Roll of the Dice: Newsletter Challenge (January 2013)

I've solved it for n=1 and n=2 but I got distraced by some string and a feather after that
Del

It's amazing how some string and a feather can distrace a cat - or even distract it.

I think you're all over-thinking this. 'A' is the first to roll and first with the opportunity to lose. As 'B' survives (50/50), 'A' gets another chance (50/50) which is half of half a chance. 25% chance 'A' wins.

One needs to assume that by 'improve' the player must roll the die so that a larger number of pips shows up in each subsequent roll compared to that same player's previous roll. Rolling the same or lower number would be a loss unless the other player also rolls the same or lower number from his previous turn. They both roll the die each turn until one get a higher number than their previous turn while the other gets the same or lower number than their previous turn.

Regardless of the number of sides of the die, there is approximately a 50-50 chance that the subsequent number is higher than the first number. The greater the number of sides of the die, the closer the approximation comes to 50-50. But for both players, the probability is the same. So - under this scenario - I'd say simply that the probability being the same for both players, it would be 50-50 for Player A to win.

.

An alternate reading of the question is that each player must roll a higher number pips that the other player has just rolled. So that if Player A rolls N1 pips, then Player B must roll N2 pips where N2>N1. Then Player A must roll N3>N2, and so on. It seems to me that the probability is still 50-50, and I'll go think about the math for a while.

You wrote, "Regardless of the number of sides of the die, there is approximately a 50-50 chance that the subsequent number is higher than the first number."

I think that the answer is no.

If you do a simple examination of all the possibilities for N=2 and N=3 you will see different results.

If N=2 there are only 4 possible scenarios for a die after 2 consecutive rolls. The first column is roll # 1 and the second column is roll # 2:

Roll 1 - Next Roll
0 0
1 1

0 1
1 0

1 0
0 1

1 1
0 0

Since every combination is equally valid, that gives the odds a 1 in 4 chance of having two consecutive rolls where the second roll is greater in value than the first a 1 in 4 chance.

However, with a 3-sides die the odds are actually 1 in 3 that the next roll will be higher.

Roll1 Next Roll
0 0
0 0
1 1

0 0
0 1
1 0

0 1
0 0
1 0

0 0
1 0
0 1

0 0
1 1
0 0

0 1
1 0
0 0

1 0
0 0
0 1

1 0
0 1
0 0

1 1
0 0
0 0

As N increases the probability of the second roll increases slightly until N reaches a number where the die is approximately a sphere, which the outcome becomes a little dicey since it becomes impossible to discriminate one side from another. :)

So, if N is 4 the probability of the second roll increases to 1 in 2.667, if N = 5 it is 1 in 2.5, and so on.

Knowing the first three probabilities should allow us to use something Lagrange interpolation to determine the probability for subsequently higher values of N.

My point is that the value of N determines all the possible outcomes for 2 consecutive rolls and as you see that as N increases the possibility for the next roll to be of greater value than the first increases in a non-linear fashion. In no case are all possible values for N the same.

If the curve of N were extrapolated on a graph, it would approach an asymptote of 50% (1 out of 2), but never quite reach it.

The answer to the original question should be the ratio of 1 to the integral sum from 1 to N = N-1, then divide that sum by the square of N.

I probably could express that better if my browser supported the math equations CR4 provides, but Safari does not.

For example, if N = 6, then the sequence is:

Probability = (N-1 + N-2 + N-3 + N-4 + N-5 + N-6) / N^2

So, the ratio is 1 : Propability

Easier than PI. :)

A simpler way to express the probability map, rather than binary numbers is (i.e., 5-sided die):

# 1 2 3 4 5
1 0 1 1 1 1
2 0 0 1 1 1
3 0 0 0 1 1
4 0 0 0 0 1
5 0 0 0 0 0

Where the row represents the number of the first roll and the column is the possible numbers for the second roll. 0 = the number is less than and a 1 = the number is greater than the first roll.

It is clearer to see the sequence that develops and how the integral sum equation I cited is the correct answer to the challenge question.

