ThatDonGuy
ThatDonGuy
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billryan
January 21st, 2024 at 7:50:11 AM permalink
For those of you looking for a quick introduction to game theory, the 2023 Stanford Math Tournament had a round that taught an introduction to some basic concepts.

Click Here to download it.
DRich
DRich
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January 21st, 2024 at 9:43:31 AM permalink
Quote: ThatDonGuy

For those of you looking for a quick introduction to game theory, the 2023 Stanford Math Tournament had a round that taught an introduction to some basic concepts.

Click Here to download it.
link to original post



I learned most of my game theory from John Horton Conway books about 30 years ago.
At my age, a "Life In Prison" sentence is not much of a deterrent.
aceside
aceside
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January 21st, 2024 at 9:54:26 AM permalink
I learned everything from casinos. This pdf file is very mathematical. Does it really help gamblers? I doubt.
100xOdds
100xOdds
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January 21st, 2024 at 10:27:36 AM permalink
Quote: aceside

I learned everything from casinos. This pdf file is very mathematical. Does it really help gamblers? I doubt.
link to original post


Thx. I d/l it but haven't read it yet and probably won't now.

And this simplified wiki is no help to me:
https://simple.m.wikipedia.org/wiki/Game_theory
Craps is paradise (Pair of dice). Lets hear it for the SpeedCount Mathletes :)
ThatDonGuy
ThatDonGuy
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January 21st, 2024 at 1:37:52 PM permalink
Quote: aceside

I learned everything from casinos. This pdf file is very mathematical. Does it really help gamblers? I doubt.
link to original post


Not really. "Game Theory" and the probability theory used in gambling are two different things. However, considering the popularity of things like the "Easy Math Problems" thread, I thought it would be of interest to a few people here.

It could be useful in gambling, in a limited way. Here's an example:
You and one other person are playing the following game:
Both of you have a coin. Each of you writes down a positive integer in red on one side, and a positive number in blue on the other.
Each of you know the other's red and blue numbers.
You both secretly choose a side of your coin - red or blue - and then reveal them.
If they are the same color, you win; if they are different colors, your opponent wins.
The amount paid is the number on the loser's coin, in dollars.
Assuming both players' strategies are to play red a particular fraction of the time (not necessarily the same for each player), what is each player's best strategy?
aceside
aceside
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January 21st, 2024 at 2:59:53 PM permalink
Let me understand your game first. Here is an example.
For player 1, red/black = 3/5.
For player 2, red/black = 9/6.

You state “if they are the same color, you win; if they are different colors, your opponent wins.”

In the above example, how much do you win if you win?
ThatDonGuy
ThatDonGuy
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January 21st, 2024 at 3:16:32 PM permalink
Quote: aceside

Let me understand your game first. Here is an example.
For player 1, red/black = 3/5.
For player 2, red/black = 9/6.

You state “if they are the same color, you win; if they are different colors, your opponent wins.”

In the above example, how much do you win if you win?
link to original post


Assuming Player 1 is the "same color wins" player:
Player 1 wins 9 if both choose red, or 6 if both choose black
Player 2 wins 5 if P2 chooses red and P1 chooses black (since P1's number is 5), or 3 if P2 chooses black and P1 loses red.

I think I understand your confusion: one player (in this case, you) is the "same color wins" player, and the other (in this case, your opponent) is the "different colors wins" player.
aceside
aceside
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January 21st, 2024 at 9:15:56 PM permalink
Player 1 may win either 9 or 6, but may lose either 5 or 3, in each game. This means that Player 1 is a long-term winner if the sides of these two coins are randomly selected. No strategy is needed for Player 1 at all.

However, if Player 2 selects a black=6 side all the time but Player 1 still randomly selects a red/black=3/5 side, Player 1 should win less money, in the long term.
Last edited by: aceside on Jan 21, 2024
ThatDonGuy
ThatDonGuy
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aceside
January 22nd, 2024 at 5:00:00 PM permalink
In a game like this, usually both sides calculate what is called a Nash Equilibrium - namely, what strategy would return the same result regardless of what the opponent did.
In general terms, let S (for "same") be the player who wins if the colors match, and D (for "different") be the player who wins if the colors differ.
Let Sr and Sb be the red and blue numbers on S's coin, and Dr and Db the red and blue numbers on D's coin.

Calculate S's strategy:
Let p be the probability that S plays their red number.
If D plays red, S's EV = p (Dr) + (1 - p) (-Sb) = p (Dr + Sb) - Sb
If D plays blue, S's EV = p (-Sr) + (1 - p) (Db) = p (-Db - Sr) + Db
These are equal when p (Dr + Sb + Db + Sr) = Db + Sb, or p = (Sb + Db) / (Sb + Sr + Db + Dr).
S's strategy is to play red with probability (sum of the two blue numbers) / (sum of all four numbers), and blue with probability (sum of the two red numbers) / (sum of all four numbers).

Calculate D's strategy:
Here, let p be the probability that D plays their red number.
If S plays red, D's EV = p (-Dr) + (1 - p) (Sr) = p (-Sr - Dr) + Sr
If S plays blue, D's EV = p (Sb) + (1 - p) (-Db) = p (Sb + Db) - Db
These are equal when p (Sr + Sb + Dr + Db) = Sr + Db, or p = (Sr + Db) / (Sb + Sr + Db + Dr).
D's strategy is to play red with probability (S's red + D's blue) / (sum of all four numbers), and blue with probability (S's blue + D's red) / (sum of all four numbers).

In the case where S's numbers are red 3 and blue 5, and D's are red 9 and blue 6
S's strategy is to play red 11/23 of the time, and blue 12/23 of the time
D's strategy is to play red 9/23 of the time, and blue 14/23 of the time
The EV for S is (11/23 x 9/23 x 9 - 11/23 x 14/23 x 3 - 12/23 x 9/23 x 5 + 12/23 x 14/23 x 6) = 897 / 529, or about 1.69565.
Of course, since the EV for D is -897 / 529, chances are D will refuse to play.
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