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Ace2
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May 26th, 2024 at 12:48:45 PM permalink
Now for a truly easy math puzzle

Using a 00 roulette wheel, whatís the expected amount of spins to see the first repeated number?

Repeated does not mean consecutive
Last edited by: Ace2 on May 26, 2024
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aceside
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May 26th, 2024 at 3:31:01 PM permalink
1482/38 = 39
Ace2
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May 26th, 2024 at 4:01:37 PM permalink
Quote: aceside

1482/38 = 39

link to original post

39 would be the maximum possible spins to see the first repeat. Incidentally, the chance of that happening is 38!/38^38 or about one in 2.0571 quadrillion. So disagree

Incidentally again, that probability can be calculated with 99.8% accuracy by taking (2 π 38)^.5 / e^38 =~ 1 / (2.0616 * 10^15)
Itís all about making that GTA
ThatDonGuy
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May 26th, 2024 at 6:25:54 PM permalink
I did a brute force calculation, and got:

729,379,411,907,019,390,404,919,209,269,696,370,543,494,910,507 / 86,740,040,640,511,330,566,403,603,474,709,827,322,252,250,248
which is about 8.4088

Wizard
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May 26th, 2024 at 9:58:41 PM permalink

8.408797212

Solved in Excel
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Ace2
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May 26th, 2024 at 10:32:24 PM permalink
TDG and Wizard,

Your answer looks good and itís pretty easy to solve this in excel.

However, I believe the most elegant, efficient, and eloquent solution is to take the integral from zero to infinity of:

(((x/38) + 1) / e^(x/38))^38 dx
Itís all about making that GTA
aceside
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May 26th, 2024 at 10:46:56 PM permalink
Hi, I still haven't figured out how to calculate this; however, I am particularly interested in this modified puzzle:
Using a 00 roulette wheel, whatís the expected amount of spins to see two consecutive 7's?
Ace2
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aceside
May 26th, 2024 at 10:57:42 PM permalink
Quote: aceside

Hi, I still haven't figured out how to calculate this; however, I am particularly interested in this modified puzzle:
Using a 00 roulette wheel, whatís the expected amount of spins to see two consecutive 7's?
link to original post

38 + 38^2 = 1,482 spins
Itís all about making that GTA
aceside
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May 26th, 2024 at 11:01:14 PM permalink
Let me change a little:
Using a 00 roulette wheel, whatís the expected amount of spins to see two consecutive same numbers, any number?
Ace2
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May 26th, 2024 at 11:03:35 PM permalink
Quote: aceside

Let me change a little:
Using a 00 roulette wheel, whatís the expected amount of spins to see two consecutive same numbers, any number?
link to original post

Iíd say 39
Itís all about making that GTA
aceside
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May 26th, 2024 at 11:18:19 PM permalink
Let me make it more complicated:
Using a 00 roulette wheel, whatís the expected amount of spins to see two consecutive 7's or any two 7s that are gapped by at least one 0 or 00? In other words, the 0 and 00 numbers do not break the 7-7 streak.
ThatDonGuy
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May 27th, 2024 at 7:46:30 AM permalink
Quote: aceside

Let me make it more complicated:
Using a 00 roulette wheel, whatís the expected amount of spins to see two consecutive 7's or any two 7s that are gapped by at least one 0 or 00? In other words, the 0 and 00 numbers do not break the 7-7 streak.
link to original post



Let E(n) be the expected number of spins needed, given that you currently have n 7's since the last non-0/00 spin
E(0) is the desired result

E(1) = 1 + 35/38 E(0) + 2/38 E(1)
38 E(1) = 38 + 35 E(0) + 2 E(1)
36 E(1) = 38 + 35 E(0)
E(1) = 19/18 + 35/36 E(0)

E(0) = 1 + 37/38 E(0) + 1/38 E(1)
1/38 E(0) = 1 + 1/38 E(1)
E(0) = 38 + E(1)
= 38 + 19/18 + 35/36 E(0)
1/36 E(0) = 38 + 19/18
E(0) = 38 x 37 = 1406

aceside
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May 27th, 2024 at 9:59:28 AM permalink
Quote: ThatDonGuy

Quote: aceside

Let me make it more complicated:
Using a 00 roulette wheel, whatís the expected amount of spins to see two consecutive 7's or any two 7s that are gapped by at least one 0 or 00? In other words, the 0 and 00 numbers do not break the 7-7 streak.
link to original post



Let E(n) be the expected number of spins needed, given that you currently have n 7's since the last non-0/00 spin
E(0) is the desired result

E(1) = 1 + 35/38 E(0) + 2/38 E(1)
38 E(1) = 38 + 35 E(0) + 2 E(1)
36 E(1) = 38 + 35 E(0)
E(1) = 19/18 + 35/36 E(0)

E(0) = 1 + 37/38 E(0) + 1/38 E(1)
1/38 E(0) = 1 + 1/38 E(1)
E(0) = 38 + E(1)
= 38 + 19/18 + 35/36 E(0)
1/36 E(0) = 38 + 19/18
E(0) = 38 x 37 = 1406


link to original post


Great! Is it possible to calculate this using the following recurrence relation in Excel? We list all possibilities and then average the result out, like this:

Number of Spins, Outcome String, Respective Probability
1, None, 0;
2, 77, (1/38)^2;
3. X77 or 707, (37/38)(1/38) +(1/38)^3;
4. XY77 or X707 or 7007, (37/38)(1/38) +(1/38)^3+(1/38)^3;
... and so on.

