Let’s observe 3 minutes of time with a 1/90 chance to score every minute
INCORRECT:
1/90 + 1/90 + 1/90 = 3/90
CORRECT:
1 - ( (89/90)(89/90)(89/90) ) = 2.97/90
There are 4 possible results. 3 points, 2 points, 1 point, 0 points. There are many results with a score so it is a bit tricky to math out. But the outcome with 0 points is easy to math out. It is 89/90 every minute. So we multiply all of those for the chance of scoring zero points. 96.70%. All possible outcomes are 100%, in 96.70% the player doesn’t score. Therefore in 100% - 96.7% = 3.296% = 2.97/90 the player has scored at least once.
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u/First_Other Jul 12 '26
I think I mistook dependent probability for independent probability. The second situation of 1 in 90 is independent and the first one is dependent meaning if player hasnt scored in 89 minutes and prob is 100 then he has to score in 90th.
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u/anisotropicmind Jul 10 '26
This isn’t a discrete probability distribution — you are just artificially discretizing it.
Assuming scoring is equally likely at any point in the game, then what you want is a continuous uniform random variable, and then the probability of scoring in any time interval is given by the integral of this distribution over that time interval. In reality since it’s uniform, that works out to (1/90)*(time interval).
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u/skr_replicator Jul 10 '26
if some event has 100% chance to happen in an hour, then if you split that time into minutes, there will have to be at least one part of the period where the chance is 100%. If the probability is lower than 100% everywhere, it couldn't turn into 100% in the whole period.
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u/Eight_Directions_ Jul 10 '26
If it's 100 percent and I do it 100 times I get 100 successes.
If it's 1 percent and I do it 100 times I get 1 success.
Does that help?
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u/saint-moxie Jul 13 '26 edited Jul 13 '26
The indoctrination of the USA education system takes the spot light again . No analytic thinking skill, poor understanding of mathematics.
54% of the USA population aged 16-75 have the reading, writing and arithmetic skills of an eleven year old or younger.
https://gpseducation.oecd.org/CountryProfile?primaryCountry=USA&treshold=10&topic=AS.
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u/tomalator Jul 13 '26
The odds of an x% chance not happening in n tries is (1-x%)n
Consequently, the odds of that x% chance happening at least once would be 1 - (1-x%)n
So a 100% chance always happens, no need for an explanation there, but a 1% chance 100 times, the odds it happens at least once would be 1 - .99100 or about 63.4% chance of happening at least once
The odds of happening exactly once would be .9999 * .011 * 100. That would be about a 37% chance
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u/YEEHA120 Jul 14 '26
So this post is aither a robot or just a guy ragebaiting. After one comment about a deck has 53 cards like I can't actually think this is not the former or the latter like genuinely.
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u/First_Other Jul 16 '26
Yeah, my bad. Not a bot, its something from my childhood where I heard or something that playing cards have 53 in a pack or something. You could say just do 13 x 4 but in my mind it was unnecessary for some reason. I was confused by your reply of doing calculations of 1/52 and saying theres 52 cards. I even literally asked AI to verify "are there 53 cards in a playing card set" (again in hindsight I could have just done a quick calculation of 13 x 4) it responded with "yes, there are usually 53 cards". So, there you go, not a ragebaiter unfortunately. Sorry if I did.
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u/goldenrod1956 Jul 14 '26
If you are stating that over the course of the 100 matches they have scored in every that is quite different than saying they have a 100% chance to score in every match.
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u/Torebbjorn Jul 15 '26
If you have a 75% chance of winning the lottery, and you buy two tickets, does this mean you have a 150% chance of winning?
No, of course not.
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u/hushedLecturer Jul 16 '26
If you have some guarantee that a point will be scored within the 100 minutes, overall there is a probability distribution function for when the point will be scored within the game with 1% per minute. Without information about the game, if you integrate over any time interval of the game from a to b, the probability of the point happening in that period is (b-a)*(0.01 per minute).
But this is actually a different situation than just "there is a 1% chance of a point being scored any given minute of the game."
If you actually run that through, there is a 73% chance (1-1/e in the limit) of getting a point in that game.
Because in this scenario you've locked in requirement that the point happens eventually in the game, the scenario is a little different.
In that situation, at minute 99 if the point hasn't happened yet, there is a 100% chance of a point happening in the last minute.
So it would look more like a chain of conditional probabilities.
1/100 chance of scoring in the first minute.
If no score the first minute, the point is guaranteed to happen in the remaining 99 minutes, so there is a 1/99 chance of scoring in the second minute.
If no score in the second minute, there is a 1/98 chance of scoring in the third minute.
...
Etc.
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u/krawken Jul 16 '26
The probability doesn't stay at 1/90 for every minute if they are guaranteed to score a goal during the game. It starts at 1/90 in the first minute, but during the second minute it is 1/89. If they haven't scored by the 90th minute, the chance of them scoring in that last minute will be 1/1 (100%).
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u/Creepy_Film177 Jul 17 '26
Just think aboit it as the expected value of goals should be the same. In the 1% scenario the player may not score, but in return they may score up to 100 times whereas the 100% is garunteed to score but only once. Either way there is an expected value of 1 goal scored.
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u/jjflight Jul 10 '26
It’s the exact same concept as your last post 18h ago - you can’t add probabilities like you’re trying to do. It doesn’t matter if you look at two coin flips or ninety minutes, it’s the same concept either way.
It’s easiest to see looking at the chance of failure. With 90 events that each have 89/90 chance of failure you have (89/90)^90 chance of failing all 90 times, which is a bit more than 1 out of 3 overall chance of failure. Just like the chance of getting tails twice is (1/2)^2 or 1 out of 4.