A poker room has a high-hand jackpot that pays out whenever a qualifying hand holds up as the highest hand for an hour (then resets). Since poker hands are stupid and annoying, let's model this as
There is an arbitrarily long sequence of X_i, i=1->inf iid U(0,1). There are n hands drawn per hour, evenly spaced. An X_i is a winning high hand if it's the first hand in the sequence that's the maximum of X_1 to X_(i+n).
We want to know the expected value of how many hands it will take for the high hand to show up. To be pedantic, define a variable Y that's a function of the X_i sequence where Y=1 if X_1 is the high hand, Y=2 if X_2 is the high hand etc.
What is the limit n->inf of E[Y]/n in closed form?
Wouldn't there be no limit to E[y] as n→∞ ? The larger the n, the stronger a hand you'd need. In this case the strongest "hand" is 1, and P(1)=0, so if I imagine n equal to ∞, then wherever you are in the sequence, it's always a near certainty that a hand later in the sequence is higher.
Yeah, E[Y] would obviously keep growing, but I wanted E[Y]/n which is how many hours it would take on average to hit. That's actually a decreasing function of n (and it's upper bounded from simple arguments).
Oh so play will continue for as many hours as necessary, and there is a minimum qualifying hand, which I suppose you want to leave as a variable, say q. You're asking how many hours it will take, on average, for a qualifying hand to show up?
Oh so play will continue for as many hours as necessary, and there is a minimum qualifying hand, which I suppose you want to leave as a variable, say q. You're asking how many hours it will take, on average, for a qualifying hand to show up?
For finite n it's simply 1/(1-q)/n
As n→∞, well why wouldn't it approach 0?
I think I'm still misinterpreting the question.
The qualifying part isn't very interesting, so don't worry about it. The hand has to hold up for an hour, so it has to be the current best hand and also beat the next n random numbers in the sequence. For a billion hands/hr, the odds of showing up in the first hundred are super low (one of the first hundred has to beat the next ~billion), but also the odds of showing up in the first hour have to be at least 50% (the highest number in the first 2 hours is 50% likely to be in the first hour, which would mean a winner in the first hour).
I'm glad it's clear to you, whosnext, but I'm feeling purty dumb over here.
The HH resets every hour, so as n→∞, the chance having a HH awarded in the first hour becomes 100%. At the same time, E[Y] grows to ∞. The winner of the hour is 100% to be at the "end" of ∞ hands, ie at the end of the hour, so the expected #hours for the winner (not qualifier) to show up is exactly one. That seems clear from reasoning even though mathematically all I've produced is an indeterminate form ∞/∞.
I feel like the question might be more interesting for finite n. But if TC is asking this (or more likely, quizzing us), there must be more to it than what I just wrote, so it's safe to say I still don't get the question.
I'm glad it's clear to you, whosnext, but I'm feeling purty dumb over here.
The HH resets every hour
In the actual promo, once somebody's won- had a hand that holds up as the high hand for an hour, they pay the money, then they start back looking for a completely new high hand.
I think you're thinking about "making a high enough hand" (which doesn't matter at all in my model), not the "having the hand hold up for the next hour" part.
Here are some results for very small N over 10,000,000 trials.
One reason I post these is in case someone comes up with an analytical method to derive these results and would find numerical results useful as verification.
N (Hands per hour)
Avg Hand # of first hand that holds up for an hour
Avg Hand # of first hand that holds up for an hour / N
1
1.718
1.7184
2
2.482
1.2408
3
3.255
1.0851
4
4.032
1.0080
5
4.808
0.9616
6
5.588
0.9314
7
6.367
0.9096
8
7.150
0.8937
9
7.926
0.8807
TomC posted above that the analytical result for N=1 is e-1.
For a question of pure probability and not strategy, with poker hands it is easier to use an atomic range than using a real distribution (0,1).
Discretize the possible hands and then you know what the minimum qualifying hand is, you can calculate how often it will happen and how long that will take, and how often it will hold for a given number of future hands etc.