I'm thinking, for this purpose, it's probably better to change the question to this: using a 00 roulette wheel, whatís the expected amount of spins to see two consecutive 7's or any two 7s that are gapped by at least one 0 or 1? In other words, the 0 and 1 numbers do not break the 7-7 streak.
ThatDonGuy
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May 27th, 2024 at 11:57:25 AM permalink
Quote: aceside

Great! Is it possible to calculate this using the following recurrence relation in Excel? We list all possibilities and then average the result out, like this:

Number of Spins, Outcome String, Respective Probability
1, None, 0;
2, 77, (1/38)^2;
3. X77 or 707, (37/38)(1/38) +(1/38)^3;
4. XY77 or X707 or 7007, (37/38)(1/38) +(1/38)^3+(1/38)^3;
... and so on.

I'm thinking, for this purpose, it's probably better to change the question to this: using a 00 roulette wheel, whatís the expected amount of spins to see two consecutive 7's or any two 7s that are gapped by at least one 0 or 1? In other words, the 0 and 1 numbers do not break the 7-7 streak.
link to original post


In theory, you could, but the combinations get unwieldy as you add spins.
A set consists of:
Zero or more groups of ((zero or more 0/00s) followed by (zero or one 7s) followed by (zero or more 0/00s) followed by a non-0/00/7),
followed by (zero or more 0/00s) followed by a 7 followed by (zero or more 0/00s) followed by another 7.
aceside
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May 28th, 2024 at 6:46:00 AM permalink
I still havenít fully understood this set of two equations:
E(1) = 1 + 35/38 E(0) + 2/38 E(1),
E(0) = 1 + 37/38 E(0) + 1/38 E(1).
Is it possible to simulate this situation to see the solution of 1406 is correct or not?

The problem is: a 0 or 00, or any combination of these two does not break the 7-7 streak. This information is not described in the above two equations.
ThatDonGuy
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May 28th, 2024 at 3:41:56 PM permalink
Quote: aceside

I still havenít fully understood this set of two equations:
E(1) = 1 + 35/38 E(0) + 2/38 E(1),
E(0) = 1 + 37/38 E(0) + 1/38 E(1).
Is it possible to simulate this situation to see the solution of 1406 is correct or not?

The problem is: a 0 or 00, or any combination of these two does not break the 7-7 streak. This information is not described in the above two equations.
link to original post


Yes, it is.
E(n) is the expected number of spins needed when you have n 7s in the current streak.
When you have one 7, you are at E(1); if the next spin is 1-6 or 8-36 (probability 35/38), you are now at zero 7s, or the E(0) state; if it is 0 or 00 (probability 2/38), you are still at one 7, or the E(1) state; if it is 7 (probability 1/38), you are done spinning.
In other words, the expected number of spins needed when you have one 7 in the streak is the sum of:
1 (the next spin)
35/38 x the expected number when you have no 7s in the streak
2/38 x the expected number when you have one 7 in the streak
1/38 x 0

I did simulate this, and got somewhere around 1406.
Wizard
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May 29th, 2024 at 12:54:14 PM permalink
Cards in a 52-card deck are turned over in a shuffled deck one at a time until the first queen appears.

What is more likely to be turned over as the next card, the queen of spades or king of spades?
Last edited by: Wizard on May 29, 2024
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Ace2
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May 29th, 2024 at 2:29:57 PM permalink
Quote: Wizard

Cards are turned over in a shuffled deck one at a time until the first queen appears.

What is more likely to be turned over as the next card, the queen of spades or king of spades?




King of spades will be next with 10/19 probability
Itís all about making that GTA
Wizard
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May 29th, 2024 at 2:50:56 PM permalink
Quote: Ace2



King of spades will be next with 10/19 probability

link to original post



I have a feeling there is some misunderstanding. First, I should have stated it's an ordinary 52-card deck. Second, the question is not asking which will happen next, but what is more likely to be turned over on the next card.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
charliepatrick
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May 29th, 2024 at 3:35:13 PM permalink
Obviously sometimes it won't be either the Qs or Ks as the next card, so we'll ignore those and only look at ones where either one or the other is.
We can then ignore all the other cards in the deck and only consider the order in which the King of Spades and four Queens appear.
Thus we have five objects, "Ks" "Qs" "Qx" "Qy" "Qz" which can be in any order
There are 120 of these (assuming the other Queens are different)
Now consider the position of the King of Spades and the position of the Queen of Spades in the five critical cards
(1 2) Ks Qs Qx Qy Qz will be neither (6 ways)
There are six ways to order the three non-Spade Queens, and similar logic applies to all the other examples.
In this case, it's impossible for either the Ks or Qs to be the "next" card; so we'll call this outcome as "will be neither".

(1 3) Ks Qx Qs Qy Qz will be Qs (6 ways)
In this case the Ks has already appeared, then a non-Spade Queen, then the Queen of Spades.
(1 4) Ks Qx Qy Qs Qz will be neither (6 ways)
In this case the Ks, then a non-Spade Queen, then another non-Spade Queen: hence "neither".
(1 5) Ks Qx Qy Qz Qs will be neither (6 ways)
(2 1) Qs Ks Qx Qy Qz will be Ks (6 ways)
(2 3) Qx Ks Qs Qy Qz will be Ks (6 ways)
(2 4) Qx Ks Qy Qs Qz will be Ks (6 ways)
(2 5) Qx Ks Qy Qz Qs will be Ks (6 ways)
(3 1) Qs Qx Ks Qy Qz will be neither (6 ways)
(3 2) Qx Qs Ks Qy Qz will be Qs (6 ways)
(3 4) Qx Qy Ks Qs Qz will be neither (6 ways)
(3 5) Qx Qy Ks Qz Qs will be neither (6 ways)
(4 2) same logic, will be Qs as (3 ?) (6 ways)
(5 2) same logic, will be Qs as (3 ?) (6 ways)
(4 1, 4 3, 4 5, 5 1, 5 3, 5 4) will be neither (36 ways)

Qs : (1 3) (3 2) (4 2) (5 2)
Ks : (2 1) (2 3) (2 4) (2 5)

This loigic seems to imply, given the next card is either the King or Queen of Spades, it's 50:50.