Given your example, for any random throw there are 10 chances the number will be higher (10 Xs) and 10 chances the number will be lower (10 Ys). Since you don't know what the initial throw is, the long term average for the second throw is as likely to produce a higher number as it is to produce a lower number.

# 1 2 3 4 5

1 0 X X X X

2 Y 0 X X X

3 Y Y 0 X X

4 Y Y Y 0 X

5 Y Y Y Y 0

With a 6-sided die, there are 36 possible combinations of two throws, N1 and N2. There are 15 chances that the pair of throws N1, N2 yields N2>N1 and 15 chances that the pair of throws yields N2<N1.

The problem with your 5 sided die (or any analysis based on your principle) is that there is no number higher than 5, so if the first roll is a 5 then there can never be a roll higher than 5.

The puzzle stipulates the second roll must be greater than the first roll, not greater than or equal to the first roll, so your table is not quite correct for the problem.

Because of that, no matter how many sides the die has it will always have less than 50% probability of producing a higher number on the second throw.

Because of that, no matter how many sides the die has it will always have less than 50% probability of producing a higher number on the second throw.

Yes, that's true. Which is why I said earlier the probablity was 'approximately' 50%. I think the key point, though, is that the long-term probabilities are the same for either player so on average either player has a 50-50 chance of winning.

In a realistic one-on-one game the players would have to agree that rolling the same number yields a draw or re-roll. (Player B would be dumb to accept a game where his first roll has a greater than 50% chance of him losing.)

In a gambling hall the 'house' would win whenever the same or smaller number is thrown - but the 'house' would need to offer a greater than 1:1 payout for a win to make the game attractive.

How do you interpret previous roll? I would take that as the previous roll is the roll the same player made before, not his/her partner's last roll.

Dumb knows no barriers. My best friend witnessed someone actually play Russian Roulette with a semi-automatic pistol in front of him. That someone promptly lost the biggest gamble in his short life.

How do you interpret previous roll?

That's one of the points I made in my first response (#2 above), that there are 2 ways to interpret the question:

One needs to assume that by 'improve' the player must roll the die so that a larger number of pips shows up in each subsequent roll compared to that same player's previous roll.

OR

An alternate reading of the question is that each player must roll a higher number pips that the other player has just rolled.

"How do you interpret previous roll? I would take that as the previous roll is the roll the same player made before, not his/her partner's last roll."

AH, in this context, please tell me where I am wrong in post #26 because I still think I am correct.

Why is it that every time, before the actual rules are agreed upon, someone has to launch into stick figure math that makes nonsense?

Sorry it is confusing.

The subsequent matrix in a followup post should be more clear:

# 1 2 3 4 5 6
1 0 1 1 1 1 1
2 0 0 1 1 1 1
3 0 0 0 1 1 1
4 0 0 0 0 1 1
5 0 0 0 0 0 1
6 0 0 0 0 0 0

Rows represent the first roll of a 6 sided die.

Columns represent the second roll.

A 0 means the second roll is less than or equal to the previous roll.

A 1 means the second roll is greater than the previous roll.

You can expand or contract the matrix to conform to any number that represents N.

The point of the matrix will prove that the odds of rolling a second roll that is greater than the last is always less than 50%.

You can calculate the exact odds using the equation I cited in my post.

Does that help?

Yes, I agree with the math, as long as it does not matter who rolls the "second" roll.

Upon further review, the player was down by contact before crossing the goal line.(Me.)

I would like a clarification that your math gives the exact odds for the next player to roll as long as the summation is carried out?

Wild guess: 1/e.

Good answer!

2 equal players faced with the same odds of achieving a goal, would yield a 50/50 chance of winning for either player....unless one was really lucky...I've seen stuff....If on the other hand player B could win by default, in that case player A would have a slightly greater chance of losing, but that would not be equal and would suck, unless of course, you were player B, who could win without really ever taking a chance because player A had eliminated his or her self before player B ever had to roll an improved roll....

