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The Problem: Provided there are 50 different dice with 50 sides. What is the probability any two dice or more will roll the same number?


My Attempt:

So, I know the probability of two different dice rolling the same number is 1/50. Since the first roll of die will always roll to a unique number and so the match is determined by the roll of the second die.

Applying same logic for successive probability, I see a series forming

0 + 1/50 + 2/50 + 3/50 + 4/50...

Solving the series I get the answer 49/2.

Please correct the series or Math if I'm doing it incorrectly.

If my Math is correct (Probably not), I don't understand how can the probability of a match is so high, please provide an explanation for this behavior.

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    $\begingroup$ A hint: instead of thinking about the probability of matches, think about the probability that there are no matches among the 50 dice with 50 sides each. $\endgroup$ – EdM Sep 24 '19 at 12:25
  • $\begingroup$ @EdM without doing the math, I'm thinking the probability would be low. $\endgroup$ – Abbas Sep 24 '19 at 12:33
  • $\begingroup$ So does that mean my Math is correct? There's a first, if it is. $\endgroup$ – Abbas Sep 24 '19 at 12:33
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    $\begingroup$ That’s why it can help to do the math. Your series has a problem as it gives a probability greater than 1. But without the math explicitly think about what would have to happen for each of the 50 dice to come up with different numbers, even with 50 possible numbers each. $\endgroup$ – EdM Sep 24 '19 at 12:38
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    $\begingroup$ Note that the probability of 2 dices having the same number is $50 \times (1/50)^2$, the probability of 3 dices having the same number is $50 \times (1/50)^3$. So, regarding your math, the series does not represent the probability of 50 dices having repeated numbers. $\endgroup$ – Ertxiem - reinstate Monica Sep 24 '19 at 12:38
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I believe you misunderstood the nature of probability measures. Recall that probability measures are restricted to the range $[0,1]$, so your answer cannot be correct. As a general rule for further problems, whenever you see probabilites greater than $1$ or lower than $0$ that should ring an alarm bell.

If we assume the result on each die to be independent events, then as you correctly said, the probability of two dice rolls resulting in the same number is $\frac{1}{50}$. This is however not the same as having 3 dice resulting in the same number, as this is $\frac{1}{50^3}*50$. Can you correct your series now?

--Edit---

One way of going around the problem (and my preferred way of resolution) is to think about the problem with complements.

Note that the total probability can be written as the sum of the probabilities of 3 disjoint macroevents

$$1 = P(\text{all rolls are different)} + P(\text{all rolls are the same)} + P(\text{something in between)} $$

We are interested in the last two summands, i.e.

$$1 - P(\text{all rolls are different)}$$

This probability can be calculated in the following way. Suppose the rolls are performed one after the other (this can be done as the rolls are independent). We consider a success if we roll a number different from any previous number.

At the first roll, this probability is certain, that is $\frac{50}{50}$. At the second roll, we will roll a different number with probability $\frac{49}{50}$. At the third roll, this probability will be $\frac{48}{50}$ and so on.

At the end, we are left with

$$1 - P(\text{all rolls are different)} = 1 - \frac{50!}{50^{50}} \approx 1 - 3.42*10^{-21} \approx 1$$

which makes sense. On 50 rolls of 50 sided dice, it is virtually certain you will have at least two numbers being equal.

For a practical visualization, check out this pseudo-random number generator with these parameters. You will see that it's almost impossible to obtain a roll where all numbers are different.

In fact, we can do even better. We know that the random variable that measures the number of times we have to roll 50 dice until they result in all different numbers (if the next time we throw all 50 dice is independent of the previous time) follows a geometric distribution with parameter $p=\frac{50!}{50^{50}}$. For this random variable, theory shows that the expected value of this random variable is $\frac{1}{p}$.

Doing the math, we see that for all numbers to be different, we would have to roll on average all 50 dice about $2.92*10^{20}$ times. Even if you had been rolling a batch of 50 dice every second since the universe was born, you would still have to repeat this process about 1000 times, on average, to have all different numbers.

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  • $\begingroup$ Thanks for your answer, I see the correction in results, but I still don't understand how this equation was created. Can you write a step by step to create the equation? Or a link? $\endgroup$ – Abbas Sep 24 '19 at 12:50
  • $\begingroup$ Edited with a solution $\endgroup$ – Easymode44 Sep 24 '19 at 13:21
  • $\begingroup$ Thanks a lot, the explanation cleared things up for me. $\endgroup$ – Abbas Sep 24 '19 at 17:29
  • $\begingroup$ You're welcome. I have added some extra information and tools. $\endgroup$ – Easymode44 Sep 25 '19 at 9:15

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