Match52
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December 3rd, 2020 at 3:42:02 PM permalink
A 52 space grid is laid out A-K four times:

Row 1 - A 2 3 4 5 6 7 8 9 10 J Q K
Row 2 - A 2 3 4 5 6 7 8 9 10 J Q K
Row 3 - A 2 3 4 5 6 7 8 9 10 J Q K
Row 4 - A 2 3 4 5 6 7 8 9 10 J Q K

A deck of 52 cards is dealt out entirely onto this grid, Ace to King, beginning with Row 1, without replacement.

>The odds of drawing/matching an Ace on the first card are 1/13.

>What are the odds of matching each successive card, without replacement, WITH THE CONDITION THAT no prior card has matched the space it was placed on the grid?

That is, what are the odds to match each card, 2, 3, 4, all the way to the 52nd card - which would have to be a King to match, where there have been no prior matches?

We came up with this solution
Draw 1 = 1/13
D2-D52 = 1/13*12/13^(D-1)

Can you confirm whether this is correct or not?

Thank you very much for your time!
Last edited by: Match52 on Dec 3, 2020
rsactuary
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December 3rd, 2020 at 4:28:20 PM permalink
does the suit count? I assume not given your calculation of Draw 1.

Also, I assume you didn't mean to show two 8s in each row?
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December 3rd, 2020 at 4:42:29 PM permalink
Quote: Match52

A 52 space grid is laid out A-K four times:

Row 1 - A 2 3 4 5 6 7 8 8 10 J Q K
Row 2 - A 2 3 4 5 6 7 8 8 10 J Q K
Row 3 - A 2 3 4 5 6 7 8 8 10 J Q K
Row 4 - A 2 3 4 5 6 7 8 8 10 J Q K

A deck of 52 cards is dealt out entirely onto this grid, Ace to King, beginning with Row 1, without replacement.

>The odds of drawing/matching an Ace on the first card are 1/13.

>What are the odds of matching each successive card, without replacement, WITH THE CONDITION THAT no prior card has matched the space it was placed on the grid?

That is, what are the odds to match each card, 2, 3, 4, all the way to the 52nd card - which would have to be a King to match, where there have been no prior matches?

We came up with this solution
Draw 1 = 1/13
D2-D52 = 1/13*12/13^(D-1)

Can you confirm whether this is correct or not?

Thank you very much for your time!



This can't be correct since the next card drawn would have 4/51 which doesn't follow your solution.
Math is the only true form of knowledge
unJon
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December 3rd, 2020 at 4:54:25 PM permalink
Ignore
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
CrystalMath
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December 3rd, 2020 at 5:09:30 PM permalink
You're solution comes pretty close, but it is easy to disprove on matching the first 2.

The first card is not an ace, so there is a 4/52 chance it was a 2 and a 44/52 chance of anything besides A or 2.
The probability of matching the first 2 is 4/52*3/51 + 44/52*4/51
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December 3rd, 2020 at 5:10:09 PM permalink
Quote: Match52



>What are the odds of matching each successive card, without replacement, WITH THE CONDITION THAT no prior card has matched the space it was placed on the grid?

That is, what are the odds to match each card, 2, 3, 4, all the way to the 52nd card - which would have to be a King to match, where there have been no prior matches?



Maybe I've had a stroke, but I couldn't seem to understand what you are asking. Do you mean "what are the chances of matching the nth card with the stipulation that none of the n-1 cards have matched their position?"

I agree that the odds of matching an ace on the first card is 1/13.

However the odds of D2, matching the 2nd card given that the first card was not an Ace? I think there are two scenarios:

- the first card was not an ace or two, and the 2nd card is a two. 44*4/(52*51) =176/2652
- the first and 2nd card are both twos c(4,2)/c(52,2) = 6/1326

So the total probability of D2 is the sum of the two terms above which is 188/2652 =.07089. Whereas your formula would yield 1/13 *12/13 = 12/169 = 0.071006.