My gut feeling is that it isn't 50:50, so perhaps the logic has an error in it!!
Ace2
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May 29th, 2024 at 4:23:33 PM permalink
Quote: charliepatrick

Obviously sometimes it won't be either the Qs or Ks as the next card, so we'll ignore those and only look at ones where either one or the other is.
We can then ignore all the other cards in the deck and only consider the order in which the King of Spades and four Queens appear.
Thus we have five objects, "Ks" "Qs" "Qx" "Qy" "Qz" which can be in any order
There are 120 of these (assuming the other Queens are different)
Now consider the position of the King of Spades and the position of the Queen of Spades in the five critical cards
(1 2) Ks Qs Qx Qy Qz will be neither (6 ways)
There are six ways to order the three non-Spade Queens, and similar logic applies to all the other examples.
In this case, it's impossible for either the Ks or Qs to be the "next" card; so we'll call this outcome as "will be neither".

(1 3) Ks Qx Qs Qy Qz will be Qs (6 ways)
In this case the Ks has already appeared, then a non-Spade Queen, then the Queen of Spades.
(1 4) Ks Qx Qy Qs Qz will be neither (6 ways)
In this case the Ks, then a non-Spade Queen, then another non-Spade Queen: hence "neither".
(1 5) Ks Qx Qy Qz Qs will be neither (6 ways)
(2 1) Qs Ks Qx Qy Qz will be Ks (6 ways)
(2 3) Qx Ks Qs Qy Qz will be Ks (6 ways)
(2 4) Qx Ks Qy Qs Qz will be Ks (6 ways)
(2 5) Qx Ks Qy Qz Qs will be Ks (6 ways)
(3 1) Qs Qx Ks Qy Qz will be neither (6 ways)
(3 2) Qx Qs Ks Qy Qz will be Qs (6 ways)
(3 4) Qx Qy Ks Qs Qz will be neither (6 ways)
(3 5) Qx Qy Ks Qz Qs will be neither (6 ways)
(4 2) same logic, will be Qs as (3 ?) (6 ways)
(5 2) same logic, will be Qs as (3 ?) (6 ways)
(4 1, 4 3, 4 5, 5 1, 5 3, 5 4) will be neither (36 ways)

Qs : (1 3) (3 2) (4 2) (5 2)
Ks : (2 1) (2 3) (2 4) (2 5)

This loigic seems to imply, given the next card is either the King or Queen of Spades, it's 50:50.


My gut feeling is that it isn't 50:50, so perhaps the logic has an error in it!!

link to original post



=
Thatís essentially the logic I used. It would be a 60/60 split except when first card is KS and the second is QS. For those six permutations itís a ďtieĒ since neither will come after the first queen, which makes it 60-54-6 in favor of KS

However, this is supposedly the answer to a different problem



Itís all about making that GTA
unJon
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May 29th, 2024 at 4:29:49 PM permalink
Quote: Wizard

Cards in a 52-card deck are turned over in a shuffled deck one at a time until the first queen appears.

What is more likely to be turned over as the next card, the queen of spades or king of spades?
link to original post



If the Ks and Qs are both still to come after the first queen is turned over, then itís equally likely that the next card is Qs or Ks. So the question is whether itís more likely that the Qs is the first queen or that the Ks was turned over before reaching the first Q.

The chance that the first queen turned over is Qs is clearly 25%.

The chance that you will turn over Ks before turning over a Q is 20%.

So itís more likely that the Ks is still to come than the Qs when the first queen is turned over.

Therefore itís more likely that the next card is Ks vs Qs.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard
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May 29th, 2024 at 4:56:43 PM permalink
Keep those answers coming.
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unJon
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May 29th, 2024 at 5:12:22 PM permalink
Quote: Wizard

Keep those answers coming.
link to original post



Is mine wrong?

I thought of another way to conceptualize but it gets to the same result:

From a fresh deck, the chance that the Ks is right after the first Q is 1/51. The chance that the Qs is right after the first Q is 1/51 given the first Q is not the first Q so 1/51 * 3/4 = 1/68.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
gordonm888
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May 29th, 2024 at 5:14:08 PM permalink
Quote: charliepatrick

Obviously sometimes it won't be either the Qs or Ks as the next card, so we'll ignore those and only look at ones where either one or the other is.
We can then ignore all the other cards in the deck and only consider the order in which the King of Spades and four Queens appear.
Thus we have five objects, "Ks" "Qs" "Qx" "Qy" "Qz" which can be in any order
There are 120 of these (assuming the other Queens are different)
Now consider the position of the King of Spades and the position of the Queen of Spades in the five critical cards
(1 2) Ks Qs Qx Qy Qz will be neither (6 ways)
There are six ways to order the three non-Spade Queens, and similar logic applies to all the other examples.
In this case, it's impossible for either the Ks or Qs to be the "next" card; so we'll call this outcome as "will be neither".

