Need a quick double-check on probability formula for 100-sided die

  • #1

Main Question or Discussion Point

Hey all, I have an interesting probability problem. I think I know what the correct formula is to solve it, but my probability is quite rusty and I could use some advice. Here's the problem:

Imagine that you have a 100-sided die, and you're rolling to determine if event X happens. Event X requires two conditions to be filled; first, you must roll between 1 and 50 on the die (boundaries included). Once you have done that, you must roll a second time, and that result must be between 1 and 60 (boundaries included). So, Roll One must be between 1 and 50, and Roll Two must be between 1 and 60. You have two chances to roll Roll One. That is to say, if your first attempt on Roll One is not 1 through 50, you may take a second attempt. This attempt is taken only if you fail on the first attempt at Roll One. If you succeed at either of these attempts, Roll Two is taken. Roll 2 may only be taken once.


So here's the way I'm looking at this: I'm calling the probability of success on a single roll of Roll One "Y", and the probability of success on Roll Two "Z". I believe that means that the probability of passing Roll One and Roll Two on the first attempt is simply (YZ). From that, I'm calculating the probability of failing Roll One once, but passing Roll One and Roll Two on the reroll to be the failure chance of the first Roll One times the success chance of the reroll: (1-YZ)(YZ). If I'm right so far, then I think the overall chance of success should be the probability of success on the first Roll One times the probability of success on the reroll: (1-YZ)(YZ)^2.

Is this correct? And for bonus points, what would the formula be for calculating the probability of event X being fulfilled N times in a row?

Thanks!
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hey Captnhaggis! Welcome to PF! :smile:
So here's the way I'm looking at this: I'm calling the probability of success on a single roll of Roll One "Y", and the probability of success on Roll Two "Z". I believe that means that the probability of passing Roll One and Roll Two on the first attempt is simply (YZ). From that, I'm calculating the probability of failing Roll One once, but passing Roll One and Roll Two on the reroll to be the failure chance of the first Roll One times the success chance of the reroll: (1-YZ)(YZ). If I'm right so far, then I think the overall chance of success should be the probability of success on the first Roll One times the probability of success on the reroll: (1-YZ)(YZ)^2.
Sorry, but this is very confused. :redface:

You seem to be making the probabilities smaller (by multiplying them) when they should be getting bigger.

You need to write P() for probabilities, with the symbol for the event inside the bracket …

then you can add the Ps, or multiply them, and we can see what is going on.
 
  • #3
Oh, right, sorry! :-p Been a long time since I took probability, I forgot the way things are formatted.

What I'm saying is this:


P(X occurs without reroll) = P(Roll One)*P(Roll Two)

And

P(X occurs on the reroll) = (1-(P(Roll One)*P(Roll Two)))*(P(Roll One)*P(Roll Two))

Thus

P(X) = (1-(P(Roll One)*P(Roll Two)))*(P(Roll One)*P(Roll Two))^2


I'm not sure if I can make it much simpler than that. :-D And I believe you're correct, the overall probability of success should be the sum of the probability of each success condition, not the product. I'll have to revise that!

Hope I've made my question clearer!
 
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  • #4
tiny-tim
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P(X occurs without reroll) = P(Roll One)*P(Roll Two)
No, it's just P(Roll One), isn't it?
P(X occurs on the reroll) = (1-(P(Roll One)*P(Roll Two)))*(P(Roll One)*P(Roll Two))
The first bracket I understand, you got it from the previous (albeit wrong) equation, but the rest I don't understand at all. :redface:
 
  • #5
The reason that I say P(X without reroll)=P(Roll One)*P(Roll Two) is because event X requires that Roll One be 1-50, and that Roll Two be 1-60.

It's like flipping a coin twice and betting that it will be "heads" both times. Each flip has only two possibilities, both of which are equally likely (50% chance of "heads"). Therefore, the probability that the two flips will both be "heads" is (50%)*(50%)=25%. As proof, if you flip the coin twice, there are only four possible outcomes: Heads/head, heads/tails, tails/heads and tails/tails. Since heads/heads is 1 possible outcome out of 4, the probability of heads/heads is 25%.

The difference in this case is that instead of two 50% chance coin flips, we have one 50% chance die roll, and one 60% chance die roll. Unless my probability is completely wrong, the probability that the first roll will be 50 or under, and the second roll will be 60 or under is equal to the product of the two probabilities (50% * 60% = 30%).

The complication comes in when we state that if the first roll is not 50 or below, it may be re-rolled once. This changes the overall probability of the system, since it allows for the possibility of overall success even if the first roll is a failure. I'm calculating the probability of overall success after having re-rolled to be the base probability of success (50% * 60%) multiplied by the failure chance of the first roll (since the re-roll doesn't happen unless the first roll fails), which is 100%-(50%). That gives (50% * 60%)*(100%-50%)

With all of that in mind, the overall chance of succeeding should be the sum of all the success conditions: probability of success without the re-roll plus the probability of success with the re-roll. That gives (50% * 60%)+((50% * 60%)*(100%-50%) = 45%
 
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  • #6
tiny-tim
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hmm … I've been misunderstanding the terminology :redface: … let me start again …
P(X occurs without reroll) = P(Roll One)*P(Roll Two)
Yes, that is correct. :smile:
P(X occurs on the reroll) = (1-(P(Roll One)*P(Roll Two)))*(P(Roll One)*P(Roll Two))
No, that first bracket is wrong.
 

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