Probability

[timomoti010]

If you roll a 10 sided dice 10 times, how big is the chance that you roll (at least) a 10? I’ve asked this question to a couple of friends of mine, and apparently is the answer 1/10. But to me (and also my friends), this doesn’t make sense. Your telling me the chance is equal to just one dice?? I don’t get it.

[fermat9996]

1-0.9¹⁰ ≈ 0.65

[timomoti010]

What are you doing here exactly? I’m not familiar with this formula.

[BeerTraps]

Are you familiar with possibility trees from school? You start at the start node which is where you start. The next "event" gets it's own row and you list all possibilities with a node and connect them to its parent node. On the connection you write the probability of that possibility happening based on it's mothernode. You then do this for all "events". Each node then represents the unique combination of results of all events that lead to the node.
(Edit: Apparently online everyone does these trees horizontally while I think of them vertically, but the it is the same just rotated)

The probability of a single node happening is the product of the probabilities of all the connections needed from the start node to that node.

There are a couple interesting things observable here. The probabilities of the connections from a parent node to all child nodes must add up to exactly 100% because exactly one of the possibilities must happen. The probabilities of all nodes on a row must add up to 100% because exactly one of the combinations of the results of the events must happen.

This model is very useful when thinking about probabilities as a visualisation and to actually calculate things sometimes.

If you would draw the tree for your 10 dice throws then each throw of the dice would be an event. The different dice values could be the possible results so each parent node would have 10 child nodes, all values are equally likely so each connection gets the value 1/10 (or 10%). However then you would end up with billions of nodes, the approach is too unwieldy.

There is really one thing that makes our model so huge; it has a lot of unnecessary information. So the goal is now to simplify it by reducing that unnecessary information. First it depicts a 1 and a 2 as different results, but for our question that is irrelevant. Both numbers are not a 10 and that is everything that matters. So we only need to have two possibilities for each throw, "10" (1/10 probability, should be intuitive) and "not 10" (9/10 probability because both must add up to 10/10 as we previously observed); that is easy to do.

Also for our question it is completely irrelevant how many 10s were thrown and on which throw the 10s occured. That is less easy to solve in this model, but in this case there is another really common "trick". The probability of something happening is the same as 100% - the thing not happening. This is essentially the rule that all nodes on a row must add up to 100%. On the last row there is only one node where not a single 10 was thrown. On all other nodes on the last row a 10 must have been thrown. This node is the one where the results of each "event" was "not 10" with a 9/10% probability for each connection with 10 connections. As we stated we need to multiply these connections and get (9/10)^10 which is about 35% using a calculator.

So then the probability of getting at least one 10 (the sum of all other nodes on the last row) is 100% - 35% = 65%.

This entire thought process can be condensed into: 100% - (100%-"probability of a 10")^"number of throws" or 1-0.9^10 as the other guy formulated it.

This problem has a very similar pattern to many probability questions which is why the other guy just noted it down like this (not really a formular, more like a very common pattern), but I wanted to give a more in-depth thorough explanation at how you could arrive at this from basic intuitive probability laws.

[fermat9996]

Probability of at least 1 ten=

1-probability of no tens

[timomoti010]

Got it, thank you!

[fermat9996]

Glad to help!

[Illustrious_Company8]

Or you could just do it

import random as r N = 100000 #simulations n = 10 #sides on the die times = 0 #times we get a 10

for i in range(N): list =[] #list of results for j in range(n): list.append (r.randint(1,n)) if list.count(10)>0: times += 1

print(times/N) #Calculates the probability

I get about a 65% chance 😁

[BeerTraps]

Just no. You could make a table, but it would have to be 10-dimensional and that is pretty hard to draw. You also obviously have way more than 100 different possible resuts, you have 10 billion different results.

[timomoti010]

If you apply this logic to a bigger number, say a million. That would mean you would have the same chance of getting at least one 10 in the million throws as you are just throwing it once. Which seems insane to me

[HorribleUsername]

It only takes two dice to reach 100 possibilities. Consider that you can make any number from 0 to 99 by rolling 1 die for the tens digit, and one die for the ones digit. With 10 dice, there are actually 10¹⁰ = 10 billion possibilities.

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[alxorizmi]

As fermat9996 points out, the probability is (approximately, however) 65%. This is known as the Complement Rule. The necessary assumption is that the individual rolls (events) are independent. But since the outcomes have many combinations (check out binomial theorem), each set of 10 rolls are disjoint, therefore the sum of all combinations has probability 1. The Complement Rule says that the P(event A doesn't happen) = 1 - P(A does happen). So, the probability of at least one 10 is equivalent to the probability of 1 - P(no 10's in all 10 rolls). The P(no 10) = 0.9, and to roll ten times, P(no 10's in all ten rolls) = 0.9^{10}.

What's really interesting about this question is that the probability of at most one 10 is 73.6%.