(1 3) Ks Qx Qs Qy Qz will be Qs (6 ways)
In this case the Ks has already appeared, then a non-Spade Queen, then the Queen of Spades.
(1 4) Ks Qx Qy Qs Qz will be neither (6 ways)
In this case the Ks, then a non-Spade Queen, then another non-Spade Queen: hence "neither".
(1 5) Ks Qx Qy Qz Qs will be neither (6 ways)
(2 1) Qs Ks Qx Qy Qz will be Ks (6 ways)
(2 3) Qx Ks Qs Qy Qz will be Ks (6 ways)
(2 4) Qx Ks Qy Qs Qz will be Ks (6 ways)
(2 5) Qx Ks Qy Qz Qs will be Ks (6 ways)
(3 1) Qs Qx Ks Qy Qz will be neither (6 ways)
(3 2) Qx Qs Ks Qy Qz will be Qs (6 ways)
(3 4) Qx Qy Ks Qs Qz will be neither (6 ways)
(3 5) Qx Qy Ks Qz Qs will be neither (6 ways)
(4 2) same logic, will be Qs as (3 ?) (6 ways)
(5 2) same logic, will be Qs as (3 ?) (6 ways)
(4 1, 4 3, 4 5, 5 1, 5 3, 5 4) will be neither (36 ways)

Qs : (1 3) (3 2) (4 2) (5 2)
Ks : (2 1) (2 3) (2 4) (2 5)

This loigic seems to imply, given the next card is either the King or Queen of Spades, it's 50:50.


My gut feeling is that it isn't 50:50, so perhaps the logic has an error in it!!

link to original post



Charlie's analysis is very good and gets us a long way there, but I believe he may have overlooked an additional consideration


By ignoring all the cards other than Ks and the four Queens, Charlie has used combination math to calculate the probability that the Ks and/or the Qs are still eligible to be picked and assigned outcomes that are sometimes dependent upon which of the two cards, Ks or Qs, are next in line to be picked first.

But there are other cards, and when one or both of the Ks and Qs are eligible to be drawn after a Queen has appeared, their probability of being drawn is 1/(52-n) where n is the number of cards that were drawn to make the first Q appear. Thus the sequences of (Ks,Qs,Qx,Qy,Qz) have probabilities and outcomes that need to be adjusted to accommodate the fact that when more cards are required to reach the first Q then there are fewer cards left and hence a greater probability of picking either the Qs or Ks if they are still eligible.

The way that this affects Charlies analysis (I think) is this: for those sequences of Charlie's in which the Ks comes out before any of the Queens, they will have an expectation for 'n' slightly larger than those sequences in which the cards drawn before the first queen emerges do not include the Ks. Thus those sequences that have a slightly lower expectation for (52-n) and a slightly higher expectation for 1/(52-n) are those in which the Ks is not eligible to be picked.

Thus, based on Charlie's calculations and perturbing them for this extra bit of logic, I claim that the Qs will be picked slightly more often than the Ks. The quantitative analysis is left as a problem to . . . the WOV math club.
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May 29th, 2024 at 5:31:42 PM permalink
Quote: unJon



Is mine wrong?
link to original post



I'd prefer to see an exchange of ideas on this before I declare the right answer.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
unJon
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May 29th, 2024 at 5:51:10 PM permalink
Quote: Wizard

Quote: unJon



Is mine wrong?
link to original post



I'd prefer to see an exchange of ideas on this before I declare the right answer.
link to original post



Ok in that case I stand by my answer and the ex-ante probabilities in my second post.
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TaxrBux
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May 29th, 2024 at 8:41:27 PM permalink
I ran a simulation, which seems to be converging on


50/50
10000000 trials, 384969 results, 192543 Qs, 192426 Ks

It's the same if I use a 5-card deck, which indicates the other 47 cards are irrelevant:
1000000 trials, 399326 results, 199618 Qs, 199708 Ks

So 40% of the time, Qs or Ks will be the second card in the deck, 20% each. 25% of the time the Qs is second, the Ks will be first. On the other hand, 25% of the time Qs is 3rd, Ks will be first, make the Qs the next card. These numbers correspond to the (1 2) and (1 3) notation from the original poster who proposed 50/50.
ChesterDog
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May 29th, 2024 at 9:11:00 PM permalink
Quote: Wizard

Cards in a 52-card deck are turned over in a shuffled deck one at a time until the first queen appears.

What is more likely to be turned over as the next card, the queen of spades or king of spades?
link to original post




My Excel result has the queen of spades and king of spades equally likely with probabilities of about 0.01923077 each.

Ace2
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May 29th, 2024 at 10:20:33 PM permalink
Quote: TaxrBux

I ran a simulation, which seems to be converging on


50/50
10000000 trials, 384969 results, 192543 Qs, 192426 Ks

It's the same if I use a 5-card deck, which indicates the other 47 cards are irrelevant:
1000000 trials, 399326 results, 199618 Qs, 199708 Ks

So 40% of the time, Qs or Ks will be the second card in the deck, 20% each. 25% of the time the Qs is second, the Ks will be first. On the other hand, 25% of the time Qs is 3rd, Ks will be first, make the Qs the next card. These numbers correspond to the (1 2) and (1 3) notation from the original poster who proposed 50/50.

link to original post

Using a 5-card deck, how are you treating KS first, QS second?
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charliepatrick
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May 30th, 2024 at 1:51:28 AM permalink
I've read the question and my understanding of the question is that you draw cards and stop when the first Queen comes out. Then you peek at the next card, which could be Ks, Qs or something else. Which is more likely, the Ks or Qs?

