Can You Improve Your Dice Roll?

• The first player who fails to improve on his next roll loses.

Image Generated by Gemini Pro

Cole Frederick recently wrote that this interesting problem has stumped him for a while. I certainly can’t solve it, but I want to try to model it a simple version of it, with Netica.

https://sciencespectrumu.com/how-many-cards-until-you-get-an-ace-5f5a10a265e3

I am also stumped by anything resembling a general solution, but I decided to try modeling a concrete version of this with a standard normal 6-sided die. It turned out to be a bit trickier than I thought.

Consider just one round, where you roll d1 and then d2, hoping to improve on the first roll d1. What are your chances of improving?

If you roll a 1 (with d1), then you’re very likely to improve on the second roll (d2, about 83% chance), but if you first rolled a 6, there is no need to roll again. Even the second player doesn’t need to roll either. You’ve lost.

More likely to lose than to win on the first round

So much for just one round. We need to consider all six rolls (the maximum that can occur) and compute the probability of an ever ascending sequence of rolls. Let’s look at the other extreme. What is the probability that each and every roll improves on the previous one? This would mean the sequence of rolls was ascending from 1 to 6. Very unlikely!

The probability that d6 > d5 > d4 > d3 > d2 > d1 is very low indeed!

Let’s build a model with some intermediary nodes, one between two rolls, and ask: is the next roll better than then previous one?

We need six equations just like the one above, between each successive roll of the next dice.

I2(d1, d2) = (d2 > d1) ? yes : no       // improve iff d2 > d1

I3 (d3, d2) = (d3 > d2) ? yes : no

I4 (d3, d4) = (d4 > d3) ? yes : no 

I5 (d4, d5) =  (d5 > d4) ? yes : no

I6 (d5, d6) =  (d6 > d5) ? yes : no     // first failure d(k+1) =< d(k)

Our model asks the question: Will his next roll improve on the previous one? His improvement must eventually fail. So we count the number of failures. Each I node is just a pairwise comparison between two rolls. It asks the question: is the second roll an improvment?

We count the number of failures (the number of NO states) for each local pair-wise comparison and compute their probabilities.

Here is another way of modeling the same problem with a survival-function. Notice that the chances of survival decrease very rapidly.

Each of the better nodes form an increasingly improbable cumulative chain. This is a more global approach

We needed 5 equations that test if each sequence is increasing:

better1 (d1, d2) =

   increasing (d1, d2 )? yes : no       // d2 > d1 ?

better2 (d1, d2, d3) =  

   increasing (d1, d2, d3) ? yes : no       // d3 > d2 > d1 ? 

better3 (d1, d2, d3, d4) = 

   increasing (d1, d2, d3, d4) ? yes : no   // d4 > d3 > d3 > d1 ?

better4 (d1, d2, d3, d4, d5) = 

   increasing (d1, d2, d3, d4, d5) ? yes : no   // d5 > d4 > d3 > d2 > d1 ?

better5 (d1, d2, d3, d4, d5, d6) = 

   increasing (d1, d2, d3, d4, d5, d6) ? yes : no  // d6>d5>d4>d3>d3>d2>d1 ? 

There are at least two natural Bayesian-network decompositions of this puzzle. One model treats the process locally: each roll is compared with the preceding roll, and the run ends at the first failed improvement. The other model treats the process globally: it asks whether the first two rolls are increasing, whether the first three rolls are increasing, and so on. The first model is a stopping-time model; the second is a survival-function model. They produce the same distribution for the run length, but they make different aspects of the problem visible.

Either way, we can summarize our result this way, with a simple model:

Check each adjacent pair left to right; stop at the first failure. Each K is exactly the probability of a strictly increasing sequence of that length.

We are now ready to answer the initial question. We can strip away the parent nodes and just keep the run-length nodes and ask: what is the probability that A wins, given that he goes first?

The nodes of interest before compiling equations

We just need a closed form version of the equation for k1 and k2. ChatGPT gave me this:

P(k1) =

   binomial(6, k1) / 6^k1
   -
   (k1 < 6 ? binomial(6, k1 + 1) / 6^(k1 + 1) : 0)

P(k2) =

   binomial(6, k2) / 6^k2
   -
   (k2 < 6 ? binomial(6, k2 + 1) / 6^(k2 + 1) : 0)

P (A_Wins | k1, k2) = 

   k1 > k2 ? true : false      

The updated network:

The first player is more likely to lose than the second player

Player A goes first. If both players last equally long, A loses, because A is the first to face the next required improvement. The first player is disadvantaged because ties go to the second player.

SOURCES & REFERENCES

[embed]How Many Cards Until You Get an Ace? I Thought the Answer Would Be Simple — It Wasn’t!sciencespectrumu.com