As those of you who are familiar with the mechanic will know, the Too Fat Lardies games rely on a card-based activation mechanism, and in most of those, the activation sequence is ended when a Tea Break (or whatever it's named for that particular ruleset) card is drawn.
So. What's the likely percentage of cards that come up before the Tea Break card comes up?
This seems quite tricky at first glance, but here's one way. The Tea Break card could be anywhere with equal probability. If it's first of N cards, then no useful cards get drawn before it, and if it's last, then all of the (N-1) useful cards get drawn. Simple intuition says the answer, the average probability, is therefore 50%.
More rigorously (and so we can wave similar results around with aplomb if we need to later!) - if the Tea Break card is drawn at position M, there are M-1 cards before it, which is (M-1)/(N-1) of the whole deck of useful cards. So the average is the sum of all those (M-1)/(N-1), with M ranging from 1 to N, divided by the N possible positions.
(And here I wish I had a maths font.)
The sum of all the numbers from 0 to N is, by a handy mathematical co-incidence, N*(N+1)/2 (check it if you don't believe me). So the sum of all the individual (M-1)/(N-1) for M from 1 to N (i.e. M-1 from 0 to N-1) is ((N-1)*N/2)/(N-1), which boils down to N/2. And if you divide that by N to get the average, you get 1/2, or 50%. Yay.So, the probability of any random card appearing before the Tea Break card is 0.5, or 50%. Let's look at what this means for the question of blinds.
At the start of the game, there are three cards in the deck: Allied Blinds, Axis Blinds and the inevitable Tea Break. And there's a 50% chance that your Allied blinds card will come up before the Tea Break.
Let's activate an Allied platoon and its Big man, so now our deck contains Allied Big Man 1, Allied Platoon 1, Allied Blinds, Axis Blinds and Tea Break. Chances of the Allied Blinds coming up and you getting to move the rest of your blinds? Still 50%. But what are the odds on Allied Platoon 1 getting to activate?
Well - in order for no part of the platoon to activate, neither the Platoon card nor the Big Man card must come up. The probability of either of those events not happening is 1/2, so the probability of both cards not coming up is 1/2 * 1/2, or 1/4, 25%. So the chance of some part of Allied Platoon 1 getting activated is 3/4 - 75%. If there were no Big Man card for it, it'd be 1/2, 50%, the same as for the blinds.
In fact, given most units in IABSM tend to come with an associated Big Man, once you've deployed a unit off blinds, its chance of getting activated either by itself or its Big Man, is 75%, compared to only 50% if it stays on blinds. Admittedly, unless it's a level 3 Big Man, all its sections won't necessarily activate if it's a platoon, but...
In other words, the chance of units remaining on blinds getting activated doesn't decrease, but it is statistically better to deploy if you want to improve the odds of at least one section of a platoon (say) activating.
This article's homework? Prove, with a reasonable degree of rigour, that adding a second Tea Break card means you draw on average 2/3 of the deck before the second Tea Break comes up.
I did stats as part of an economics degree and still can't work out stuff like this.
ReplyDeleteIts 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.
Maths people can tell you why - I can't.
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 card
ReplyDeleteThink 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.
ReplyDeleteThat 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.
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).
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.
Here's the kicker: the 'expected return on investment' is exactly the same, however many tickets you buy. WTF???
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.
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
(999,999x$0 + 1x$100,000)/1,000,000 = $0.10 = 10 cents on the dollar.
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).
A $2 investment (just to show it's not a case of the extremes being special cases)? 999998x$0/1000000+2x$100000/1000000
= $0.2 = 20 cents (out of $2 invested) i,e, 10 cents on the dollar. Lotteries really are losing propositions.
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.
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.
ReplyDeleteSpeaking 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).
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!