# eli5 number probability

155 views
0

I have a truly burning question.
If you pick a number 1-10 100 times completely randomly and each number has a 10% chance of being picked each time (so picking a 7 once doesn’t decrease the likelihood of picking it on the next try) why won’t you end up with 10 of each number?

In: 1 You would end up with approximately 10 of each number. But the chances of getting *exactly* 10 of each number are still very small. Because probability doesn’t predict what will happen for exactly n trials. It predicts what will happen over *many* trials, where *many* means, really, the limit as n goes to infinity. The more and more times you repeat the random trial of picking a number from 1 to 10, the closer and closer you’ll get to having 10% of the total from each number. But you only get *exactly* 10% of each in the limit, with infinitely many trials (which is of course impossible to do in real life) It may be easier to see what’s happening by taking a simpler case. Take a fair coin with a 50% chance of landing (H)eads and 50% chance of landing (T)ails and flip the coin 4 times.

After 4 flips, there are 16 possible outcomes:

HHHH / HHHT / HHTH / HHTT / HTHH / HTHT / HTTH / HTTT / THHH / THHT / THTH / THTT / TTHH / TTHT / TTTH / TTTT

Given the same reasoning as in your question, we would expect that for a coin flipped with equal chance of outcomes, we would have 2 Heads and 2 Tails after each flip. In reality, only 6 of the possible 16 cases have this configuration.

This same principle can be applied to your original question. I won’t do the math on it because the total number of possible outcomes is (literally) 10^100 but it works the same way.

Hope this helps. When you pick a number with a 10% chance, the chances of picking the second number, for each number, is not 10% anymore, it’s 10% x 10% which makes it a 1% chance. To ELY5, you pick 1 and then you have 10 choices for the second number, you pick 2 and then 10 choices for the second number… We have 10 choices for the first number and each choice has 10 choices for the second number. 10×10=100. So the chances of picking a 7 and then any number is 1%. Chances of picking a 7 and then a 7? 1%. By picking more numbers, you make more probabilities. Each step you add to pick a number gives you 10 new possibilities per each possibility you had before. So, if you pick a 7, the chances of picking a number between 1-10 is also 1%, however, the chances of picking 7 and 7 again is 1%, but the chances of picking 7 and then any number but 7, is 9%, which is much higher than 1%. (PS: If you want to pick the number 7 a hundred times, the probability of it would be 1 in 10^100) > If you pick a number 1-10 100 times completely randomly and each number has a 10% chance of being picked each time (so picking a 7 once doesn’t decrease the likelihood of picking it on the next try) why won’t you end up with 10 of each number?

It may be easier to see how this works with smaller numbers. You flip a coin twice. It’s either heads or tails, 50% probability, and the two flips are independent. Your question is “why don’t you end up with exactly 1 heads and 1 tails?”

Flip the coin once. Suppose (for the sake of argument) that it lands on heads. You flip it again. There is nothing to influence the coin to land tails here – it’s still 50-50 whether it lands heads or tails.