Probability of losing if a 5 comes up (dice problem)

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The probability of losing in this dice game is 1, as the player continues rolling until they roll a 5, which results in a loss. The probability of rolling a 5 on any given roll is 1/6, while the probability of not rolling a 5 is 5/6. Over an infinite number of rolls, the likelihood of never rolling a 5 approaches zero, meaning the player will eventually lose. The discussion emphasizes that the only way to avoid losing is to roll something other than a 5, but this probability diminishes with each roll. Thus, the conclusion is that the probability of ultimately losing is certain.
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Homework Statement


A player goes on rolling one six sided normal die (unbiased) infinite times. But if he gets a 5 then he loses and stops rolling anymore. However until and unless he gets a 5, he continues to roll the die.What's the probability of losing??


Homework Equations


Look, I know that the probability of a 5 turning up is 1/6. But in an infinite number of rolls this can be huge (if the 5 doesn't turn up)!:smile: I mean, we need the probability of losing and not a 5 turning up. So maybe I'll get an infinite G.P.?? You lose when you get a 5. I am really messed up with this problem so, please would anyone help me?? I will be really pleased.
 
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Not sure if it is the right way of reasoning on the problem, but I would say it's still 1/6.

I mean, if you consider the frequentist approach to probability, that 1/6 is not an inner part of the dice, but it's something that arises after a number of rolls n that tends to infinity. Hence, this should be the answer to your question.

Btw, I advice you to wait for more qualified posters. :smile:
 
agnibho said:

Homework Statement


A player goes on rolling one six sided normal die (unbiased) infinite times. But if he gets a 5 then he loses and stops rolling anymore. However until and unless he gets a 5, he continues to roll the die.What's the probability of losing??


Homework Equations


Look, I know that the probability of a 5 turning up is 1/6. But in an infinite number of rolls this can be huge (if the 5 doesn't turn up)!:smile: I mean, we need the probability of losing and not a 5 turning up. So maybe I'll get an infinite G.P.?? You lose when you get a 5. I am really messed up with this problem so, please would anyone help me?? I will be really pleased.

What is the probability that the first toss is not 5? What is the probability that the second toss is not 5? In general, what is the probability that all of the first n tosses are not 5?

RGV
 
If the game is that 'he rolls a single die until he gets a 5 and then he loses', the probability of losing is 1! You say, "we need the probability of losing and not a 5 turning up" but, according to what you say the only way to lose is to roll a 5.
And, apparently, he is not allowed to stop rolling until he gets a five- there is no way to get a 5 except by losing!

As far as the numbers are concerned, you say "I know that the probability of a 5 turning up is 1/6. But in an infinite number of rolls this can be huge". No, it is not. The only way he can not lose on a given roll is to roll anything other than a 5 and the probability of that is 5/6. The probability of rolling a string of 'not fives' n times is (5/6)^n. Multiplying a number less than 1 by itself repeatedly makes it smaller not larger. The limit, as n goes to infinity, of (5/6)^n is 0.
 
HallsofIvy said:
If the game is that 'he rolls a single die until he gets a 5 and then he loses', the probability of losing is 1! You say, "we need the probability of losing and not a 5 turning up" but, according to what you say the only way to lose is to roll a 5.
And, apparently, he is not allowed to stop rolling until he gets a five- there is no way to get a 5 except by losing!

As far as the numbers are concerned, you say "I know that the probability of a 5 turning up is 1/6. But in an infinite number of rolls this can be huge". No, it is not. The only way he can not lose on a given roll is to roll anything other than a 5 and the probability of that is 5/6. The probability of rolling a string of 'not fives' n times is (5/6)^n. Multiplying a number less than 1 by itself repeatedly makes it smaller not larger. The limit, as n goes to infinity, of (5/6)^n is 0.

In my post I had asked the OP to (essentially) perform the computation you did here, guided by a short sequence of simple questions to answer. I had been hoping the OP would do that, but he/she did not respond with answers to my questions, so I don't know whether or not he/she ever got the point.

RGV
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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