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January 16

+10% and -10%

you can increase something by 10% to make it bigger. if you then take the new larger result and multiply it by 90% do you get the original? 212.96.61.236 (talk) 06:03, 16 January 2015 (UTC)[reply]

The tricky part with percentages is the base. If you increase 100 by 10%, you get 110, then...

...if you then decrease it by 10% of the original base, that's 10% of 100, so 10 less than 110 = 100.

...if you then decrease it by 10% of the new base, that's 10% of 110, so 11 less than 110 = 99.

Something else interesting is that while which base you use matters, which order you do the operation does not. If you decrease 100 by 10%, you get 90, then...

...if you then increase it by 10% of the original base, that's 10% of 100, so 10 more than 90 = 100.

...if you then increase it by 10% of the new base, that's 10% of 90, so 9 more than 90 = 99.

StuRat (talk) 06:14, 16 January 2015 (UTC)[reply]

I've amended the question. So, your answer is "no, increasing something by 10% by mutliplying by 1.1 and then removing one-tenth of the new larger result by multiplying by 90% does not get you the original. 212.96.61.236 (talk) 06:15, 16 January 2015 (UTC)[reply]

Correct, that's the "new base" case. I'll mark this Q resolved. StuRat (talk) 06:20, 16 January 2015 (UTC)[reply]

Thanks. So actually +10% then -10% is in all cases exactly 99% of original (not just for starting from 100). right? So increasing 45 by 10% which is 4.5 then decreasing the result (49.5) by 10% results in 99% of the original? +10% then -10% is the same as 99% of the original? It could be used as a "trick" (if another party doesn't notice) to reduce anything by exactly 1%? Not that I would ever do this. 212.96.61.236 (talk) 07:43, 16 January 2015 (UTC)[reply]

Well, yes, but you forgot to say "then -10% of the new base". Remember, it's ambiguous without that.

Also note that a larger percentage makes for a much larger discrepancy, so +50% then -50% (of the new base) gives you 75% of the original, as does -50% then +50% (of the new base). StuRat (talk) 08:05, 16 January 2015 (UTC)[reply]

Resolved

Only brain-dead people ignorant on logarithms use percent. Let e^{0.01 x} be pronounced x degrees and written x° . Note that x°y° =(x+y)° , and that one degree is approximately equal to one percent: 1° = 1.01005 and 1% = 1.01000. Use degrees, and life becomes pleasant: If you increase by 10° and then decrease by 10° you get 0°. Bo Jacoby (talk) 22:38, 22 January 2015 (UTC).[reply]

Modified Snakes and Ladders strategy

If you have a standard snakes and ladders board with approximately equal numbers of snakes and of ladders, all with start points roughly equally spaced. And you add an element of strategy by permitting, at each turn, a player to choose whether to roll one die or two dice, thus having a choice of range of 1..6 linearly or 2..12 non-linearly. Is there a sensible approach to the choice?

All I can come up with is don't use 2 dice when 6, 7 or 8 away from a snake. Anything better? -- SGBailey (talk) 22:53, 16 January 2015 (UTC)[reply]

My immediate thought is that rolling a single die doesn't improve things, but only postpones a potential trip up a ladder or down a snake. In fact, if rolling two dice allows you to move directly forward by the total of the two, without landing on an intermediate point (or allows you to pick one of two possible intermediate points if the two dice show different numbers), it is probably better to always roll two dice. I think (for a "sufficiently large board") the optimum strategy is simply always to roll two dice, because that gives you the highest expected number of steps forward. For boards that are not "sufficiently large", though, I suspect you have to consider the graph of the individual board, and I'm not sure that there's a general answer. RomanSpa (talk) 23:29, 16 January 2015 (UTC)[reply]

There is a little on a possible strategy for a modified form of the game in our article on Snakes and Ladders, but not much on the unmodified game, which is basically a game of chance. RomanSpa (talk) 23:37, 16 January 2015 (UTC)[reply]

But surely when you are (say) 4 away from a snake you want to roll a number other than a 4 and throwing two dice will get that mostly and jump you past it. If on the other hand you are seven away from a snake, 2 dice are very likely to land you on it. Thus one dice will take you closer to it and with the next throw you use two dice. There must be some way of analyzing this. -- SGBailey (talk) 00:12, 18 January 2015 (UTC)[reply]

And if you are 1 away from a snake, (and then the next 11 clear), you certainly want to use 2 dice...Naraht (talk) 00:26, 18 January 2015 (UTC)[reply]

I think we need to be clear about what it means to roll two dice. Are the dice distinguishable, with the rule that in the event of rolling (m,n) you must first move m steps to land on a square, then move a further n steps? Or is the rule like backgammon, where you must first move either m or n steps at your option, landing on one of two intermediate squares before advancing a further n or m steps? Or is the rule simply that you immediately advance (m+n) steps, without landing on any intermediate square? The choice of rule affects the choice of strategy. RomanSpa (talk) 10:42, 18 January 2015 (UTC)[reply]

"simply that you immediately advance (m+n) steps, without landing on any intermediate square" -- SGBailey (talk) 22:34, 18 January 2015 (UTC)[reply]

We know from our article on the game that the average number of throws required to finish the game is 39.6 if the Milton Bradley configuration is used. If we assume (as was the case in my family when I was growing up) that winning requires (a) that you complete the game first, and (b) that nobody else subsequently completes the game in the same round (this avoids arguments of the form "X won because he got to move before Y"; if both X and Y complete the game after the same number of die rolls, a draw is declared), the unmodified game is a game of pure chance, with symmetry between the players. If there is an optimal strategy to reduce the number of rolls following the introduction of a second die, it is rational for every player to follow it; the game remains a game of pure chance with symmetry between the players. That is, following a change in rules there will be a temporary advantage to players who discover the optimal strategy earlier than their rivals, but this will disappear once everyone has learnt the optimal strategy. RomanSpa (talk) 10:58, 18 January 2015 (UTC)[reply]

I think it's a fairly complicated problem in its full generality. For example, it is not necessarily the case the snakes are bad, since they can give you a second chance of reaching a particularly good ladder. Regardless, it is easy to construct a board which cannot be completed at all if 1 is always chosen, or if 2 is always chosen (but can be completed if 1 or 2 is chosen where appropriate).

There are about

2^{n}

different strategies. For each such strategy, finding the expected time to reach the finish can be found by solving a linear system. I don't know of an efficient way to find the optimal strategy for an arbitrary board. But one can try a greedy method: Start with a random viable strategy, then repeatedly go over each square, and see if switching the strategy for this square improves the expected time (I believe this can be done more efficiently than solving the whole linear system from scratch). Continue until there is no improvement by changing a single square. In theory I guess you could end up in a strictly local optimum, but the board would have to be particularly pathological for this to happen. -- Meni Rosenfeld (talk) 23:09, 18 January 2015 (UTC)[reply]

What's

n

there? RomanSpa (talk) 00:12, 19 January 2015 (UTC)[reply]

I guess it will be the number of squares (that you have to roll dice to leave) on the board. -- SGBailey (talk) 07:05, 19 January 2015 (UTC)[reply]

Right. Though the number of strategies can be reduced if we remember that there are some squares you can never start your turn at, and hence, have no association decision. -- Meni Rosenfeld (talk) 13:17, 19 January 2015 (UTC)[reply]