There's another problem and that is you continue until you either find Ks, Qs or the end of the deck. The second table, in the enclosed spoiler, considers this idea.
I still think the chances are equal and you only have to consider the five key cards. The only difference between what comes as the peeped card, and which comes next if you continue is the non-key cards imbetween. So nevertheless here is another look at it only looking - just look as what comes first, and then second (and sometimes) third. Others don't matter.
Only looking at the peek card..
KsQsnoneTotal
Qs?anyanyany
6
0
18
24
Qx?anyanyany
18
18
36
72
KsQs?anyany
0
0
6
6
KsQx?anyany
0
6
12
18
24
24
72
120

Keep going until you either find Ks or Qs (the only situation you can't is if you see Ks then the first Q = Qs). In the other situations you've either already seen Ks, so will eventually find Qs; or the first Queen was Qs, so you will eventually see Ks, or the first Queen was something else, in which case it's 50:50 whether you see Ks or Qs.
I get that this gives Ks as more likely at Pr=10/19.
KsQsendTotal
Qs?anyanyany
24
0
0
24
Qx?anyanyany
36
36
0
72
KsQs?anyany
0
0
6
6
KsQx?anyany
0
18
0
18
60
54
6
120


Wizard
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May 30th, 2024 at 5:57:38 AM permalink
The correct answer is 50/50. Looks like Charlie gets credit for the first correct response.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
unJon
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May 30th, 2024 at 6:01:51 AM permalink
Quote: Wizard

The correct answer is 50/50. Looks like Charlie gets credit for the first correct response.
link to original post



Hmmmm.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
TaxrBux
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May 30th, 2024 at 6:48:47 AM permalink
Quote: Ace2

Quote: TaxrBux

I ran a simulation, which seems to be converging on


50/50
10000000 trials, 384969 results, 192543 Qs, 192426 Ks

It's the same if I use a 5-card deck, which indicates the other 47 cards are irrelevant:
1000000 trials, 399326 results, 199618 Qs, 199708 Ks

So 40% of the time, Qs or Ks will be the second card in the deck, 20% each. 25% of the time the Qs is second, the Ks will be first. On the other hand, 25% of the time Qs is 3rd, Ks will be first, make the Qs the next card. These numbers correspond to the (1 2) and (1 3) notation from the original poster who proposed 50/50.

link to original post

Using a 5-card deck, how are you treating KS first, QS second?
link to original post


The "next card" is not the Ks or Qs, so it's counted as a trial, but not a result. The results are only counted when the next card is one of the interesting ones.
gordonm888
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May 30th, 2024 at 7:07:06 AM permalink
Quote: Wizard

The correct answer is 50/50. Looks like Charlie gets credit for the first correct response.
link to original post



Well that's been shown by simulation. But simulation is a method that provides answers without insight, it's a lazy approach to math problem solving. It would be nice to see an analytic mathematical proof.

Mathematicians have shown that the Riemann Hypothesis appears to be true over many millions of "trials." But it has not been proven.

Clearly, the solution of 50/50 for 'relative frequency of Ks and Qs following the first queen' would not be true if the deck had only one suit, i.e., spades. What if the deck has 2 suits or 5 suits? Does the 50/50 relative frequency depend upon there being 4 suits?
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unJon
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May 30th, 2024 at 7:58:18 AM permalink
Quote: gordonm888

Quote: Wizard

The correct answer is 50/50. Looks like Charlie gets credit for the first correct response.
link to original post



Well that's been shown by simulation. But simulation is a method that provides answers without insight, it's a lazy approach to math problem solving. It would be nice to see an analytic mathematical proof.

Mathematicians have shown that the Riemann Hypothesis appears to be true over many millions of "trials." But it has not been proven.

Clearly, the solution of 50/50 for 'relative frequency of Ks and Qs following the first queen' would not be true if the deck had only one suit, i.e., spades. What if the deck has 2 suits or 5 suits? Does the 50/50 relative frequency depend upon there being 4 suits?
link to original post



Charlieís post had logic to it under assumption you can ignore cards other than queens and Ks. I still find it very counterintuitive. I just ran his logic through a 2 suited deck and itís also 50/50:


KQsQ - no winner
KQQs - Qs winner
QQsK - Qs winner
QKQs - Ks winner
QsQK - no winner
QsKQ - Ks winner
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
charliepatrick
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May 30th, 2024 at 8:13:02 AM permalink
^ Thanks - yes I'm old school so the exams always asked for your workings, so I always try to include them in my answers. Also it's nice when one sees a short-cut or an "out of left field" approach that's easier to understand than complicated formulae. Since it's in spoiler tags it also gives an insight into the approach I used.

In this case I also came up with a similar, but alternative approach. Interestingly it also gave the same answer for the "next card" only, but a different answer (which agreed with another answer in the thread) where you carried on until you did find an other card. What seems counterintuitive is these two answers would be different.
TaxrBux
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May 30th, 2024 at 8:20:01 AM permalink
Quote: gordonm888

Quote: Wizard

The correct answer is 50/50. Looks like Charlie gets credit for the first correct response.
link to original post



Well that's been shown by simulation. But simulation is a method that provides answers without insight, it's a lazy approach to math problem solving. It would be nice to see an analytic mathematical proof.

Mathematicians have shown that the Riemann Hypothesis appears to be true over many millions of "trials." But it has not been proven.

Clearly, the solution of 50/50 for 'relative frequency of Ks and Qs following the first queen' would not be true if the deck had only one suit, i.e., spades. What if the deck has 2 suits or 5 suits? Does the 50/50 relative frequency depend upon there being 4 suits?
link to original post



For 2 suits, there are only 6 combos. You can add as many jokers (non-spade K, any other non-Q rank) as you want; if a joker is positioned directly after the first Q, that shuffle is not one of interest. The number of jokers before the first Q or in between any of the cards of interest, and the order of the jokers is irrelevant. To convince yourself of this, consider any random permutation of jokers, and any random distribution of the jokers before, between, and after the cards of interest. Whatever that shuffle is, there are exactly 6 ways that the cards of interest can be arranged in the non-joker positions. The same is true for every joker permutation. So for the shuffles where a card of interest directly follows the first Q, the number of joker permutations are equal for each arrangement of cards of interest, and therefore do not affect the probabilities we are looking for.

