Reroll Token Randomizer

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infinitejester00
Posts: 2
Joined: 22 March 2015, 05:11

Reroll Token Randomizer

Post by infinitejester00 »

Is anyone else noticing that often times reroll tokens produce the exact same results? About 90% of the time rerolls produce almost all the same numbers, sometimes on up to three dice at a time.
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Cyan100
Posts: 3
Joined: 06 February 2024, 08:56

Re: Reroll Token Randomizer

Post by Cyan100 »

I had this happen a lot with my physical copy with dice, too. So I really think that's only a probability/bias thing. Make a list of your results from a few hundred rolls and we can see if there is something off.
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Earthboundia
Posts: 74
Joined: 01 August 2022, 04:53

Re: Reroll Token Randomizer

Post by Earthboundia »

It happens to me a lot in-person as well. I think it's a case of perception bias. I notice it less now, but I rely much less on re-rolls than I used to.
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Sparge42
Posts: 15
Joined: 07 September 2022, 14:40

Re: Reroll Token Randomizer

Post by Sparge42 »

For what it’s worth, (I think) the probabilities are:

Rerolling one die and getting the original value: 1-in-6
Number of outcomes while rolling one die = 6^1 = 6
Number of favorable outcomes = 1

Rerolling two dice and getting the original two values: 1-in-18
Number of outcomes while rolling two dice = 6^2 = 36
Number of favorable outcomes = 2

Rerolling three dice and get the original three values: 1-in-36
Number of outcomes while rolling three dice = 6^3 = 216
Number of favorable outcomes = 6

Rerolling three dice (of three different values) and having at least two of the original three values come up is: 1-in-9
Number of outcomes while rolling three dice = 6^3 = 216
Number of favorable outcomes = 24

Disclaimer: I am not a mathematician, nor do I play one on television.
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Jellby
Posts: 1452
Joined: 31 December 2013, 12:22

Re: Reroll Token Randomizer

Post by Jellby »

Sparge42 wrote: 28 April 2024, 21:17 Rerolling two dice and getting the original two values: 1-in-18
Number of outcomes while rolling two dice = 6^2 = 36
Number of favorable outcomes = 2
Except if the original two values are identical, then the number of favorable outcomes is only 1. The need to consider special cases like these is what complicates the calculations.
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