The piece of information you have overlooked is that when the previous cards do not match their grid spaces there is some chance that they have depleted the number of cards that would match the nth grid space.
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Match52
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December 3rd, 2020 at 5:29:42 PM permalink
Suits do not count and thanks! I fixed the 8 - SMALL FONT!
Match52
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December 3rd, 2020 at 5:31:05 PM permalink
The next card would not be 4/51 because we have to account for the probability that the first card was a 2 (so there will only be 3/51)
Match52
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December 3rd, 2020 at 5:32:52 PM permalink
So the total probability of D2 is the sum of the two terms above which is 188/2652 =.07089. Whereas your formula would yield 1/13 *12/13 = 12/169 = 0.071006.

PRECISELY! So what are the odds of a 3 on 3? We need to provide for multiple possibilities
1) The Game Ended on Ace
2) The Game Ended on 2
3) There are 4 threes
4) There are 3 threes
5 There are 2 threes

How would you factor this one?
I believe once you determine the answer, it will match the formula posted!
Match52
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December 3rd, 2020 at 5:47:35 PM permalink
CrystalMath your solution is right on except we have to provide for this

**WITH THE CONDITION THAT no prior card has matched the space it was placed on the grid?

So your solution has to be multiplied by 1- the chance there was an Ace --that is what the formula provides
Match52
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December 3rd, 2020 at 5:51:07 PM permalink
The formula for 2 would be 4/52*3/51 + 48/52*4/51 BUT we have to provide for the probability that an Ace was drawn which would end the probability of a 2 ever being drawn AT ALL (since no further cards are drawn if there is a match on any previous) - so we have to multiply by 1 minus the chance of an Ace. - The formula provides for this!
Match52
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December 3rd, 2020 at 5:53:50 PM permalink
The formula for 2 would be 4/52*3/51 + 48/52*4/51 BUT we have to provide for the probability that an Ace was drawn which would end the probability of a 2 ever being drawn AT ALL (since no further cards are drawn if there is a match on any previous) - so we have to multiply this probability by 1 minus the chance of an Ace. - The formula provides for this!
Match52
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December 3rd, 2020 at 5:56:49 PM permalink
I am sorry if I was unclear (and If I am posting incorrectly). I sincerely appreciate the help and I am not sure I am correct but I don't think the solutions provided account for the probability that a previous card matched (and thus ended the contest).

Except for this nuance, the formula for 2 would be 4/52*3/51 + 48/52*4/51 BUT we have to provide for the probability that an Ace was drawn which would end the probability of a 2 ever being drawn AT ALL (since no further cards are drawn if there is a match on any previous) - so we have to multiply this probability by 1 minus the chance of an Ace. - The formula provides for this!

Does this make sense/prove out? I can provide the rest of the math or an excel
CrystalMath
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December 3rd, 2020 at 6:03:18 PM permalink
Quote: Match52

The formula for 2 would be 4/52*3/51 + 48/52*4/51 BUT we have to provide for the probability that an Ace was drawn which would end the probability of a 2 ever being drawn AT ALL (since no further cards are drawn if there is a match on any previous) - so we have to multiply this probability by 1 minus the chance of an Ace. - The formula provides for this!



To get a match on 2, there are two ways to get there:
1. First card is not an ace and not a 2, then second card is a 2. =44/52*4/51
2. Both cards are 2. =4/52*3/51

This matches what I said before.
I heart Crystal Math.
Match52
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December 3rd, 2020 at 6:08:06 PM permalink
Absolutely and thank you again sincerely. The other factor is that if there is an Ace on the first draw, there are no subsequent draws.

**WITH THE CONDITION THAT no prior card has matched the space it was placed on the grid**

So we have to factor in the probability of an Ace on the first card ending the game.
That means the formula for 2 is (4/52*3/51 + 48/52*4/51 ) * (1-4/52).

For 3 we have to add in the probability of a Ace (which ends the draws on Ace) or a 2 (which ends the draws on 2) and so on
Match52
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December 3rd, 2020 at 6:32:29 PM permalink
You phrased this better than I had

"what are the chances of matching the nth card with the stipulation that none of the n-1 cards have matched their position?"

That is exactly the question but I believe the solution for 2 is (48/52*4/51 + 4/52*3/51) * (1-4/52).

We have to provide for the stopping of the game if there is an Ace.

