Specifically I’m asking about this situation that came up during a D&D game:
Say I’m rolling a 20 sided die, and I do not want to roll a 1. I know that the odds of rolling a 1 are 1/20.
I know that the chances of rolling a 1 twice in a row is (1/20 * 1/20), which is far a lower occurrence.
Say then, before I rolled my “real” roll, I rolled the die again and again until I landed on a 1, then proceeded to roll my “real” roll, would I have reduced the odds of rolling a 1 to (1/20 * 1/20), given that I’ve just rolled a 1 prior?
This is the logic I’m having trouble reasoning about and I’d appreciate it if anyone could clarify what is or is not accurate about the assumptions being made in this scenario.
In: Mathematics
No. Each die roll is an independent event. It doesn’t depend on past rolls.
Some people erroneously use this strategy to pick lottery numbers, thinking that picking a number that hasn’t come up in recent draws has a higher probability of being drawn.
Nothing you do before rolling the die changes the outcome of rolling that die. The chance is 1/20 to get a 1 every time you roll.
When rolling a dice, each roll is totally independent of the rolls before it. So with your first “real” roll, the odds of getting 1 are 1/20 as you say. Then when you roll again, your odds of getting a 1 are still 1 in 20. The odds don’t change because you did lots of fake rolls in between.
If you want to improve your luck in D&D I’m afraid the best option is just to not piss off the DM. ;p
No, you wouldn’t reduce the odds in any way.
Each individual roll is not being influenced in any way by any previous roll, they’re all separate events. The die does not have any kind of “memory” that stores previous roll information.
Also, in D&D, it is considered impolite to roll your dice outside of your turn/being asked to by the DM. The noise and activity tend to distract people from the narrative/events being discussed.
Say you are going to roll the dice twice. Right now you have a 1/20 X 1/20 chance of rolling two 1’s. Now you roll one dice and it lands on a 1. You pick up the dice again ready to roll the 2nd time. You are no longer going to roll the dice twice so you dont have 1/20 X 1/20 chance of hitting two 1’s. You are only rolling the dice once so you have a1/20 chance of hitting a 1. What happens before you roll the dice doesn’t influence future rolls.
This is already well answered, so I’ll just chime in that this is known as “The Gambler’s Fallacy” in statistics. You see it to hilarious degrees in casinos.
People wait to join “hot” craps tables where 7’s aren’t being rolled, pre-roll dice, and think they’re “due” to win a blackjack hand.
I will say, as a former casino manager, that the thought that the odds will change from roll to roll, hand to hand, spin to spin is how casinos make most of their money.
*”To be safer, always bring a bomb with you when you fly. What are the odds of two bombs on the same plane?”*
Your thinking has the same problem as this faux advice. Your decision to bring a bomb has no influence at all on what other people will do. Your other rolls have no influence on what will happen in that particular roll.
The conditional probability of rolling a 1 given that you just rolled a 1 is the same as the probability of rolling a 1 given that you just ate a glazed donut. It’s still 1/20. The die has no memory of the past.
The chance of rolling a 2 and then a 1 are also (1/20)*(1/20). Same with every other number you roll before the 1. Just another way of showing the two rolls are independent.
No, each roll is a separate event, and thus each roll is 1/20. The combined odds of rolling a 1 twice in row is 1/400. However, once you’ve rolled the die the first time and got a 1, you eliminated 380 possible pairings, thus returning your odds to 1/20.
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