The problem definition does not explicitly state what constitutes 'improve'. We assume it is a higher number, but there is nothing in the problem definition that actually says this. All we know is that 'improve' is a difference, but not the direction of that difference. In this light, how would you change your answer? Or would you?

I don't think that changes anything as long as the definition for "improve" does not include the same number as the last role.

Ambiguity alert. Does "the previous roll" mean the same player's previous roll, or the immediately preceding roll of the other player? (Seen also by Usbport.)

There seems to be a perennial problem with unclear newsletter challenges.

If that were the case player A would lose when they rolled the die because it would not be an improvement on a previous roll....

Huh????

Then it must be their first roll. You can't have a second roll without the first. :)

Indeed, then that would mean that each roll must be an improvement on the previous roll made by that particular player.......

Huh???

I think I get your point, but I am going to side with the idea that the very first role is exempt from the comparison to the previous roll.

My reasoning is that it is impossible to compare a very first roll with a previous that does not exist, so it is only logical to assume the first roll must be exempt.

I think there is a specific term for that, but the term escapes me. It is like a preexisting condition or state that is intrinsically understood as part of the problem.

It's like a low level software requirement document. There are some things that are simply not needed in the description of requirement document because they are deemed to be understood by the reader. For instance, a low level requirement document may contained detailed information, but it is reasonable to simply state that the code will be written in C. There is no need to include a complete description and rules for the C language because it is understood that engineer/reader already grasps that.

Didn't Roger Pink have a formula simular to this about choosing door Number one, two and three. And after you choose and change your answer it changes the whole statistical matter.

Can't you apply something simular here.

The rules are ambiguous. A is a con man, and B is a noob. A wins especially since by rule, he rolls first. As I stated earlier, they need to quit gambling, and get back on the loom, silly boys.

To me, it is clear that, to stay in the game after the first roll, each subsequent roll must better the previous roll of the other player. I see no injustice in this as long as they play an equal number of times, alternating starters.

I'm not sure of these tables some of you use but my approach to the logic is:

The first roll can hit any of N numbers, say it is 3. The second roll must hit any of the N-3 higher digits left or lose.

If the first roll is in the lower half of N, there will be more higher numbers available than smaller or equal so B has a better than 50% chance of staying in the game. If B now picks in the high end, say N-1, there is little chance of A beating him. If A started with a high number, the picture is quite different.

I don't know if I can push this to gain a pre-game set of odds or not, I'm doubtful that I have the training and/or ability.

There might turn out to be some recursive formula, perhaps quite thorny; but I suspect one will have to draw the whole tree.

Are we searching for a formula for each N, or for a total probability distributed over all N? (Another ambiguity.)

If the rules are interpreted as you state, that each roll of the die must be an improvement on the other players roll, then player A would have an advantage in that he or she gets a free pass on the first roll and can not lose, whereas player B does have a chance of losing on the first roll, therefore the odds would be in player A's favor....If, on the other hand, each player must best their last roll, then the odds become even....

But your rules cannot stand, under them, if A picks the N card at first go, B wins. After B throws any number, A is now required to throw and improve on her first pick and she cannot do it and in failing to do so, loses the game. Under my rules, A wins, as she should after picking the top number because B cannot beat it.

A and B are both losers for rolling die, when they should be making textiles.

After A has her throw, B's probability of advancing to the next round is (n-1)/2n, where n is the number of sides >2, it approaches 1/2 as n gets bigger.

So, A's chance of winning that round is(n+1)/2n

To be continued.

Probability still remains 50%. Just by introducing a condition, the chances get loaded to one side after a few attempts. Otherwise probability of A or B winning remains 50%.

AROOGAH AROOGAH... OVERTHINK ALERT.
Del

Player A wins 5 times out of 6.

Probability A wins: PA=(1/2)/1! + (3/4)/3! + (5/6)/5! + (7/8)/7! + ... = 5/6

Probability B wins: PB=(2/3)/2! + (4/5)/4! + (6/7)/6! + (8/9)/8! + ... = 1/6

We have to decide which interpretation is correct.