Does this seem correct?
CrystalMath
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December 3rd, 2020 at 7:24:52 PM permalink
Quote: Match52

You phrased this better than I had

"what are the chances of matching the nth card with the stipulation that none of the n-1 cards have matched their position?"

That is exactly the question but I believe the solution for 2 is (48/52*4/51 + 4/52*3/51) * (1-4/52).

We have to provide for the stopping of the game if there is an Ace.

Does this seem correct?



I can see your thinking, but your formula is incorrect. For the first part, you must calculate the conditional probabilities that the first card is a 2 or not given that the selected card was not an ace. The conditional probability of not being a 2 is 44/48 and the conditional probability of it being a 2 is 4/48. When you put these in, you will see that it simplifies to my previous solution:
(44/48*4/51 + 4/48*3/51) * (1-4/52)
=(44/48*4/51 + 4/48*3/51) * (48/52)
=(44/52*4/51 + 4/52*3/51)
=0.070889894

I also ran a simulation of 5 billion rounds, and get the following results:
Card Number Probability
10.0769153747058823
20.0708959829411765
30.0653428296078431
40.0602346490196078
50.0555321645098039
60.0512043825490196
70.0472218988235294
80.0435617060784314
90.0401810262745098
100.0370697531372549
110.0342054219607843
120.0315671260784314
130.0291355278431373
140.0274571731372549
150.0253012731372549
160.0233237278431373
170.0215010392156863
180.0198182801960784
190.018274458627451
200.0168560174509804
210.015547888627451
220.0143461131372549
230.0132320680392157
240.0122098090196078
250.0112663558823529
260.0103982415686275
270.00980000215686275
280.00903099823529412
290.00832462294117647
300.00767371509803922
310.00707545450980392
320.0065239737254902
330.00601478705882353
340.0055511431372549
350.00511578823529412
360.00472130901960784
370.00435657647058824
380.00402158039215686
390.00371179588235294
400.00349901235294118
410.00322404392156863
420.00297092843137255
430.00274013098039216
440.00252641921568627
450.00232914549019608
460.00214757019607843
470.00198080176470588
480.00182649901960784
490.00168572
500.00155507490196078
510.00143643392156863
520.00132484529411765
I heart Crystal Math.
ssho88
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December 4th, 2020 at 7:00:54 AM permalink
Quote: CrystalMath

I can see your thinking, but your formula is incorrect. For the first part, you must calculate the conditional probabilities that the first card is a 2 or not given that the selected card was not an ace. The conditional probability of not being a 2 is 44/48 and the conditional probability of it being a 2 is 4/48. When you put these in, you will see that it simplifies to my previous solution:
(44/48*4/51 + 4/48*3/51) * (1-4/52)
=(44/48*4/51 + 4/48*3/51) * (48/52)
=(44/52*4/51 + 4/52*3/51)
=0.070889894

I also ran a simulation of 5 billion rounds, and get the following results:

Card Number Probability
10.0769153747058823
20.0708959829411765
30.0653428296078431
40.0602346490196078
50.0555321645098039
60.0512043825490196
70.0472218988235294
80.0435617060784314
90.0401810262745098
100.0370697531372549
110.0342054219607843
120.0315671260784314
130.0291355278431373
140.0274571731372549
150.0253012731372549
160.0233237278431373
170.0215010392156863
180.0198182801960784
190.018274458627451
200.0168560174509804
210.015547888627451
220.0143461131372549
230.0132320680392157
240.0122098090196078
250.0112663558823529
260.0103982415686275
270.00980000215686275
280.00903099823529412
290.00832462294117647
300.00767371509803922
310.00707545450980392
320.0065239737254902
330.00601478705882353
340.0055511431372549
350.00511578823529412
360.00472130901960784
370.00435657647058824
380.00402158039215686
390.00371179588235294
400.00349901235294118
410.00322404392156863
420.00297092843137255
430.00274013098039216
440.00252641921568627
450.00232914549019608
460.00214757019607843
470.00198080176470588
480.00182649901960784
490.00168572
500.00155507490196078
510.00143643392156863
520.00132484529411765




Prob D3 = P(XX3) + P(3X3) + P(X33) + P(333) = (44 * 43* 4 + 4 * 44* 3 + 44 * 4* 3 + 4 * 3 * 2)/52/51/50 = 8648/132600 = 0.065218702, which is very close to your simulation results.

