Re: [math-fun] playing card color guessing game
Hmm, strategy seems clear: guess the majority color remaining, if any. So the answer can be expressed in terms of all possible shufflings, the N = 52!/(26!)^2 balanced bit strings of length 52, since for each one we know the expected payoff with this strategy (a messy sum): E = Sum_{x balanced bit string length 52} (1/N) * Sum_{1 <= k <=52} etc. —Dan Allan Wechsler wrote: ----- I know how to analyze this, but I haven't tried to actually do the work. But I'm going to make a wild-ass stab based on an intuition. Is it $32.87? On Wed, Dec 26, 2018 at 11:50 AM Thane Plambeck <tplambeck@gmail.com> wrote:
the usual 52-card playing deck is shuffled and then revealed one-by-one to a player who guesses the color (red or black) of each card prior to its being revealed. the player earns one dollar for each card whose color he guesses correctly; there is no penalty for being wrong. what is a fair entry price to play? ----=-
I agree that it seems to make the most sense to be greedy. But I haven't proven it yet! There is something about the logic of this game that keeps tripping me up. On Dec 26, 2018 3:18 PM, "Dan Asimov" <dasimov@earthlink.net> wrote: Hmm, strategy seems clear: guess the majority color remaining, if any. So the answer can be expressed in terms of all possible shufflings, the N = 52!/(26!)^2 balanced bit strings of length 52, since for each one we know the expected payoff with this strategy (a messy sum): E = Sum_{x balanced bit string length 52} (1/N) * Sum_{1 <= k <=52} etc. —Dan Allan Wechsler wrote: ----- I know how to analyze this, but I haven't tried to actually do the work. But I'm going to make a wild-ass stab based on an intuition. Is it $32.87? On Wed, Dec 26, 2018 at 11:50 AM Thane Plambeck <tplambeck@gmail.com> wrote:
the usual 52-card playing deck is shuffled and then revealed one-by-one to a player who guesses the color (red or black) of each card prior to its being revealed. the player earns one dollar for each card whose color he guesses correctly; there is no penalty for being wrong. what is a fair entry price to play? ----=-
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I think the key thing is that the state of the cards don’t depend on your bet, so there’s no hard in being greedy - no point in sacrificing for the future. Cris
On Dec 26, 2018, at 1:42 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I agree that it seems to make the most sense to be greedy. But I haven't proven it yet! There is something about the logic of this game that keeps tripping me up.
On Dec 26, 2018 3:18 PM, "Dan Asimov" <dasimov@earthlink.net> wrote:
Hmm, strategy seems clear: guess the majority color remaining, if any.
So the answer can be expressed in terms of all possible shufflings, the N = 52!/(26!)^2 balanced bit strings of length 52, since for each one we know the expected payoff with this strategy (a messy sum):
E = Sum_{x balanced bit string length 52} (1/N) * Sum_{1 <= k <=52} etc.
—Dan
Allan Wechsler wrote: -----
I know how to analyze this, but I haven't tried to actually do the work.
But I'm going to make a wild-ass stab based on an intuition. Is it $32.87?
On Wed, Dec 26, 2018 at 11:50 AM Thane Plambeck <tplambeck@gmail.com> wrote:
the usual 52-card playing deck is shuffled and then revealed one-by-one to a player who guesses the color (red or black) of each card prior to its being revealed. the player earns one dollar for each card whose color he guesses correctly; there is no penalty for being wrong. what is a fair entry price to play? ----=-
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Cris, that's pretty convincing, thanks. C(0,0) = 0 C(0,n) = n C(1,1) = 3/2 The simplest nontrivial case seems to be C(1,2). If you make the greedy guess, you have a 2/3 chance of being right and getting 3/2 + 1, and a 1/3 chance of being wrong and getting 2. 2/3 * 5/2 + 1/3 * 2 = 7/3. If you make the perverse guess, you have a 1/3 chance of being right and getting 3, and a 2/3 chance of being wrong and getting 3/2. So the total value is 2; the greedy guess is 1/3 better. On Wed, Dec 26, 2018 at 3:44 PM Cris Moore <moore@santafe.edu> wrote:
I think the key thing is that the state of the cards don’t depend on your bet, so there’s no hard in being greedy - no point in sacrificing for the future.
Cris
On Dec 26, 2018, at 1:42 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I agree that it seems to make the most sense to be greedy. But I haven't proven it yet! There is something about the logic of this game that keeps tripping me up.
On Dec 26, 2018 3:18 PM, "Dan Asimov" <dasimov@earthlink.net> wrote:
Hmm, strategy seems clear: guess the majority color remaining, if any.
So the answer can be expressed in terms of all possible shufflings, the N = 52!/(26!)^2 balanced bit strings of length 52, since for each one we know the expected payoff with this strategy (a messy sum):
E = Sum_{x balanced bit string length 52} (1/N) * Sum_{1 <= k <=52} etc.
—Dan
Allan Wechsler wrote: -----
I know how to analyze this, but I haven't tried to actually do the work.
But I'm going to make a wild-ass stab based on an intuition. Is it $32.87?
