## Dicey Dice 2

Posted by DP on February 3, 2014 – 9:23 am

There are three fair dice, each numbered 1 through 6; one Blue, one Red, and one Yellow.

You and your opponent will each pick one die and roll. Highest number wins*.

*Special cases:

If there is a tie, and the values are EVEN: Blue beats Red, Red beats Yellow, and Yellow beats Blue.

If there is a tie, and the values are ODD; Switch dice with each other and re-roll.

A Red ‘1′ beats a Blue ‘6′, a Blue ‘1′ beats a Yellow ‘6′, and a Yellow ‘1′ beats a Red ‘6′.

Is there an advantageous strategy to picking your initial die?

### 5 Responds so far- Add one»

### Post a reply

« Dicey dice
444 »

February 4th, 2014 at 9:46 pm

Consider playing with the red and blue dice. In a given roll, there are 17 (= 15+3-1) ways that the blue die would win, 16 (=15+1) ways that the red die would win and 3 ways for an odd tie.

It’s now already obvious that an initial blue player is most likely to beat the an initial red player. Blue has a clear advantage in the first round, and it is relatively rare for a second or more round to be played.

Let P(B) = 17/36, P(R) = 16/36 and P(T) = 3/36 denote the probabilities that the blue die wins, the red die wins or there is an odd tie – each case is for a single roll.

Let P(Blue) and P(Red) denote the probabilities that the initial blue and red players eventually win.

Then P(Blue) = P(B) + P(T)P(Red) and P(Red) = P(R) + P(T)P(Blue)

Substitute the second of those into the first and rearrange to get

P(Blue) = (P(B) + P(T)P(R))/(1 – P(T)^2)

Plugging in the numbers => P(Blue) = ((17/36) + (3/36)(16/36))/(1 – (3/36)^2)

=> P(Blue) = 660/1287 and P(Red) = 1 – P(Blue) = 627/1287

Identical results follow for the other dice. So your strategy is to let the other player choose the first die. If he picks B,R,Y then you pick Y,B,R in that order.

February 5th, 2014 at 8:20 am

Thanks Chris. I was curious about the results, especially because the probability of winning (say with Blue) is only slightly better than losing (say to Red). Exactly 1/36 better. And with the probability of a tie being 3/36 and the advantage switching on the re-roll, I just wasn’t sure what would happen.

I had to keep switching the rules of the problem to where the advantage switched on the re-roll. Otherwise the problem would have been even more simple, and not really worthy of even posting. Maybe it is silly enough it didn’t really need to be on here to begin with, but it’s been slow and I had some time a couple of days ago.

At first, in my head I imagined some infinite series where after several re-rolls the probabilities become (mathematically) 50% for each. I know that wasn’t worded very elegantly. Hopefully you follow what I mean. If not, here’s maybe what I mean:

P(BlueWinsVsRed) = 17/36 + 3/36 (16/36 + 3/36 [17/36 + 3/36 {16/36 + ... }])

Maybe that’s what you have, just with some identities included. Anyway, thanks for acknowledging this one. Keep ‘em coming!

February 5th, 2014 at 12:19 pm

Hi DP. I appreciate that you were posting the problem to keep the site alive – so thank you for the problem.

Your infinite series is correct. But the infinite series contains itself, so:

P(BlueWinsVsRed) = 17/36 + 3/36 (16/36 + 3/36 [17/36 + 3/36 {16/36 + ... }])

= 17/36 + 3/36 (16/36 + 3/36 P(BlueWinsVsRed)

and that is what I got immediately after substituting my P(Red) into my P(Blue), although I didn’t show that intermediate result.

February 5th, 2014 at 2:27 pm

Thanks for the tip. I just wrote it out and came up with the same values as yours.

February 5th, 2014 at 6:09 pm

I’m a nit. I could have used P(Red) = 1 – P(Blue) earlier. Then using

P(Blue) = P(B) + P(T)P(Red), and substituting

=> P(Blue) = P(B) + P(T)(1 – P(Blue))

=> P(Blue) = (P(B) + P(T))/(1 + P(T))

= ((17/36) + (3/36)/(1 + 3/36) = 20/39

=> P(Blue) = 20/39 and P(Red) = 19/39.

I also hadn’t noticed that 660/1287 and 627/1287 reduced to 20/39 and 19/39.