Challenge Questions

seems to me the best stratigy is to "roll the dice over it's edge" on your turn to 3 unless the 4 is showing and in that case roll it to the 2...... this seems to allow the most latitude for adjustments around the final roll .... at least when I played out many possiblities on an excel spreadsheet.

It's sort of like the bat throw where you work your fist to the top of the bat and have the greatest flexibility for adjustments if you've "paced yourself" during the process.

.....wasn't much good at probabilities in college so can't answer the second part of the question..... by the way Irvin this is a great challenge question...

The die has 4 other sides visible. If the players can move to see all sides, or if they are allowed to rotate the die around its vertical axis to see all sides, all they have to do is locate the side with the lowest number, and turn that side up. When sum is 27 and the 1 is on top, or the sum is 28, the other player cannot make a move without the sum going over 28.

According to Wikipedia, the sides are usually numbered so that opposite sides total 7. This knowledge allows a player to determine the sides he or she cannot see, so rotating the die would not be needed.

If the 6 is the first number to show, then the 2 is visible because the one is covered up. When the 2 is turned up, it exposes the 1. The players alternate between the one and the two until the sum equals 27 or 28. The progression of sums would be: 6, 8, 9, 11, 12, 14, 15, 17, 18, 20, 21, 23, 24, 26, 27. The one is showing, and the first player wins.

This is just one example of this strategy in action. I haven't yet worked out the math to express this as a formula instead.

The observations on the possibilities are good, but the strategy seems flawed - if you presented me with 24 (and one on top) I would roll the die to place 4 on top (giving 28) and you would lose.

Unless there is some crucial simplification, this looks to me to be a horribly complicated problem that needs to be worked backwards, classifying each position as winnable or not winnable. Maybe a pattern will emerge by the time you turn back to a total of 14, but I can't yet see how this might work. In the meantime, I'm hoping someone will come up with the simplification...

"The die has 4 oth----"

Thank you!

a) What is the winning strategy?

- the only winning strategy i could think of is to be the player not to roll the dice or not to be the player to hold the dice first. That way you're not the first one to have something to add up to and you got to have the chance of having the smaller number to start with.

b) What is the probability for the player throwing a dice to win the game if both players make the best possible moves?

- its the 1 out 6 chance. that is when he got to roll the dice at 1

I think the question is a little confusing with the terms "throwing the dice", "turns the dice over one edge", "flipping the dice over one edge". Secondly, which one of the above moves the players make alternatively?

It would be better if you can give an example with 5-6 moves of dice (need not be the best moves), may be starting with 5 on top. I guess this is a straight forward question.

OK here is my take on the game (Hope it helps you to understand how it works):-

Player 1 throws the die with initial possible outcomes 1, 2, 3, 4, 5, or 6 all possible.

Let us say for argument's sake a 4 is showing.

Now if 4 is on top then 3 is on the bottom (As opposite faces sum to 7 on a standard die)

Now Player 2 must roll the die about 1 straight edge essentially putting one of the available 4 numbers on top. This means that player 2 must leave either 1, 2, 5 or 6 on top (3 and 4 are not possible because 4 is on top and 3 is on the bottom and player 2 must only roll once about one edge).

Let's say he chooses to leave 6 on top (1 is now on the bottom and Total score is now 10).

Now player 1 must roll the die about 1 straight edge essentially putting one of the available 4 numbers on top. This means that player 1 must now leave either 2, 3, 4 or 5 on top.

Let's say he chooses to leave 5 on top (2 is now on the bottom and Total score is now 15).

They continue alternately until the player who makes the total 29 or larger is the looser.

My basic strategy would obviously be for me to try and reach a total of exactly 28 at some stage on my roll. Regardless of what numbers are available to my opponent's roll, he will now loose for certain as any number available on the die makes him/her the one to exceed 28.

Another winning move for me is to reach a total of 27 while showing a 1 or 6 on top.