KsQsQh
KsQhQs - Qs wins
QsKsQh - Ks wins
QsQhKs
QhKsQs - Ks wins
QhQsKs - Qs wins

So for the 2 suit case, it's also 50/50. For the n suit case, P(Ks in position 2) = P(Qs in position 2) = 1/(n+1). When Ks is in position 2, P(Qs in position 1) = 1/n, and when Ks is in position 1, P(Qs in position 3) = 1/n. So again, the probabilities are equal. I think if we asked whether the As, Ks, or Qs is more likely, we would again get equal probabilities. I think as long as the trigger to select the "cut" point does not exhaust the cards of interest, the probabilities for the next card being a specific one will be equal. Examples would be the one suit case or the next card after the 4th Q.
Wizard
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May 30th, 2024 at 11:10:56 AM permalink
I can think of three ways to prove the probabilities are equal. In fact 1/52 for both cards.

Proof #1

Column 1 = Position of first Queen
Column 2 = Probability first Queen is in this position. If c is the position of the first Q, then this probability is permut(48,c-1)/permut(52,c).
Column 3 = Probability next card is Queen of Spades. This is (3/(52-c))*(1/4).
Column 4 = Product of columns 2 and 3
Column 5 = Probability next card is King of Spades. This is always 1/52.
Column 6 = Product of columns 2 and 5

Note the sum of both columns 4 and 6 is 1/52.

Position of first Q Probability first card queen Probability next card is Q of spades Product next card is Q of spades Prob next card is Ks Ks product
1 0.076923 0.014706 0.001131 0.019231 0.001479
2 0.072398 0.015000 0.001086 0.019231 0.001392
3 0.068054 0.015306 0.001042 0.019231 0.001309
4 0.063888 0.015625 0.000998 0.019231 0.001229
5 0.059895 0.015957 0.000956 0.019231 0.001152
6 0.056072 0.016304 0.000914 0.019231 0.001078
7 0.052415 0.016667 0.000874 0.019231 0.001008
8 0.048920 0.017045 0.000834 0.019231 0.000941
9 0.045585 0.017442 0.000795 0.019231 0.000877
10 0.042405 0.017857 0.000757 0.019231 0.000815
11 0.039376 0.018293 0.000720 0.019231 0.000757
12 0.036495 0.018750 0.000684 0.019231 0.000702
13 0.033758 0.019231 0.000649 0.019231 0.000649
14 0.031161 0.019737 0.000615 0.019231 0.000599
15 0.028701 0.020270 0.000582 0.019231 0.000552
16 0.026374 0.020833 0.000549 0.019231 0.000507
17 0.024176 0.021429 0.000518 0.019231 0.000465
18 0.022104 0.022059 0.000488 0.019231 0.000425
19 0.020153 0.022727 0.000458 0.019231 0.000388
20 0.018321 0.023438 0.000429 0.019231 0.000352
21 0.016604 0.024194 0.000402 0.019231 0.000319
22 0.014997 0.025000 0.000375 0.019231 0.000288
23 0.013497 0.025862 0.000349 0.019231 0.000260
24 0.012101 0.026786 0.000324 0.019231 0.000233
25 0.010804 0.027778 0.000300 0.019231 0.000208
26 0.009604 0.028846 0.000277 0.019231 0.000185
27 0.008496 0.030000 0.000255 0.019231 0.000163
28 0.007476 0.031250 0.000234 0.019231 0.000144
29 0.006542 0.032609 0.000213 0.019231 0.000126
30 0.005688 0.034091 0.000194 0.019231 0.000109
31 0.004913 0.035714 0.000175 0.019231 0.000094
32 0.004211 0.037500 0.000158 0.019231 0.000081
33 0.003579 0.039474 0.000141 0.019231 0.000069
34 0.003014 0.041667 0.000126 0.019231 0.000058
35 0.002512 0.044118 0.000111 0.019231 0.000048
36 0.002069 0.046875 0.000097 0.019231 0.000040
37 0.001681 0.050000 0.000084 0.019231 0.000032
38 0.001345 0.053571 0.000072 0.019231 0.000026
39 0.001056 0.057692 0.000061 0.019231 0.000020
40 0.000813 0.062500 0.000051 0.019231 0.000016
41 0.000609 0.068182 0.000042 0.019231 0.000012
42 0.000443 0.075000 0.000033 0.019231 0.000009
43 0.000310 0.083333 0.000026 0.019231 0.000006
44 0.000207 0.093750 0.000019 0.019231 0.000004
45 0.000129 0.107143 0.000014 0.019231 0.000002
46 0.000074 0.125000 0.000009 0.019231 0.000001
47 0.000037 0.150000 0.000006 0.019231 0.000001
48 0.000015 0.187500 0.000003 0.019231 0.000000
49 0.000004 0.250000 0.000001 0.019231 0.000000
Total 1.000000 0.019231 0.019231
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
charliepatrick
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May 30th, 2024 at 12:41:32 PM permalink
I agree with the idea and that it's 1/52, but not that the Pr (Ks) is always 1/52. For instance if the first Queen is in position 49, there's no way to have a Ks afterwards as the rest of the cards are Queens.