If each player has to improve on his own throw, there is a premium on picking a low number to start and then picking the lowest of higher remainder, this repeated till there are no more. Picking a high number to start limits the number of possible plays before the top number is reached.

If each player must throw a higher number than his opponent to stay in the game, there may be an advantage for the first to throw, or for the second, but if an equal series is played with alternating starts, there is no overall advantage.

There needs to be a clarification of the rules: I read the rules to mean that each subsequent roll of the die must be a higher number than the previous roll, regardless of A or B rolling it. Thus if player A rolls a 6 (N=6) on his first turn, player B has a 100% chance of losing (player A has 100% chance of winning in this case) based on he rolls the die and the best he can do is tie with A.

Certainly you do mean die singularly, and not plural.

Also if player A rolls a 5 (N=6), then B has only a 16% chance of winning (not losing). Here A has an 83.3% chance of winning. And so on ad nauseum. Besides, I am busy and lazy, and do not have the time or inclination to derive the mathematical formula for this.

This is all thought out very nicely, and the chances of winning based on the rolls, well calculated. The way I see though it is always 50%. Before any rolls are made the odds are even, as there are only 2 players it is the equivalent of a coin toss. 1:2

What are the chances that player A wins? 50-50.

Algorithm to calculate answer:

Compute random numbers (0 to 1) until number is less than previous number.

Record number of times until this happens

Repeat many times. Odd iterations are A wins, Even iterations are B wins.

Record number of A wins versus total.

Matlab program indicates A wins .8333

Histogram of number of iterations:

Bug in program, Probability A wins should be 0.6322. Sorry about that.

And this is my final answer!

This looks like approximately the right answer for N=∞.

After having read all the responses to this question, all I can say is that you have all left out one single thing. That is luck. I say the luckiest guy wins. If you have no luck, stay out of the game. On more serious note, 50 - 50 is correct.

does bad luck count? can I still play?

There is another possibility that has not been addressed, the die may not necessarily be numbered, and the quality of what is best could change with time, for instance, say the die was directional and you were navigating a course where directional changes were necessary, at every turn the die was rolled to determine which direction to proceed, whether the direction chosen was an improvement or not, may be open to question, and the player with the most convincing ability to argue a point would have an advantage....

or not....

Hey, I can see New Jersey from up here!

My take on the rule of this game:

If Player A rolls x, then Player B must roll x+1 or greater (if possible), otherwise Player A wins. If Player B is successful, this roll becomes the new x, and Player A must now roll x+1 or greater (if possible). This continues until one or the other fails to succeed (or succeeds in failing).

My take on the rules are that the players must "improve on the previous roll". There could not be a winner without 4 rolls of the dice/die, although there could be a loser, or even 2 losers after only 2 rolls. Still before any rolls are made the chance of player A winning is 50-50. Both players have an equal chance of winning, add another player and that is a different story. The chance of a coin toss being heads is always 50-50.

My contention is that for any N-sided die, there is a 1-in-N chance that the first roll will be the highest possible value for N, and Player A wins immediately. 1 roll, game over. If the first roll is N-1, Player B has only a 1-in-N chance of beating Player A. If he does, 2 rolls, game over. Hence, the shortest possible contest could be decided in 1 roll of the die.

This of course does not address the original question of probability...

So if using a N-sided die that has an odd number of sides, player A has to potentially roll 1 more time than does Player B. I still maintain before the first roll the chances of Player A winning are 50-50, after that first roll the odds change unpredictably, no matter how you read the rules.

True, true. But since Player A has the chance of winning outright on the first roll, his probability of winning will always be more than 50%, although it approaches 50% the more sides there are on the die.

So in some sort of weird way the losest number on any given die should Zero giving the first roll of a zero an istant loser, as it would not improve on any "understood scoring" prior to the first role. This would change things just a smidge.

No, rolling the lowest possible number (let's say zero) means that your opponent has the best chance of beating your roll. However, Player A can still win if Player B subsequently rolls the lowest possible number (another zero). In a tie, the earlier number is the winner. Remember, we're talking about the first roll of the die. On the first roll, a zero is not neccessarily an automatic loss.

it approaches 50% the more sides there are on the die.