(My opponent now can only roll a 2, 3, 4 or 5, all of which makes him/her loose)

Another winning move for me is to reach 25 using a 3 or a 4.

Also a winning move for me is reaching 24 with a 3 or a 4 and so on.

My gut feeling right now regarding the probability is that player 1, who throws the die, should have zero probability of winning if they both play the best possible strategy.

My reason for thinking this is that player 2 has the ability (albeit limited to 4 different numbers each time) of controlling the sum in his/her favour after each play.

If the first player(A) throws the dice with a '1' on the side, the second player(B)should turn the dice so the '1' is on top ? A is forced to keep providing this opportunity, and so A looses (he gets the initial throw of >1 plus additional points >1). Maybe....I think dice can be left or right handed, and I haven't figured what happens if the initial throw doesn't produce a '1' on the side face. Don't go first sounds better to me (2/3 chance of '1' on the side).

Obviously the solution requires backtracking. I present here my analysis, which I hope it's not ridiculously wrong, as I spent too few time for it! At least it might provide some insight to others to reach the solution.

So, let's consider two players A and B, and say A is to play. When is he certain to win? Its easy to see that if player B has left him a total of 26, 25, 24 or 23, A can then make the sum 28 virtually winning the game.

But also if he gets a 27 and 1 is playable (i.e. on one of the four vertical sides), or he gets 22 and 6 is playable he still can win. To summarize, the winning sum A wishes to see before his turn is:

27&1, 26, 25, 24, 23, 22&6

Going back one step, B is supposed to avoid leaving one of the above to A. To make story short, B wishes to see one of those before his turn:

20&1, 19, 18, 17, 16, 15&6

in which case he can leave either 21 or (22 & not6) to A, virtually B wining the game.

I have no time to check it more thoroughly, but it seems the pattern is the same, only decreased by 7 on every step backwards. In this manner:

A wins if before his turn he he finds: 13&1, 12, 11, 10, 9, 8&6

B wins if before his turn he he finds: 6&1, 5, 4, 3, 2, 1&6

The first and last cases are invalid of course, so the last line becomes:

B wins if before his turn he he finds: 5, 4, 3, 2

So, if A (who apparently is the one who initially casts the dice) throws 1 or 6 loses, otherwise he wins, i.e. the one who starts the game has a 4/6 chance to win the game.

I think it is a good method, but there seems to be an important bit of the detail that is not always fully considered.

For example, if you are left with 25 and there is a 3 or a 4 showing, you cannot turn to 28.

I think also the the person who throws the die is not the person who makes the first move to an adjacent number.

I've been through to see if I can establish a good pattern, but nothing emerged. It's quite likely I have made a mistake in the tracing, so I shall go through it again this evening.
In the unlikely case that I have everything correct, and assuming that both players make the optimum moves, the person who first throws the die will lose unless the thrown die settles showing 1 (i.e. he has a 1/6 probability of winning).

For example, if you are left with 25 and there is a 3 or a 4 showing, you cannot turn to 28.

Indeed. Silly me...

No, but you could turn to 27 and cover the one so your opponent could not stay below 29.

G.A., as that looks right to me. Perhaps (1,2) to (1,6) should also be marked to clarify the repeat structure for totals of 26 and lower, but that is just a niggle.

I think the grid is correct, even though the text was confusing. In order to win you have to turn the die so that your opponent starts from a condition described by a squares with a cross, and that if you start from a position without a cross you can always turn the die to do this

GA - Nice job. Seems that the best stategy at this point is to have your opponent roll the die.

Jim,

You're bang on the money. GA from me too.

Excellent Solution and Excellent Challenge

Nice Job Jim (GA)

Also a nice Wikipedia-proof challenge

a)Strategy is the end game from 21 to 28 in leaving your opponent no move to make the 28 total or leaving him with only a move that will go over. Turning the winning spots of the die down and out of play for your opponent.

Was there more to come?

Best strategy

At any point between 20 and 22 (inclusive) make the largest turnover possible for the have best possible chance of winning.

