Poker dice probability straight3/31/2023 ![]() Right now I am pretty sure my method of counting is flawed given that I am allowing for a roll to count as a lesser hand, but I'm also not sure why exactly nor how to remedy that problem. Additionally, the numbers I get for rolling 6 total dice make a lot of sense, and rolling an extra die is strictly better than rerolling a die, so P(Reroll Pair) must be between 90.74% and 98.46%. For example, if I roll a hand of, reroll my and get another, that's a failure case, so rerolling a die shouldn't be somehow guaranteed to get you a pair. It's not too surprising that an error could result in this after thinking about it, since the number of ways to reroll into a particular hand should be the same, but there is a simple counter-example where even after rerolling your dice, you could still fail to get a Pair. Well, not totally ridiculous, but they are always double the probability of getting that hand without a reroll. I then add the final quotient from #3 to P(Hand) and I get. Step 2's Result / 6 => probability that rerolling the right die resulted in the right face coming up Step 1's Result / 5 => probability that we rerolled the right die ![]() ![]() Of rolls that could be rerolled into this hand / total number of possible rolls => probability of getting the roll that led to this reroll (1/7776 * number of rolls) Then, to calculate the final probability, I take the original probability of getting that hand P(Hand), and then I calculate the probability of a rerolled hand as the number of rerolls that resulted in that hand divided by: the total number of possible 5 die rolls you can get times 5 (the number of dice in the hand) times the number of faces on the die. Those hands are then tracked separately from the original hands. So, for every possible roll, I iterate through all the dice and assign them a new die face, then check to see what hands the new roll counts towards. To calculate the probability of getting a certain hand after a reroll, I added the probability of getting the hand on your first roll to the probability that any roll could have a single die rerolled and result in that hand. This is proving harder to do than I thought though. I have P(Straight) so all I need to do is calculate P(Reroll Straight). I envisioned the probability of getting, for example, a Straight to be P(Straight) + P(Reroll Straight). I have written a program that will just iterate through every possible die roll and determine which hands it qualifies for, then spits out the probabilities of getting that hand based on how many different combinations of dice could create it. ![]() Because hands are not exclusive, they add up to >100%.Ĭalculating the odds of getting a hand with extra rolls is pretty easy. Here is a list of the hands I am counting, and have put in the description what hands they map onto in traditional poker dice, along with the probabilities of getting them using the system above. For example, a Five of a Kind counts as a Four of a Kind, Three of a Kind, Two-Pair, Pair, and Flush. ![]() To make things a bit more difficult, I am counting a hand of poker dice as any hand lower than it that it qualifies for a Pair is any hand where two die faces share a value. I want to know the probabilities of getting poker dice hands (plus a "Flush" which is all odd or even) when accounting for rerolls and extra rolls. ![]()
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