tag:blogger.com,1999:blog-4167684119478977136.post3286665968146831659..comments2019-12-11T04:29:14.414+00:00Comments on Trouble At T'Mill - a wargaming blog: Probability for Wargamers 5 - The Tea Break CardMike Whitakerhttp://www.blogger.com/profile/02165272678144625943noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-4167684119478977136.post-66907216422199918642012-10-25T20:23:56.928+01:002012-10-25T20:23:56.928+01:00Great post Mike, and some great comments as well. ...Great post Mike, and some great comments as well. From experience, it usually "feels" (a horribly subjective term) that you're making progress in a game when you've activated at least one platoon, even where the other platoons may be on Blinds. It is a very finely balanced decision though, not least because Blinds convey a number of movement and protection advantages. <br /><br />Speaking from experience of playing and umpiring, I've seen some terrific use of Blinds in a game (units moving fast and close assaulting from Blinds) and some cases of really bad luck (Blinds barely moving).<br /><br />All in all, it's just part of the fascination with the rule mechanic, but its certainly fun to try and calculate a 'winning' approach!Sidney Roundwoodhttps://www.blogger.com/profile/14795563060856586670noreply@blogger.comtag:blogger.com,1999:blog-4167684119478977136.post-70226881620962775922012-10-25T11:24:43.384+01:002012-10-25T11:24:43.384+01:00Think of the deck being reversible (which it is: y...Think of the deck being reversible (which it is: you just turn it over (face up)). Then it becaomes obvious one TB card in the deck of N cards being at position M is just as probable as its being at Position N-M (which is position M in the reversed deck). Taken overall, then, the 'statistical expectation' is that the TB card will be at half way. In other words that M will equal N-M, i.e. M=N/2. Note though that this 'expectation' is really only a mathematical way of describing the symmetry I've already mentioned: that the TB card is as likely to be M cards from the bottom as it is from the top.<br /><br />That symmetry extends to as many cards as you like. Consider the deck from the top until you get to the first TB card (at M), then think of the remainder of the deck as the deck of N-M cards. The 'statistical expectation' is, as seen before, that the second TB card will divide the remaining deck in equal halves (I think we have to bear in mind here that this gets a bit weird with N-2 (All minus the TB cards) not being a multiple of 3, but never mind: the maths still works). The symmetry again is preserved if were we mentally were to flip the deck over, in which case the two TB card buried within switch roles, as it were. The only way that symmetry will work is if M and (N-M)/2 were equal. And that is true if M=N/3 or 3M=N. <br /><br />By induction, then, you can infer the 'statistically expected' distribution of any number of TB cards within a deck. All of which can be very misleading: the probability that the TB cards will indeed be so distributed would be exceedingly low (usually).<br /><br />That bit about 1 million tickets vs 1 ticket - not quite true. Each ticket, supposing it has a unique number, is equally likely to win the lottery. Think: suppose you bought every ticket in a lottery of a million tickets. You would win for certain. Except that you wouldn't because the total costs of the tickets exceeds that of the prize. The more tickets you buy, the more likely you will collect the prize, but that increased likelihood comes at a price.<br /><br />Here's the kicker: the 'expected return on investment' is exactly the same, however many tickets you buy. WTF???<br />Consider a lottery in which there are 1 million tickets, going for a dollar each, and the prize (we'll suppose there's just one, but it makes no diffference) is $100,000. <br /><br />You buy 1 ticket for a buck. There's a million to one chance you'll pick up 100,000 dollars; 999,999 chances you lose your dollar. Your return on investment, statistically speaking is <br />(999,999x$0 + 1x$100,000)/1,000,000 = $0.10 = 10 cents on the dollar.<br /><br />Suppose you bought all million tickets. That will cost you a million dollars. You win the $100,000 - guaranteed. Your return on investment? Ten cents on the dollar (you got $100,000 but it cost you a million).<br /><br />A $2 investment (just to show it's not a case of the extremes being special cases)? 999998x$0/1000000+2x$100000/1000000<br />= $0.2 = 20 cents (out of $2 invested) i,e, 10 cents on the dollar. Lotteries really are losing propositions.<br /><br />All of which might - or might not - have anything to do with war gaming... but I think it is useful to know what a losing proposition looks like, and they do turn up in wargames. Archduke Piccolohttps://www.blogger.com/profile/15533325665451889661noreply@blogger.comtag:blogger.com,1999:blog-4167684119478977136.post-7356938852858299242012-10-24T16:10:45.489+01:002012-10-24T16:10:45.489+01:00I'm in the same league as Phil: a degree in ec...I'm in the same league as Phil: a degree in economics but rather weak in stats. In any case Thanks for the explanation I actually and intuitevely posted in the Yahoo Group tbe idea that a unit with its card and the BM has a higher lilkelihood of being activated than a blind with one card. However, the advantages of being in blinds are significantly higher than being deployed plus the fact that a blind card has higher optionality value (at least innitially) as the player can activate more units with a single cardAnibal Invictushttps://www.blogger.com/profile/00574972963418062956noreply@blogger.comtag:blogger.com,1999:blog-4167684119478977136.post-22983994839770692662012-10-24T15:13:58.100+01:002012-10-24T15:13:58.100+01:00I did stats as part of an economics degree and sti...I did stats as part of an economics degree and still can't work out stuff like this.<br /><br />Its like the lottery ticket thing - you can buy a million lottery tickets but have no better chance (apparently) of winning the lottery than a person with 1 ticket. <br /><br />Maths people can tell you why - I can't.Phil Broedershttps://www.blogger.com/profile/18218127480258279732noreply@blogger.com