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Topic: The Primrose Path Paradox.
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Rock Brentwood

Posts: 116
Registered: 6/18/10
The Primrose Path Paradox.
Posted: Nov 20, 2013 8:28 PM
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A classic situation seen in old Westerns is that a cowpoke loses a game of poker, pulls out the gun and tells the opponent that they're going to keep playing until he wins back his money.

Ignoring the question of why he didn't just use the gun to take the money (which, in light of what's about to be described would actually be MORE fair and honest!) the questions are -- how long on average would it take to win back the money, and what are the changes it will eventually happen?

(A) If the game is p < 50% for the perpetrator, then the amount of time (on average) would be infinity, while the chances on eventually winning it all back would be p/q, where q = 1 - p. If he gets N rounds behind, then it becomes (p/q)^N. There is a point where it drops under 50% (the "cut your losses" point).

(B) If the game is p > 50%, then the amount of time on average would be 1/(p - q) rounds, and the probability of eventually winning everything back is 100%. If he's N rounds down, then it becomes N/(p - q) for time and 100% for probability.

(C) If p = 50%, the average time is infinity, but the probability of eventually making it all back is 100%. Thus, the tip of the iceberg of the paradox.

But there's more to it than that.

So, now change venues. Suppose you're playing a game where your probability of winning is p, and of losing is q (p + q = 1, p > 0 and q > 0) and you lose the first time.

For probabilities p in the range 1/3 < p < 1/2, there is better than a 50% chance on eventually breaking even. The only difference here is that there is a "cut your losses" point, where the probability of eventually getting back drops below 50%.

For the case 1/3 < p < 1/2, the "cut your losses" point occurs where N > 1/log_2(q/p) = 1/log_2((1 - p)/p).

For a game where the margin is close to 50-50, such as p = 30/(30 + 31), q = 31/(30 + 31), one would have N = 1/log_2(31/30) ~ 30 ln(2) ~ 21.

So, by making p close to 50-50, you can be strung along to very deep levels before reaching to "cut your losses" point.

The paradox for p = 1/2 and even the quasi-paradox for (1/3 < p < 1/2) not only catches the feel and spirit that every obsessive gambler feels, experiences, but that it is the very trap underlying the addiction itself.

Moreover, it seems to be a case of a more general situation of this type, that I call the "Primrose Path Paradox". This is the situation where
(a) at each point, the rational choice is to step forward, but
(b) where each forward step brings you closer to ruin, overall, and
(c) by the time the likelihood or inevitability of ruin becomes clearly recognizable, it's too late to turn back.



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