The idea for this article was inspired by a tweet from WatchItPlayed’s Rodney Smith, while playing the game Roll For It: https://vine.co/v/hTMDn5BjJY7. In fact I just heard from Chris Leder (the designer of Roll For It), that he experienced the exact same luck on his first roll in an early prototype of the game. So what is the probability of rolling six different numbers on six dice? And are people justified in how surprised they are when they see this roll? Whether you are familiar with Roll For It or not, now is a great time to play a few rounds using the free online app at http://www.rollforitgame.com/online.html.

Before we get to the 15pt card from the video, let’s consider the 2pt card that is won by rolling two ones. What are the chances of winning this card, when you have only two dice to roll? Most gamers understand that the probability of rolling a one on a six sided die is 1/6. But we need to roll two ones to win this card. To help illustrate this, lets use the function P(x,y) to describe the probability of rolling at least x ones with y dice. So P(1,1) is the probability of rolling a one with a single die, or 1/6. And P(1,0) is the probability of rolling a one with zero dice. This is impossible, and we’ll call this P(x,y) = 0 whenever x > y. What we’d really like to calculate next is the probability of rolling two ones with two dice, or P(2,2). Note that I’m only using this P(x,y) function as a place holder for probabilities that still needs to be calculated. I don’t intend to derive or use any “magic” formulas for calculating this below.

It’s often helpful to consider rolling dice one at a time to help simplify problems like this. The first die that we roll has a 1/6 chance of being a one, which means that winning depends entirely on the other die. However, the 5/6 of the time that we don’t first roll a one, we cannot win because there’s only one die left to roll and we need two ones to win the card. Here’s how we can write this using the function P(x,y) as described above.

    P(2,2) = 1/6 * P(1,1) + 5/6 * P(2,1)
    P(2,2) = 1/6 * 1/6      + 5/6 * 0    <= simplify P(1,1) and P(2,1) as described above.
    P(2,2) = 1/36 = 2.77%

Can you imagine the 36 outcomes that this probability represents? Think about matching every value on one die with every possible value on the other: 6 * 6 = 36.

Next we can look at a slightly more complex problem. What if we have three dice that we are trying to roll two ones with. That extra die should give us a better chance at winning the card. We'll again start by accounting for the different possibilities that rolling the first die could lead to. We'll roll a one on the first die 1/6 of the time, in which case we only need one more one from the two remaining dice. The other 5/6 of the time, we'll need to roll two ones with our last two dice (and we just calculated this probability to be 1/36). Here's what this looks like using our function notation:

    P(2,3) = 1/6 * P(1,2) + 5/6 * P(2,2)
    P(2,3) = 1/6 * P(1,2) + 5/6 * 1/36   <= simplify P(2,2) as calculated above
    P(2,3) = 1/6 * P(1,2) + 5/216

We still have one more step to break down P(1,2): the probability of rolling at least a single one with two dice (Note: it's ok if we roll extra ones, since the Roll For It cards specify the minimum rather than exact dice necessary for scoring). So if the first and second dice that we roll are ones, we are successful no matter what is rolled on the third die. If the first die is a one and the second die isn't, then we still have the third die and final chance at winning the 2pt card. Here's what this looks like using the function P(x,y).

    P(2,3) = 1/6 * ( 1/6 * P(0,1) + 5/6 * P(1,1) ) + 5/216
    P(2,3) = 1/6 * ( 1/6 * 6/6 + 5/6 * 1/6 ) + 5/216	   <= simplify P(0,1) to 6/6, and P(1,1) to 1/6
    P(2,3) = 16/216 = 7.40%

So that one extra die nearly triples our chances of winning.

Next, lets move onto a couple of the harder cards: starting with the 15pt card that requires rolling all six sixes. The probability of this is P(6,6), since the P(x,y) should be the same whether you're trying to roll x ones versus x sixes. Using this function notation, we can break the problem down as we have been. We can also save ourselves some effort by ignoring the 5/6 chance of any die being a non-six, because we know that the probability of winning with any non-six die is 0.

    P(6,6) = 1/6 * P(5,5) + 5/6 * P(6,5)
    P(6,6) = 1/6 * 1/6 * P(4,4) + 5/6 * P(5,4)
    …
    P(6,6) = 1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 1/6
    P(6,6) = 1 / 46,656 = 0.00214%

Now, let's look at the example from Rodney's tweet that helped motivate this article: the 15pt card that requires rolling one die with each value from one to six. Do you think that this will be harder, easier, or the same difficulty as rolling all sixes with six dice? We can no longer use our P(x,y) function because every die needs a different value to win. However we can still break the problem apart by considering each die-roll in order. Since we need one die with each value to win, the first die that we roll can be anything: 6/6. The second die can be anything other than the value rolled on the first die: 5/6. The 1/6 of the time that the second die actually matches the first die, we lose. So we can leave this zero probability out of our calculations. The third die must be anything other than what was rolled on the first two dice, and so on.

    One Die: 	6/6
    Two Dice: 	6/6 * 5/6 + 1/6 * 0
    Three Dice:	6/6 * 5/6 * 4/6 + 2/6 * 0
    …
    Six Dice:	6/6 * 5/6 * 4/6 * 3/6 * 2/6 * 1/6
    Six Dice:	720 / 46,656 = 1.54%

So there's the answer to Rodney's question, and to our own question about whether this might happen more or less often than we expect. If we expect rolling six different numbers to happen as often as rolling six sixes, we should be very surprised. Rolling six different numbers should happen 720 times as often as rolling six sixes. Remember this next time you play Roll For It, since both cards are worth the same 15pts.

There are certainly many more probabilistic considerations within Roll For It as well as other games that I look forward to investigating in future articles. But this seems like a good place to stop for now. Please leave a comment below, if you're enjoying this series, if you have any questions, or if you have any requests for future topics. I've really enjoyed writing these first two installments, and hope people are finding them an interesting read.




3 Comments

MantaScorp says:

This is utterly incredible, but it makes my brain hurt! :D

Gary Dahl says:

I agree MantaScorp! I was thinking about this while raking leaves, and rolling two straights in a row (0.0238%) is about as likely as rolling any six of a kind with two attempts (0.0257%).

Eduardo says:

Excellent! Keep going… :)


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