Let N=posn of first Queen, then consider how many places the Ks could be.
(i) There are (N-1) possibile positions before the first Q.
(ii) After the first Queen, there are (52-N) places, but three of these contain another Queen.
(iii) Thus, of these (52-N), there are (49-N) places where the Ks might be.
(iv) So the probability the "next slot" is a possible place for Ks is (49-N)/(52-N)
(v) There are 48 possible places for the Ks (given four places are taken by Queens).
(vi) So the net probability is 1/48 * (49-N)/(52-N)

I've used a different method to work out Pr(first Queen) is in posn(N).
The Pr of the first is clearly 4/52.
The pr of the second is the product that there hasn't been a Queen yet * (4/51). I keep a running precentage using things like 1-1/52 etc.
Posn 1st QPr of thatPr next=QsPr next-KsBC (Qs)BD (Ks)
1
.076 923
.014 706
.019 608
.001 131
.001 508
2
.072 398
.015 000
.019 583
.001 086
.001 418
3
.068 054
.015 306
.019 558
.001 042
.001 331
4
.063 888
.015 625
.019 531
.000 998
.001 248
5
.059 895
.015 957
.019 504
.000 956
.001 168
6
.056 072
.016 304
.019 475
.000 914
.001 092
7
.052 415
.016 667
.019 444
.000 874
.001 019
8
.048 920
.017 045
.019 413
.000 834
.000 950
9
.045 585
.017 442
.019 380
.000 795
.000 883
10
.042 405
.017 857
.019 345
.000 757
.000 820
11
.039 376
.018 293
.019 309
.000 720
.000 760
12
.036 495
.018 750
.019 271
.000 684
.000 703
13
.033 758
.019 231
.019 231
.000 649
.000 649
14
.031 161
.019 737
.019 189
.000 615
.000 598
15
.028 701
.020 270
.019 144
.000 582
.000 549
16
.026 374
.020 833
.019 097
.000 549
.000 504
17
.024 176
.021 429
.019 048
.000 518
.000 460
18
.022 104
.022 059
.018 995
.000 488
.000 420
19
.020 153
.022 727
.018 939
.000 458
.000 382
20
.018 321
.023 438
.018 880
.000 429
.000 346
21
.016 604
.024 194
.018 817
.000 402
.000 312
22
.014 997
.025 000
.018 750
.000 375
.000 281
23
.013 497
.025 862
.018 678
.000 349
.000 252
24
.012 101
.026 786
.018 601
.000 324
.000 225
25
.010 804
.027 778
.018 519
.000 300
.000 200
26
.009 604
.028 846
.018 429
.000 277
.000 177
27
.008 496
.030 000
.018 333
.000 255
.000 156
28
.007 476
.031 250
.018 229
.000 234
.000 136
29
.006 542
.032 609
.018 116
.000 213
.000 119
30
.005 688
.034 091
.017 992
.000 194
.000 102
31
.004 913
.035 714
.017 857
.000 175
.000 088
32
.004 211
.037 500
.017 708
.000 158
.000 075
33
.003 579
.039 474
.017 544
.000 141
.000 063
34
.003 014
.041 667
.017 361
.000 126
.000 052
35
.002 512
.044 118
.017 157
.000 111
.000 043
36
.002 069
.046 875
.016 927
.000 097
.000 035
37
.001 681
.050 000
.016 667
.000 084
.000 028
38
.001 345
.053 571
.016 369
.000 072
.000 022
39
.001 056
.057 692
.016 026
.000 061
.000 017
40
.000 813
.062 500
.015 625
.000 051
.000 013
41
.000 609
.068 182
.015 152
.000 042
.000 009
42
.000 443
.075 000
.014 583
.000 033
.000 006
43
.000 310
.083 333
.013 889
.000 026
.000 004
44
.000 207
.093 750
.013 021
.000 019
.000 003
45
.000 129
.107 143
.011 905
.000 014
.000 002
46
.000 074
.125 000
.010 417
.000 009
.000 001
47
.000 037
.150 000
.008 333
.000 006
.000 000
48
.000 015
.187 500
.005 208
.000 003
.000 000
49
.000 004
.250 000
.000 000
.000 001
.000 000
Totals
.019 231
.019 231
gordonm888
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May 30th, 2024 at 2:36:03 PM permalink
Much better discussion! Sorry for being so grumpy early.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
unJon
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May 30th, 2024 at 3:05:48 PM permalink
I figured out the flaw with my initial reasoning.

Yes the Ks comes before the first Q 20% of the time and yes a Qs is the first queen 25% of the time. But 20% of that 25% of the time (5%) the Ks came out before the Q of spades, so it is equally likely that the Qs and the Ks remains in the deck to come after the first Q is turned over.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
ThatDonGuy
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May 30th, 2024 at 3:35:31 PM permalink
Quote: Wizard

Cards in a 52-card deck are turned over in a shuffled deck one at a time until the first queen appears.

What is more likely to be turned over as the next card, the queen of spades or king of spades?
link to original post


Not much sense in putting this in a spoiler if the solution is out in the open.
Anyway, here's my answer, with proof:

There are C(52,4) ways for the four Queens to appear in the deck without regard to suits.

Of these, the number where the first two Queens are consecutive = C(50,2) + C(49,2) + C(48,2) + ... + C(3,2) + C(2,2), which, by the Hockey Stick Theorem, equals C(51,3).
1/4 of these have the Queen of Spades as the second Queen (i.e. the first card after the first Queen).
P(the Queen of spades is immediately after the first Queen) = C(51,3) / C(52,4) x 1/4 = 1/52.

This leaves (C(52,4) - C(51,3)) ways to place four Queens such that the first two are not consecutive. There are 48 cards that can be the card following the first Queen, one of which is the King of Spades.
P(the King of Spades is immediately after the first Queen) = (1 - C(51,3) / C(52,4)) x 1/48 = 12/13 x 1/48 = 1/52.