Actually you can simulate a die with a very large number of sides using a random number generator. I wrote a Matlab script to do this and it gave A about a 63% chance of winning.

P, the probability that A wins, is a function of N. When N=1, P=1; and P approaches 1/2 asymptotically as N goes to infinity. Hence we seek a power law that passes through (N,P) = (1,1) and (infinity, 1/2). There are an infinite number of such laws, but any third data point selects exactly one. Clearly (2,P) = (2, 3/4); so P = 1/2 + the Nth power of 1/2.

Clearly? When N=2, A has a 50% chance of winning on the first roll. Failing that, he has a 50% chance of winning on the second roll. Thus B's total chance of winning is 25%.

A closed form solution should be possible, but I have only managed to find solutions for N=2,...,5 by proceeding roll-by-roll and adding up the probability of A winning due to various outcomes at each roll. A good programmer could have written a recursive program to do this for arbitrary N faster than I did it by hand.

N P(A wins)

2 3/4

3 19/27

4 175/256

5 2035/3125

The nature of the probability adding problem is illustrated by the figure below for N=4. Here, the size of each rectangle represents the probably of the series of rolls that led to it, green rectangles are outcomes where A wins and red outcomes where B wins. The game must terminate after no more than N rolls, thankfully.

There is an error in the figure, not on purpose.

I'd bet on this answer being correct, under the assumptions that "previous roll" means the other person's last roll and that "improve" means "be greater than". With these assumptions, the first player can win with an N on the first roll, although technically, the game doesn't end until the second player rolls a "lame duck" roll, which cannot be greater than N.

In that case, my "wild guess" in 7↑ was off only by B versus A.

As N goes to infinity, A(N) goes not to 1/2 as surmised, but to 1-1/e.

P=.6322. Thanks, I got it empirically and was wondering what that number was.

For the three sided die, there are nine possibilities. A wins five of them and B wins four of them.

This is based on each player having to better the opponent's last throw.

This is an interesting and instructive error. It is the reason that probability is hard. In principle one need but count cases, but what counts as a case is often hard to say. When N=3, if you count less than 27 cases, they must be severally weighted. E.g., if A wins with a 3 on the first roll, that counts either as 9 of 27 cases or as 1 case with a weight of 9. When N is small, the easiest way to get the cases right is count all N^2, even when the game ends in less than N^2 rolls. When N is large, you must devise a system for determining weights, which you then check by requiring your weights to sum to N^2.

Further illustrating the difficulty: N^2 should have been N^N.

I understand this, and agree that it should be as corrected in your #66. I wonder though, if this fits the problem as set.

On further thought, I concede, you are correct. I went through several thought experiments and decided that the array of number holders, N^(N+1), N^N rows by N columns, is there, but the rules of the problem prevent all of them being filled.

This is the first third of the array. winners are bold, unfilled holders are italic. The second and third parts of the array are the same except that the first digit is 2 and 3, respectively. The winning numbers are different per the rules.

1→1→1

1→1→2

1→1→3

1→2→1

1→2→2

1→2→3

1→3→1

1→3→2

1→3→3

if N=1 there is no chance for player B to win, in all other cases the chance should be (N-1)/(2N)!

1 sided dice are as common as magnetic monopoles.

Sorry!

(N-1)/(2N)

!

0% The wording of the rules state everyone loses.

I believe that halfb1t and rixter provided theoretical and experimental answers that are consistent with each other and follow these assumptions:

• that "previous roll" means the other person's last roll
• that "improve" means "be greater than"
• and that a person immediately loses when her roll does not improve on the previous roll

Aren't those assumptions consistent with the wording of the rules?

This sounds like a new take on an old problem;

Two people meet for a meal. They toss a coin, and first person getting a head has to pay the bill. Over a period of time, the first person will end up paying more - they took the initial gamble of 50%.

In this case, it's been extended. When player A gets their second roll of the dice, they are immediately first to gamble.

On the basis of that gross simplification, player B will win. Overall, the odds are the same as the meal/coin-tossing game. It's on CR4 someplace.

