![]() ![]() The probability of rolling one of a kind, another die to match it on the second roll, and then a three of a kind on the third roll is (6!/7776) x C(4, 1) x (100/1296) x (1/216) = 0.01 percent.The probability of rolling a pair, matching an additional die on the next roll, followed by matching the last two dice on the third roll is 6 x C(5, 2) x (100/7776) x C(3, 1) x (25/216) x (1/36) = 0.74 percent.The probability of rolling a pair, matching an additional pair on the next roll, followed by matching the fifth die on the third roll is 6 x C(5, 2) x (100/7776) x C(3, 2) x (5/216) x (1/6) = 0.89 percent.The probability of rolling three of a kind, matching an additional die on the next roll, followed by matching the fifth die on the third roll is 6 x C(5, 3) x (25/7776) x C(2, 1) x (5/36) x (1/6) = 0.89 percent.The probability of rolling a single die, then nothing matching this, then matching with the correct four of a kind on the third roll is (6!/7776) x (625/1296) x (1/1296) = 0.003 percent. ![]() The probability of rolling a matching pair, then nothing, then matching with the correct three of a kind on the third roll is 6 x C(5, 2) x (100/7776) x (125/216) x (1/216) = 0.21 percent.The probability of rolling three of a kind, then nothing, then matching with the correct pair on the last roll is 6 x C(5, 3) x (25/7776) x (25/36) x (1/36) = 0.37 percent.The probability of rolling four of a kind, then nothing, then matching the last die on the last roll is 6 x C(5, 4) x (5/7776) x (5/6) x (1/6) = 0.27 percent. ![]() I apologize for this undoubtedly overly pedantic and silly post (already covered by heehaww's post), but I thought there might be some value in spelling this all out in painstaking and obvious detail. In summary, the true odds (against) associated with an event that is likely to occur with probability P are given by :1 (against). And the true odds for an event that is likely to occur 25% are 3:1 against. So we confirmed that an event having 3:1 true odds (against) is likely to occur 25% of the time. Let's confirm this with our example from above by plugging in P = 0.25 into that equation for T. Some event that has T:1 odds against is likely to occur with probability P equal to 1/(T+1).Īpplying some simple algebraic manipulations, we can see that this relationship can be written "in the other direction" (that is, T in terms of P). Okay, so it is easy to go from odds to probabilities. I hope it is patently obvious if some event is 3:1 against, that means that it is likely to occur 1 time in every (3+1) total occurrences. What is the inherent underlying probability associated with this event? So what are the correct odds of selecting a spade from a regular standard deck of 52 cards?Ĭlearly, the odds are 3:1 against (again, obviously, there are three other suits that are "against" your one suit of spades).Ī probability of 1/4 or 0.25 or 25% is clearly equivalent to 3:1 odds (against).Ĭoming at it another way, suppose you are told that some event has 3:1 odds (against). Now, many people, especially in a gaming or betting environment, speak of "odds" rather than probabilities. This is easily seen and easily understood to be 1/4, since, of course, spades is one suit out of the four suits.ġ/4 = 0.25 = 25% (all of this is super obvious) Suppose you know that the probability of selecting a spade from a regular standard deck of 52 cards is 13/52 (of course). Sometimes an example makes things crystal clear. Since their are two extra cards, does that mean I need to re-calculate every poker hand? But because I am including the Jokers in this game, does the values change? For example, I've seen the probability of getting Royal Flush is ~ 0.000154%. I've seen on the internet others calculating the probabilities of all poker hands but they are based around the 52 card deck. I also had a crack at the 5 cards of same colour and got this (please correct if wrong - still an amateur) So I had a crack at the first question I asked about AAAAJ and got 0.0000632% probability. Remember the total number of possible 5-card hands that you derived above.ĭivide the first number by the second number.Īwesome guys! I'm really having a lot of fun using maths to bring this game to life, the universal language. How many ways are there to have a "qualifying" hand (5 red cards)? Let's tally it for red, and then multiply our result by 2 to include black (obviously the prob of 5 black cards = prob of 5 red cards). ![]() And the formula is what you typed above.ĭo you mean either all 5 red cards or all 5 black cards (with no jokers)? Yes, I and many other people write this as C(54,5) where the C is referred to as the "Choose" function. ![]()
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