Therefore, both the Queen of Spades and the King of Spades have probability 1/52 of being the first card after the first Queen.
Last edited by: ThatDonGuy on May 30, 2024
unJon
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May 30th, 2024 at 9:32:26 PM permalink
Hereís my intuitive (for me) way to think about this problem.

Chance Ks is next. Remove Ks from deck and shuffle it. Now that first Queen is in position N. Insert the Ks into a random spot in the shuffled deck and the chance it will be placed in position N+1 is 1/52.

Chance Qs is next. Remove Ks from deck and shuffle it. Now that first Queen (of the 3 still in the deck) is in position N. Insert the Qs into a random spot in the shuffled deck and the chance it will be placed in position N+1 is 1/52. It doesnít matter that you might put the Qs in a position less than N, making a new stopping point. It only matters if the place the Qs goes is N+1.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard
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June 2nd, 2024 at 2:45:46 PM permalink
Here is my proof #2 both probabilities are 1/52.

Let's consider the probability the Qs is next first. Take it out of the deck. There are 51! ways to arrange the other 51 cards. Then put it back in right after the first queen. So, the total combinations where the Qs follows the first queen is 51! The probability the Qs follows the first queen is 51!/52! = 1/52.

Do do that same thing, but remove the Ks. You will get the same probability of 51!/52! = 1/52.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
ThatDonGuy
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June 3rd, 2024 at 4:37:30 PM permalink
Here's a new problem - bad news: it's a "proof" problem.

Interplanetary Express serves 65 different locations throughout the solar system. Every pair of locations is served by a route that is one of four colors - red, blue, green, or yellow. It turns out that, given any three locations, the three routes that connect them to each other are not all the same color.

Somebody wants to add a 66th location, with routes to each of the other 65, and maintaining the "no single-color triangles" policy.
Prove that this is impossible.

(Hint: the fact that you can with 65 is irrelevant; in fact, I am not entirely sure that such a mapping exists.)
Wizard
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June 9th, 2024 at 9:37:03 PM permalink
Sorry Don, I don't even know where to start with your problem. It's probably out of my league.

However, here is one that is in my league.

Two logicians sit down with 13 cards, one of each rank. Aces are high. They shuffle them and each draws a card. The higher card wins.

Logician 1 has the option to offer to trade cards.
If the offer is made, logician 2 may freely accept or decline it.
If both agree to switch cards, they do. Otherwise, they each keep their own.

What is the optimal strategy for each player? What is the probability of each winning assuming both utilize optimal strategy?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
charliepatrick
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June 10th, 2024 at 12:57:02 PM permalink
I canít see logic where thereís a win-win situation, ie both players gain from the swap. I know about the 333333 444411 552222 dice game, so probably need to think more. My feeling is that by making an offer you are giving some info about which cards you might hold, and B would only swap if it was more likely that they held the lower card. I did look at swapping with either an Ace of 2, but this just flipped Ace to a loser and 2 to a winner.
djtehch34t
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June 10th, 2024 at 3:55:34 PM permalink
Quote: ThatDonGuy

Here's a new problem - bad news: it's a "proof" problem.

Interplanetary Express serves 65 different locations throughout the solar system. Every pair of locations is served by a route that is one of four colors - red, blue, green, or yellow. It turns out that, given any three locations, the three routes that connect them to each other are not all the same color.

Somebody wants to add a 66th location, with routes to each of the other 65, and maintaining the "no single-color triangles" policy.
Prove that this is impossible.

(Hint: the fact that you can with 65 is irrelevant; in fact, I am not entirely sure that such a mapping exists.)
link to original post


Assume we have a graph that satisfies the condition (call this 4-colorable). Take the 66th vertex and look at its edges. The color C1 that has the maximum number of edges must have at least ceiling(65/4) = 17 edges. Look at the subgraph with the vertices we get from those edges. This must not have any edges with C1 (aka be 3-colorable), or else we get a cycle with color C1.

Pick an arbitrary vertex in the subgraph of size 17 and repeat the argument. Color C2 must have ceiling(16/3) = 6 edges. So, this creates a 2-colorable subgraph of size 6. Repeat the argument once more to get a 1-colorable subgraph of size ceiling(5/2) = 3. But, this is a contraction, as that's the same as a cycle of the same color.
Mental
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June 10th, 2024 at 5:18:34 PM permalink
Quote: Wizard

Sorry Don, I don't even know where to start with your problem. It's probably out of my league.

However, here is one that is in my league.

Two logicians sit down with 13 cards, one of each rank. Aces are high. They shuffle them and each draws a card. The higher card wins.

Logician 1 has the option to offer to trade cards.
If the offer is made, logician 2 may freely accept or decline it.
If both agree to switch cards, they do. Otherwise, they each keep their own.

What is the optimal strategy for each player? What is the probability of each winning assuming both utilize optimal strategy?
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I assume the shuffle is random and the drawing of the cards is fair with no possibility of either player gaining knowledge of the cards during the draw. I also assume the players look at their own cards (not stated above) before L1 makes any offer.

Both player have a strategy available to them that allows them to win 50% of the time. L1 makes never makes an exchange offer. L2 accepts no offer.

If there exists a strategy that allows L1 to gain an advantage, then L2 has a perfect defense. Reject all offers. Since L1 cannot possibly gain an advantage, L1 will either (1) make no exchange offers or (2) only make the useless offer to exchange all deuces (which will be rejected by L2).
This forum is more enjoyable after I learned how to use the 'Block this user' button.
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