Usual late night disclaimer applies.

It is simple .... A shoots B .... it's not called 'die' for nothing.

A wins!

The way I see it. Player "A" has a better chance to lose only because he or she is first.

When he loses the game is over. player "B" does not have to match.

As for the percent value number I have no idea of the math.

This is if you must improve on your own roll.

The wording also make the option that one must improve the opponnets roll as well. then The answer would be "B" has a better chance to lose.

Some idiot voted you 'off topic'. I've cast a vote to negate that - it was a well expressed post that was completely on topic.

CR4 is currently suffering from some lunatic who is under the impression that admin are not aware of who it is. If they persist, it's a fair bet that they will be noticed and action taken.

I'l leave this 'on topic' (even though it isn't), so I can cite an example to admin so that they can track down whoever it is that does not comprehend how the vote system is intended to work.

The key to this puzzle is that the first player to roll the die (Player A) may roll the highest number on die and that there are no ties. I solved puzzle for simple case of 4-sided die and came up with answer that Player A has 5/8 chance of winning based on probability of Player B winning after Player A throws four possibilities. I then extrapolated to general case of N-sided die that resulted in (N+1)/2N probability of Player A winning. The probability approaches 50% when N is a large number.

Since this is an "engineering" forum, when N-> quite a large number you are no longer rolling a die, you are rolling a freaking ball (as an engineer would see it), so no one wins, because the ball will not stop rolling for a long time, and also who could tell how many dots are on the "face" that is microscopic? Of course if you are only simulating this in a computer then who cares, because reality has been supplanted by virtuality.

Well, approaching ∞ simply helps us generalize the equation for solving more iterations of potential "real world" objects (for instance, dice).

Oh, I am sure that is why we like mathematical models. I am just saying that will never be played in Vegas.

The previously specified probability (N+1)/2N is only correct if only two rolls occur. Additional elements to equation are required if more than two rolls occur which is likely for a large N!

On the statement above, its shows that players A and B uses the same N sided die, in the same environment, and in the same situation. It doenst say that if Player A win, player B losses or other way around. Just the rule is "the player who does not improve from previous roll losses". Let assume that luck, ability or even powers of Player A and B are the same: say

A = B

Let say: Total percentage of winning will be 100%. Any win of Player A and Player B will be 100%, means, if Player A wins 70% player B 30%, if A 40% then B is 60%, and so on... lets equate this:

Win of Player A + Win of Player B = 100%

Player A + Player B = 100%

A + B = 100%

BUT: A = B

Then by substituting: A + A = 100%

2(A) = 100%

dividing each side by 2: A = 100% / 2

A = 50%

So the answer will be: Player A has a 50% probability of winning against Player B.

I can't speak for other possible voters, but posts 7, 56, 59, 67, and 69 progressed to and confirmed the correct solution. Various posts afterward that continued to give incorrect information were naturally off topic. (While the question was still being explored, various wrong approaches could still be considered on-topic.)

The probability of Player A winning is 2/3.

The basis for my answer was the potential results of every roll. The numbers in the table below represent the probability of each player succeeding in rolling a number that would continue the game during that round (forgive the formatting, I couldn't get spaces to stick).

Player A - Player B

1 - .5

.25 - .125

.0625 - .03125

The odds for each player during each subsequent round are 1/4 their own odds from the previous round because there have been 2 instances of 1/2 odds since their last turn.

As this continues you see that the sum of Player A's chances approaches double those of Player B's or 2/1 in favor of Player A. This translates to a probability of 2/3 for Player A winning.

I started with figuring out the chances of the second player continuing based on the roll of the first player. I did this first for a 5 sided die, then 9, then 13 all as a check. So, for a 5 sided die for each possible roll of the first player we have the following for the second player.

(4/5+3/5+2/5+1/5+0)/5 which equals 10/5/5 or 10/5^2 or 2x5/5^2 and therefore

probability for player B is 2n/n^2 and then the probability for player A is:

1-(2n/n^2)

I made and error. The previous formula is for one round.