< Earlier Kibitzing · PAGE 286 OF 286 ·
|Jul-01-16|| ||WannaBe: Most interesting...
In theory, every flip can be tails, you keep collecting a dollar, on the otherside of the coin, it, too can be heads infinitely, too.
Of course, we know that's not likely to happen. Hummmmmm
|Jul-01-16|| ||alexmagnus: No X will so because the expected payout is 1/2*1+1/4*2+1/8*4+... =1/2+1/2+1/2+... - that is, infinite.|
|Jul-02-16|| ||Tiggler: Not the correct series <alexmagnus>.|
|Jul-02-16|| ||john barleycorn: < Tiggler: ...
"Our second puzzle involves gambling. Let me describe a new game at the casino. The rules are as follows:
The house puts one dollar in a pot. The dealer flips a coin. If the coin comes up heads, the house doubles the amount of money in the pot and the dealer flips the coin again. If the coin comes up tails, then the game is over and you win the contents of the pot. The game costs X dollars to play.
If the casino is to break even playing this game, how much should they charge — what should be the value of X? And how much would you be willing to pay to play this game?"...>
Can the player ever lose his wager? Not from the description of the game execution. Looks like a variant of "comes tails you win comes head you do not lose"
|Jul-02-16|| ||alexmagnus: <Tiggler> Why not? It is correct, at least if we assume that the casino can pay out any amount of money (a real casino would of course go bankrupt at some point). |
But if the casino says the the game stops at the point the jackpot exceeds their (finite) money reserves - then, of course, X is a finite value, and quite a small one at that. If we assume the casino has one billion dollars, then the expected payout is only $15.50 (the power of two exceeding one billion is 2^30, so the coin can be flipped up to 31 times, resulting in a 31/2 expected payout).
|Jul-02-16|| ||Tiggler: <alexmagnus> Third term is wrong.|
|Jul-03-16|| ||john barleycorn: < Tiggler: <alexmagnus> Third term is wrong.>|
I read the 3rd term as (1/8)*4 and that is pretty much the mathematical expections for winning 4$.
Again, I like to ask: "when does the player lose his wager?". According to the game description it is not clear to me.
Then of course we have the "golden rule" that you only gamble the money you can afford to lose. Of course, there is always the "utility function" which makes 2$ an initial good bet for the player.
|Jul-03-16|| ||Tiggler: <john barleycorn>
|Jul-03-16|| ||john barleycorn: < Tiggler: <john barleycorn>
The only way for the player to win 4$ with 3 coinflips is (head, head, tails)which is one of 8 possibilities so its probability is 1/8.
So the mathematical expectation is (1/8)*4. At least that is the way I learned it.
|Jul-03-16|| ||Tiggler: <The only way for the player to win 4$ with 3 coinflips is (head, head, tails)>|
This does not win $4.
|Jul-03-16|| ||john barleycorn: <Tiggler: ...
This does not win $4.>
1. the bank puts one dollar down and flips the coin. head comes.
2. the bank adds one dollar to the pot (now 2 dollar in total) and flips the coin. head comes.
3. the bank adds 2 dollars to the pot (now 4 $ in total) and flips the coin.
tails come and the player wins the 4$.
|Jul-03-16|| ||Tiggler: <3. the bank adds 2 dollars to the pot (now 4 $ in total) and flips the coin.>|
You are correct, and so was <alexmagnus>.
I solved the wrong problem: I had the bank adding $1 each time head comes. That is a much nicer puzzle imo, but not the correct one.
In that case the series is:
1/2 + (1/4)*2 + (1/8)*3 + (1/16)*4 + ...
To sum it we decompose as follows:
1/2 + 1/4 + 1/8 + 1/16 ...
. + 1/4 + 1/8 + 1/16 ...
. + 1/8 + 1/16 ...
. + 1/16 ...
= 1 + 1/2 + 1/4 + 1/8...
I thought that was the correct answer. Sorry for mixing it up
|Jul-03-16|| ||john barleycorn: Actually, 2$ would be the amount I would bet for the player when he loses on head coming. But this is not said the game description. So, any amount is ok for the player as he can never lose.|
|Jul-03-16|| ||alexmagnus: <Tiggler> Funnily, that variant of the game came to my mind too, with the same solution as you posted.|
|Jul-03-16|| ||john barleycorn: Still, my question: "When does the player lose?" There is no misreading possible as it is not mentioned in the game description.|
<If the casino is to break even playing this game, how much should they charge — what should be the value of X?>
is absolutely meaningless in that case.
|Jul-03-16|| ||beatgiant: <john barleycorn>
The player loses money if the charge X he paid is higher than the amount he won.
|Jul-03-16|| ||beatgiant: The gambling game in question is the well-known "St. Petersburg Paradox." Wikipedia has a good article on it.|
|Jul-03-16|| ||WannaBe: Here's the link, I'll save y'all from typing it in google. =)|
|Jul-04-16|| ||john barleycorn: <beatgiant: <john barleycorn>
The player loses money if the charge X he paid is higher than the amount he won.>|
Ah, thank you. so the player pays an amount X at the beginning and it is automatically collected by the casino irrelevant of the outcome of the coin flips.
|Jul-14-16|| ||al wazir: It's convention time!
As is traditional, a banquet is held at the conclusion of the Republicratic convention. Two hundred senior partisans, selected by invitation from the party's most principled princes and dignified dignitaries, stream into the ballroom where tables have been set up.
When they arrive they find that seats has been preassigned. Every setting has a name tag, and the invitees are required to sit at the place with their own tag.
Two hundred mostly elderly, increasingly hungry, slightly tipsy (from pre-banquet cocktails) guests mill around, bumping into chairs and tables, stepping on each other's toes, as they try to locate their places. Some give up the search and sit at random, only to be displaced when the rightful occupant finds his name and claims the place. A few headstrong mavericks -- most of them long-serving legislators accustomed to their own safe electoral seats -- redistribute the name tags. As a result, some tables have more tags than chairs.
Is there a better way to do it? (Cf. https://www.amazon.com/Art-Computer...)
|Jul-14-16|| ||Schwartz: Just a small amount should search at the same time, unless there's a lot of room.|
|Jul-14-16|| ||john barleycorn: get them seated by trained stuff.|
|Jul-14-16|| ||Tiggler: More cocktails! Then bring them in on wheelbarrows.|
|Jul-14-16|| ||Tiggler: Open seating is the reason Southwest Airlines boards twice as fast and saves billions.|
|Jul-15-16|| ||al wazir: <Tiggler: Open seating is the reason Southwest Airlines boards twice as fast and saves billions.> Good point. But in this scenario the seating isn't open.|
Since you mention airline seating, I should point out that some airlines *platoon* boarding passengers. Their tickets are labeled "Group 1," "Group 2," and so on, and the groups are assigned seats in such a way that Group 1 goes to the back of the plane where it won't interfere with Group 2. Then Group 2 takes the seats forward of them, and so on.
So one solution to the problem would be to give the banqueters tickets with table numbers. Alternatively, they could simply be told ahead of time what table they are assigned to, and at appropriate intervals the Maître d' could announce over the PA system "Table 1 now seating ... Table 2 now seating ...," etc. But I'm not convinced that there isn't a better way.
I recently attended an outdoor luncheon where they must have been around 200 guests, if not more. Owing to a conflict (it was a chess conflict, as a matter of fact), I arrived an hour after everyone else had been seated, and even knowing the table I was supposed to sit at I had a hell of a time finding my seat. So if everyone arrived at the same time it must have been a mess.
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