Yahtzee Probabilities and a State Matrix
Date: 12/30/98 at 21:26:47 From: Bruce A Ryan Subject: Rolling dice In playing the game Yahtzee, what are the odds of rolling 5 of a kind within three rolls of the dice?
Date: 01/01/99 at 04:54:08 From: Doctor Pat Subject: Re: Rolling dice I'll start by making some assumptions. First, the goal is only to get five of a kind. If we get two or three or four of a kind we pick up the remaining dice and roll to improve our position, and for two pair or full house we pick up (randomly) the best of two or the best hand, and so on. Given that I think the probabilities on the first roll are : P(5 of a kind) = 1/6^4 P(4 of a kind) = 25/1296 P(3 of a kind) = 125/648 P(2 of a kind) = 75/108 P(no two alike) = 5/54 (here we pick up all five and roll again) This gives a transition matrix from roll to roll equal to the matrix below: from to 1 2 3 4 5 1 5/54 75/108 125/648 25/1296 1/6^4 2 0 5/9* 10/27** 15/6^3 1/6^3 3 0 0 25/36 5/18 1/6^2 4 0 0 0 5/6 1/6 5 0 0 0 0 1 * If all three of the other dice come up the the same value, that's a transition from state 2 to state 3: p = 125/216 - 5/216 = 120/216 = 5/9. ** If all three of the other dice come up with the same value, that's a transition from state to to state 3... but with a different value: p = 75/216 + 5/216 = 80/216 = 10/27 Taking the third power of this matrix gives us the probability of ending in any of the five states. I got the following probabilities for the result after three rolls: P(no two alike) = .00079 P(two alike) = .256 P(three alike) = .4524 P(four alike) = .2447 P(five alike) = .046 If those don't add up to exactly one, the reason is roundoff error. In all, there is only about a 5% probability of getting a "Yahtzee." For more information on transition matrices, please see: Probability Transition Matrices http://mathforum.org/library/drmath/view/54265.html - Doctor Pat, The Math Forum http://mathforum.org/dr.math/
Date: 01/01/99 at 17:45:47 From: Bruce Ryan Subject: Re: Rolling dice Dear Dr. Pat, I received your answer to my question. I had asked: in playing the game Yahtzee, what are the odds of rolling five of a kind within three rolls of the dice? However, I think I misled you. My question actually pertains more to just rolling 5 dice for 5 of a kind. I probably should have written that in a game we have at our social quarters, we are allowed to roll 5 dice just 3 times for a quarter. If you roll 5 of a kind you win the built up jackpot. You are only allowed to roll the dice using all 5 dice each time you roll. In other words, if you roll 3 of a kind on your first roll, that doesn't count. You have to pick up all 5 dice and roll again. What are the odds of rolling 5 of a kind out of these 3 rolls? Thank you very much for your time. Bruce A. Ryan
Date: 01/02/99 at 02:22:48 From: Doctor Pat Subject: Re: Rolling dice Bruce, Boy, that is a lot easier. I'm glad you didn't ask that first or I would not have had the fun of doing all the other stuff. But now about the new situation: The probability, as I said in my first response, is 1/6^4 or about .00077 that you will succeed on any roll. This means that in three tries the probability is 770 out of a million that you will win on the first try. The probability that you will win on the second and third try is almost exactly the same (actually minutely less.) This means the overall probability is about two chances out of a thousand roughly (3*.00077). If it happens to someone more than once, be very suspicious. Good luck, - Doctor Pat, The Math Forum http://mathforum.org/dr.math/
Date: 01/12/99 at 10:16:14 From: Florian Subject: Dice probability - Yahtzee Hi, I'm sorry but I cannot follow your thoughts completely. How did you get the transition matrix? How does it work? How can you get the probability from this matrix? Sincerely, Florian
Date: 01/12/99 at 16:25:04 From: Doctor Pat Subject: Re: Dice probability - Yahtzee Florian, The matrix is called a state matrix, and it gives the probability that you will move from any "state" to the same or another state. In your problem the "states" would be the maximum number of dice showing the same number after each roll. If you draw five circles in sort of a pentagon and put the numbers one to five in them, you can draw a graph of what the matrix shows. If you think of the one as the position of no two alike, this is where you are before you roll. On the first roll, five things may happen: 1. You may roll five different numbers. This would be indicated on your graph by a line from (1) looping back to (1). The probability of this event is given in row 1 (from 1) column 1 (to 1) of the matrix. 2-5. You may have two alike of one or more numbers. This would be indicated on your graph by a line from (1) to (2). The probability of this event is given in row 1, column 2. The same thing happens for three of a kind (from (1) to (3)) - row 1 column 3, etc. After the first roll you may be at any one of the positions. Of course if you are at five, you quit and win. This means the circle (5) only has a loop back to itself, and the probability in row 5 is all zero except for column 5 which has probability of one. On any subsequent roll, the probability that you will move from, say, circle (3) to circle (4) is given by the matrix row 3, column 4. The matrix first row, then, gives you the probability of being in any state after one roll. A great Russian mathematician named Markov showed that if you multiply this matrix times itself, you get the probabilities for positions after two rolls, and if you raise it to the third power, you get the probabilities after three rolls. To get the actual values in each cell, I drew out a tree diagram of every possible position after three rolls and computed individual probabilities, but once I had the transition matrix, all else was easy. You can look this up using the search engine at the Math Forum. Try the search words: Markov matrix Good luck, - Doctor Pat, The Math Forum http://mathforum.org/dr.math/
Date: 05/01/2002 at 21:39:17 From: Phil Lindsey Subject: Yahtzee Odds Would you please check the answer given by Dr. Pat above? Specifically, I think that line 2 in the Transition Matrix should be as follows: To 1 2 3 4 5 From 2 0 125/216 75/216 15/216 1/216 I will be able to sleep better if I know I am not going crazy! Thank you. Phil Lindsey
Date: 05/02/2002 at 14:38:40 From: Doctor Ian Subject: Re: Yahtzee Odds Hi Phil, The first difference is that you get 125/216 for the probability of going from 2 of a kind back to 2 of a kind, while Dr. Pat gets 5/9. I think your reasoning is that we already have two of some value, so the next three dice all have to be different from this value: (5/6)^3 = 125/216. However, this doesn't take into account that if all three of the dice come up with the same value, you're no longer going from state 2 to state 2, but from state 2 to state 3. So you have to exclude the possibility that the three dice come up with the same value. That can happen in 5 ways, so the probability is 125/216 - 5/216 = 120/216 = 5/9 The second difference is that you get 75/216 for the probability of going from 2 of a kind to 3 of a kind. I think your reasoning is that we already have two of some value, so one of the next three dice has to be the same, and the other two have to be different. That can happen in three ways, so: (1/6 * 5/6 * 5/6) + (5/6 * 1/6 *5/6) + (5/6 * 5/6 * 1/6) = 3(1/6 * 5/6 * 5/6) = 75/216 However, suppose that all three dice turn up the same? Then we've also moved into state 3... but with a different value! The probability of rolling 3 of a kind with three dice is 1/216, and we have five possible values to choose from, so the transition probability is 75/216 + 5/216 = 80/216 = 10/27 Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
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