On Wed, Dec 26, 2018 at 11:50 AM Thane Plambeck <tplambeck@gmail.com> wrote:
the usual 52-card playing deck is shuffled and then revealed one-by-one to a player who guesses the color (red or black) of each card prior to its being revealed. the player earns one dollar for each card whose color he guesses correctly; there is no penalty for being wrong. what is a fair entry price to play? ----=-
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Another way to put the argument: whatever you do on this turn, your expected payoff from the optimal strategy on the remaining turns doesn’t change. So you might as well maximize your expected payoff on this turn, i.e., be greedy. It might be fun to invent a version of the game where this isn’t true - taking an extra card from the deck if you guess wrong, for instance - so that some degree of bet-hedging becomes optimal. C
On Dec 26, 2018, at 2:03 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Cris, that's pretty convincing, thanks.
C(0,0) = 0 C(0,n) = n C(1,1) = 3/2
The simplest nontrivial case seems to be C(1,2). If you make the greedy guess, you have a 2/3 chance of being right and getting 3/2 + 1, and a 1/3 chance of being wrong and getting 2. 2/3 * 5/2 + 1/3 * 2 = 7/3. If you make the perverse guess, you have a 1/3 chance of being right and getting 3, and a 2/3 chance of being wrong and getting 3/2. So the total value is 2; the greedy guess is 1/3 better.
On Wed, Dec 26, 2018 at 3:44 PM Cris Moore <moore@santafe.edu> wrote:
I think the key thing is that the state of the cards don’t depend on your bet, so there’s no hard in being greedy - no point in sacrificing for the future.
Cris
On Dec 26, 2018, at 1:42 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I agree that it seems to make the most sense to be greedy. But I haven't proven it yet! There is something about the logic of this game that keeps tripping me up.
On Dec 26, 2018 3:18 PM, "Dan Asimov" <dasimov@earthlink.net> wrote:
Hmm, strategy seems clear: guess the majority color remaining, if any.
So the answer can be expressed in terms of all possible shufflings, the N = 52!/(26!)^2 balanced bit strings of length 52, since for each one we know the expected payoff with this strategy (a messy sum):
E = Sum_{x balanced bit string length 52} (1/N) * Sum_{1 <= k <=52} etc.
—Dan
Allan Wechsler wrote: -----
I know how to analyze this, but I haven't tried to actually do the work.
But I'm going to make a wild-ass stab based on an intuition. Is it $32.87?
On Wed, Dec 26, 2018 at 11:50 AM Thane Plambeck <tplambeck@gmail.com> wrote:
the usual 52-card playing deck is shuffled and then revealed one-by-one to a player who guesses the color (red or black) of each card prior to its being revealed. the player earns one dollar for each card whose color he guesses correctly; there is no penalty for being wrong. what is a fair entry price to play? ----=-
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For 52 I get 30.0406647747139. For 1000 I get 519.321591265648. Here's a sequence by powers of ten. This can easily be extended. The last value gives an indication that there may be a quickly-converging sum. 10 6.53174603174603 100 55.7822564509274 1000 519.321591265648 10000 5062.16727352803 100000 50197.6668602967 On Wed, Dec 26, 2018 at 1:12 PM Cris Moore <moore@santafe.edu> wrote:
Another way to put the argument: whatever you do on this turn, your expected payoff from the optimal strategy on the remaining turns doesn’t change. So you might as well maximize your expected payoff on this turn, i.e., be greedy.
It might be fun to invent a version of the game where this isn’t true - taking an extra card from the deck if you guess wrong, for instance - so that some degree of bet-hedging becomes optimal.
C
On Dec 26, 2018, at 2:03 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Cris, that's pretty convincing, thanks.
C(0,0) = 0 C(0,n) = n C(1,1) = 3/2
The simplest nontrivial case seems to be C(1,2). If you make the greedy guess, you have a 2/3 chance of being right and getting 3/2 + 1, and a 1/3 chance of being wrong and getting 2. 2/3 * 5/2 + 1/3 * 2 = 7/3. If you make the perverse guess, you have a 1/3 chance of being right and getting 3, and a 2/3 chance of being wrong and getting 3/2. So the total value is 2; the greedy guess is 1/3 better.
On Wed, Dec 26, 2018 at 3:44 PM Cris Moore <moore@santafe.edu> wrote:
I think the key thing is that the state of the cards don’t depend on your bet, so there’s no hard in being greedy - no point in sacrificing for the future.
Cris
On Dec 26, 2018, at 1:42 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I agree that it seems to make the most sense to be greedy. But I haven't proven it yet! There is something about the logic of this game that keeps tripping me up.
On Dec 26, 2018 3:18 PM, "Dan Asimov" <dasimov@earthlink.net> wrote:
Hmm, strategy seems clear: guess the majority color remaining, if any.
So the answer can be expressed in terms of all possible shufflings, the N = 52!/(26!)^2 balanced bit strings of length 52, since for each one we know the expected payoff with this strategy (a messy sum):
E = Sum_{x balanced bit string length 52} (1/N) * Sum_{1 <= k <=52} etc.
—Dan
Allan Wechsler wrote: -----
I know how to analyze this, but I haven't tried to actually do the work.