Since any roll over 23 can go over 28 (23 + 6=29), and the closer to 28 results in a higher the probability of going over.

When a player is just below 23 he should turn the dice over (largest number)to come as close to 28 as possible. The next roll will most likely go over for the other player.

It seems to me that the winning strategy is to maximize the total number count on the first 2-3 turns of the die, then use smaller increments that get the opponent to turn to 22 or higher first to allow yourself the turn to 28 - a kill shot. whomever reaches 21 first will lose unless a 1 is turned, whomever gets to 22 or higher first will lose period unless the difference to 28 is not available. The critical range is thus from 15-21.

The player who rolls the dice initially could try this ; Keep turning the dice such that each running total he gets is the lowest possible number that's a multiple of 4 or 7.

All my potential test subjects are now asleep, and my brain is pooped. It seems to work - the strategy will leave the player (A) on 28, thus forcing his opponent to loose. 28 is multiples of 4 or 7, yeah ? I couldn't beat myself trying this, but maybe I'm biased.

There's some sort of madness in that method.......the problem is like some horrible version of the game Nim.

It repeats every ninth. Would that make it more suitable for impecunious golfers or for wallpaper?

Wallpaper gets my vote.

Let's say the die is rolled and it comes up 6. The next player turns it so 5 is up, total is 11. First player then turns 6 up, total is 17. Next player turns 4 up, total is 21. First player turns up 6, total is 27, 1 is not available, thus he wins. Your turn to roll.

Wallpaper won't stick to my padded walls !

I still think there's something to the Nim comparison, I just don't know what ! Player B can't push his first total to 7 or 12, but A can. I dunno where that's going, but it beats counting sheep !

....if it's plus fours all the time, golfers could use it.

Nice summary.
And it could be turned into an even (fair?) game if the not-to-be-exceeded total is 7+9i (i ≥ 1)

but don't forget there is a useful unmentioned option if N=4 (gives probability of 1/2)

No: the thrower only wins if he rolls a 4:-

6: bust

5: bust

4: win

3: other player turns up a 1

2: other player turns up a 1

1: other player turns up a 3

Although an excellent challenge, the official answer still falls short when it comes to winning strategy and understanding fully the safety points.

See Jim35848's post #11 for a complete foolproof winning strategy...

Illuminating as Jim's post #11 is, I think the winning method* is more precisely described in his post #30. I would give Jim more G.A's for this if I could do so legitimately
*Only because I hesitate to call it strategy...

Yes, I agree.

Jim gave an excelent answer, but you should not be so critical to the official answer as the question did not ask for the foolproof solution. If you play this game with someone, it is much easier to remember the "strategy" then the complete table. Like in chess, not everybody is capable to remember all the moves from a database.

I am glad you consider this challange as a good one.

Hi Irvin,

Welcome to CR4 and thank you for a great challenge question. I am sure you will enjoy participating in the discussions which can (Just like any conversational discussion) get a bit heated at times.

I am sorry if you think I am being critical of the official answer but we have a long history here on CR4 where some of the official answers are way off the mark. Have a look at the "The Cloud Shadow" http://cr4.globalspec.com/blogentry/6689/Cloud-Shadow-CR4-Challenge-08-26-08 or "Fixing the pump" http://cr4.globalspec.com/blogentry/5457/Fixing-The-Pump-CR4-Challenge-04-08-08 for examples of this.

My point is basically that the second part of your question asks the probability of a win if both players play the best possible moves. This is indeed 1/3 when both players play the entire strategy which Jim has outlined. One or both players playing the strategy of using 10, 19 and 28 as safety stops will not unfortunately yield this outcome.

And yes, I know I am being pedantic and I apologise.

Welcome again to CR4

Hi Irvin,

Just wanted to say thanks for this question - I've had a lot of fun thinking about it. Hope you have some more questions for the future !

Kris

"...Twelve Dreys of Christmas..."

This time of year, I'd think cavity nests would be more attractive anyway, so I guess it's no surprise they haven't started yet...