If a cylinder’s height is the same as its diameter, then if you take a plane cross-section of it through diameters of it’s two ends, the cross-section you end up with is a square. This suggests to me that if you were to roll this “die” exactly such that this square cross-section could be properly described as rolling back and forth across the surface upon which you rolled the “die” (impossible in practice, of course) then the chance of the die landing on circular end A would be 1/4, circular end B would be 1/4, and the chance of landing on the rounded part of the cylinder would be 1/2. (Each possibility corresponds to some side or sides of the square cross-section.)
Meanwhile, if you rolled this “die” perpendicularly to the way it was rolled in the last example, such that it can be aptly described as literally “rolling” on its rounded side, then the chance of “landing” on the rounded side is 1.
This suggests to me that the chance of landing on the rounded side on a random roll lies between 1/2 and 1. Total WAG though.
Indistinguishable , this is for you, but I’m going to use jackelope’s post to work off of:
[QUOTE=jackelope]
Ohhhh, now I get it. I think. So the very first toss can be, but is not necessarily, part of the first matching pair? In other words:
Team 1 wins with tosses of [H H] or [H T H H] (and others, of course)
Team 2 wins with [T T] or [T H T T]
Team 3 wins with [T H H] or [H T T] or [H T H T T] or [T H T H H]
And so on. Is that right? I’ll take your word for it that the eventual probabilities work out to 1/3 apiece.
And the infinitely long series that would never give an answer would be [H T H T H T H T] etc., right?..
[/QUOTE]
I honestly didn’t understand your earlier posts, but it seems to me that after the first flip, either team 1 or team 2 is eliminated. After that, Team 3 has a 50% chance of winning. So this method is as fair as Joey P’s idea; only more complicated.
You are correct in saying that either Team 1 or Team 2 are eliminated after the first flip. However, you are incorrect in saying that Team 3 afterwards has a 50% chance of winning. After the first flip, Team 3’s chances of winning are precisely half those of his uneliminated competitor’s. (This is because the first matching pair is twice as likely to be the same as the first coin as it is to be different from the first coin. To see this, let’s assume, without loss of generality, that the first coin came up H. Then, in every following pair of flips, either alternation is maintained (TH), in which case, we have to look at the next pair of flips to see who wins, or alternation is broken on the first of the pair (HH or HT), in which case Team 3 loses, or alternation is broken on the second of the pair (TT), in which case Team 3 wins. One way for Team 3 to win vs. two equally likely ways for Team 3 to lose.)
[QUOTE=fiddlesticks]
So the only people involved in the actual coin flip were the Raiders and the Falcons, and so there only a single coin flip was necessary. Had the Raiders won the flip, the Chiefs would have drafted fourth and the Falcons fifth.
[/QUOTE]
Not exactly. Had the Raiders won the toss, there would have been a second toss between the Falcons and Chiefs. Cite. That’s because only Oakland’s and Kansas City’s records could be compared for the purpose of establishing draft order.
[QUOTE=Indistinguishable]
You are correct in saying that either Team 1 or Team 2 are eliminated after the first flip. However, you are incorrect in saying that Team 3 afterwards has a 50% chance of winning. After the first flip, Team 3’s chances of winning are precisely half those of his uneliminated competitor’s. (This is because the first matching pair is twice as likely to be the same as the first coin as it is to be different from the first coin. To see this, let’s assume, without loss of generality, that the first coin came up H. Then, in every following pair of flips, either alternation is maintained (TH), in which case, we have to look at the next pair of flips to see who wins, or alternation is broken on the first of the pair (HH or HT), in which case Team 3 loses, or alternation is broken on the second of the pair (TT), in which case Team 3 wins. One way for Team 3 to win vs. two equally likely ways for Team 3 to lose.)
[/QUOTE]
OK. I was focusing too much on the HTHTHTHT scenario. Now that I go through it step-by-step, I see that if we assume that the first flip is T, then Team 1 is out. Now there is a 50% chance that Team 2 wins outright, and a 50% chance that the game continues on equal ground.
To use tennis terms, it’s as though Team 2 goes into the second flip with Advantage Team 2.
Whether or not this system gives each team an equal chance of winning, I don’t know. But I am officially conceding that my logic in my previous post was flawed.
[QUOTE=Indistinguishable]
Hm, I haven’t yet found a way to beat 8/3 tosses for one ternary digit, but I can match it, I guess. Toss the coin repeatedly until a tails comes up. Count the number of heads. If this is odd, then output A. If it’s even, then flip the coin again to decide between B and C. Each output has probability 1/3, and the expected number of tosses is 8/3. There’s no great reason to favor this over “Retry on TT”, but I like it anyway. Can anyone prove that 8/3 is the optimal for producing one ternary digit?
[/QUOTE]
I think the 8/3 is optimal; here’s my sketchy reasoning:
A three-way coin-tossing scenario is a tripartition of a complete prefix code in {H,T}[sup]*[/sup]. (That is, we toss the coin until we find one of the strings in the code; its completeness guarantees that this will always happen eventually, and the prefixness guarantees that this will always be an unambiguous identification.) For it to be fair, each of the three sets in the partition must have equal probability 1/3.
Now a greedy approach immediately gives the OP’s algorithm: Since one of these sets has total probability 1/3, it can’t have any length-1 strings, and it can have at most one length-2 string. If we assign one such string, say HH, to the set, we are left with 1/12 to account for. So we can’t have any length-3 strings, and we can have at most one length-4 string–say TTHH; and so on.
Must this be optimal? Well, what else can we do? Rather than include HH, we can wait and include, say, some length-3 strings instead; but for the same reason that we couldn’t add any length-3 strings after adding HH, we can only include 2 of them. This can’t possibly help with the expected number of tosses. More generally, we can’t ever accumulate more probability at length<n than we can with the greedy approach, because all strings have probabilities which exactly divide the probabilities of shorter strings; so by not being greedy, all we can do (I think) is increase the expected number of tosses. (This is all conditioned on one particular winner, but since the three sets are symmetric, the unconditioned expected number of tosses is the same.)
That’s right; there’s not actually any loss of information in “discarding TT” since the TT string has no information.