But I'm going to make a wild-ass stab based on an intuition. Is it $32.87?
On Wed, Dec 26, 2018 at 11:50 AM Thane Plambeck <tplambeck@gmail.com> wrote:
the usual 52-card playing deck is shuffled and then revealed one-by-one to a player who guesses the color (red or black) of each card prior to its being revealed. the player earns one dollar for each card whose color he guesses correctly; there is no penalty for being wrong. what is a fair entry price to play? ----=-
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The exact value for n=52 is 3724430600965475/123979633237026 That denominator is (52 choose 26)/4. -tom On Wed, Dec 26, 2018 at 1:48 PM Tomas Rokicki <rokicki@gmail.com> wrote:
For 52 I get 30.0406647747139.
For 1000 I get 519.321591265648.
Here's a sequence by powers of ten. This can easily be extended. The last value gives an indication that there may be a quickly-converging sum.
10 6.53174603174603
100 55.7822564509274
1000 519.321591265648
10000 5062.16727352803
100000 50197.6668602967
On Wed, Dec 26, 2018 at 1:12 PM Cris Moore <moore@santafe.edu> wrote:
Another way to put the argument: whatever you do on this turn, your expected payoff from the optimal strategy on the remaining turns doesn’t change. So you might as well maximize your expected payoff on this turn, i.e., be greedy.
It might be fun to invent a version of the game where this isn’t true - taking an extra card from the deck if you guess wrong, for instance - so that some degree of bet-hedging becomes optimal.
C
On Dec 26, 2018, at 2:03 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Cris, that's pretty convincing, thanks.
C(0,0) = 0 C(0,n) = n C(1,1) = 3/2
The simplest nontrivial case seems to be C(1,2). If you make the greedy guess, you have a 2/3 chance of being right and getting 3/2 + 1, and a 1/3 chance of being wrong and getting 2. 2/3 * 5/2 + 1/3 * 2 = 7/3. If you make the perverse guess, you have a 1/3 chance of being right and getting 3, and a 2/3 chance of being wrong and getting 3/2. So the total value is 2; the greedy guess is 1/3 better.
On Wed, Dec 26, 2018 at 3:44 PM Cris Moore <moore@santafe.edu> wrote:
I think the key thing is that the state of the cards don’t depend on your bet, so there’s no hard in being greedy - no point in sacrificing for the future.
Cris
On Dec 26, 2018, at 1:42 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I agree that it seems to make the most sense to be greedy. But I haven't proven it yet! There is something about the logic of this game that keeps tripping me up.
On Dec 26, 2018 3:18 PM, "Dan Asimov" <dasimov@earthlink.net> wrote:
Hmm, strategy seems clear: guess the majority color remaining, if any.
So the answer can be expressed in terms of all possible shufflings, the N = 52!/(26!)^2 balanced bit strings of length 52, since for each one we know the expected payoff with this strategy (a messy sum):
E = Sum_{x balanced bit string length 52} (1/N) * Sum_{1 <= k <=52} etc.
—Dan
Allan Wechsler wrote: -----
I know how to analyze this, but I haven't tried to actually do the work.
But I'm going to make a wild-ass stab based on an intuition. Is it $32.87?
On Wed, Dec 26, 2018 at 11:50 AM Thane Plambeck <tplambeck@gmail.com> wrote:
the usual 52-card playing deck is shuffled and then revealed one-by-one to a player who guesses the color (red or black) of each card prior to its being revealed. the player earns one dollar for each card whose color he guesses correctly; there is no penalty for being wrong. what is a fair entry price to play? ----=-
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-- -- http://cube20.org/ -- http://golly.sf.net/ --
Sorry, I meant no harm.
I think the key thing is that the state of the cards don’t depend on your bet, so there’s no hard in being greedy - no point in sacrificing for the future.
Cris
On Dec 26, 2018, at 1:42 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I agree that it seems to make the most sense to be greedy. But I haven't proven it yet! There is something about the logic of this game that keeps tripping me up.
On Dec 26, 2018 3:18 PM, "Dan Asimov" <dasimov@earthlink.net> wrote:
Hmm, strategy seems clear: guess the majority color remaining, if any.
So the answer can be expressed in terms of all possible shufflings, the N = 52!/(26!)^2 balanced bit strings of length 52, since for each one we know the expected payoff with this strategy (a messy sum):
E = Sum_{x balanced bit string length 52} (1/N) * Sum_{1 <= k <=52} etc.
—Dan
Allan Wechsler wrote: -----
I know how to analyze this, but I haven't tried to actually do the work.
But I'm going to make a wild-ass stab based on an intuition. Is it $32.87?
On Wed, Dec 26, 2018 at 11:50 AM Thane Plambeck <tplambeck@gmail.com> wrote:
the usual 52-card playing deck is shuffled and then revealed one-by-one to a player who guesses the color (red or black) of each card prior to its being revealed. the player earns one dollar for each card whose color he guesses correctly; there is no penalty for being wrong. what is a fair entry price to play? ----=-
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participants (4)
-
Allan Wechsler -
Cris Moore -
Dan Asimov -
Tomas Rokicki