Question : Is possible to find a general formula for Dn ?
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December 4th, 2020 at 8:16:07 AM permalink
Match52, notice that much of what you have posted (that is correct) was posted earlier by me.
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December 4th, 2020 at 8:48:14 AM permalink
Quote: ssho88


Prob D3 = P(XX3) + P(3X3) + P(X33) + P(333) = (44 * 43* 4 + 4 * 44* 3 + 44 * 4* 3 + 4 * 3 * 2)/52/51/50 = 8648/132600 = 0.065218702,



To calculate P(XX3) you must calculate that
1. "the first X is not ace AND the second X is not 2"
2. "the first two XX are not both 2"

I think you are missing the 2nd of these terms.

Quote: ssho88


Question: Is possible to find a general formula for Dn ?



Great question! Clearly it is probably possible to write a recursive formula. I imagine it will be possible to write a finite series expression for Dn,using maybe using nested summation functions " Σ "? Is there a cleaner way to write an analytical expression?????
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ssho88
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December 4th, 2020 at 9:43:34 AM permalink
gordonm888,

1. "the first X is not ace AND the second X is not 2"
2. "the first two XX are not both 2"

I think you are missing the 2nd of these terms.

I still don't get it, can you please show me how to calculate the value of D3 ?

EDITED

I think P(XX3) should be = (40*43*4 + 4*44*4)/132600 = 7584/132600

P(3X3) = 4*44*3/132600 = 528/132609
P(X33) = 528/132600
P(333) = 24/132600

Prob D3 = 8664/132609 = 0.065339367 ?
Last edited by: ssho88 on Dec 4, 2020
CrystalMath
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December 4th, 2020 at 11:46:34 AM permalink
Quote: ssho88

gordonm888,

Prob D3 = 8664/132609 = 0.065339367 ?



yes
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CrystalMath
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December 4th, 2020 at 11:48:24 AM permalink
I wrote a recursive program to calculate this. I'm going to stop here, because it really takes a lot of time doing this method. Anyhow, here are the results for the first 10 spots, which are very close to my simulation.

Card Number Probability
10.0769230769230769
20.0708898944193062
30.0653393665158372
40.0602320928371344
50.0555319563723086
60.0512058440398016
70.0472233915504426
80.0435567504101312
90.0401803752151958
100.0370708275356189
I heart Crystal Math.
USpapergames
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December 4th, 2020 at 12:06:11 PM permalink
Quote: Match52

A 52 space grid is laid out A-K four times:

Row 1 - A 2 3 4 5 6 7 8 9 10 J Q K
Row 2 - A 2 3 4 5 6 7 8 9 10 J Q K
Row 3 - A 2 3 4 5 6 7 8 9 10 J Q K
Row 4 - A 2 3 4 5 6 7 8 9 10 J Q K

A deck of 52 cards is dealt out entirely onto this grid, Ace to King, beginning with Row 1, without replacement.

>The odds of drawing/matching an Ace on the first card are 1/13.

>What are the odds of matching each successive card, without replacement, WITH THE CONDITION THAT no prior card has matched the space it was placed on the grid?

That is, what are the odds to match each card, 2, 3, 4, all the way to the 52nd card - which would have to be a King to match, where there have been no prior matches?

We came up with this solution
Draw 1 = 1/13
D2-D52 = 1/13*12/13^(D-1)

Can you confirm whether this is correct or not?

Thank you very much for your Time!



Can someone please re-explain this to me in clear English? I can come up with the correct formula (if there is 1)! All I can tell at this point is there is no way the formula provided will equal the probabilities of matching any card draws.
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December 4th, 2020 at 3:36:54 PM permalink
Quote: USpapergames

Can someone please re-explain this to me in clear English? I can come up with the correct formula (if there is 1)! All I can tell at this point is there is no way the formula provided will equal the probabilities of matching any card draws.



"what are the chances of matching the nth card with its position with the stipulation that none of the n-1 cards have matched their position?"
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USpapergames
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December 4th, 2020 at 4:47:38 PM permalink
Quote: gordonm888

"what are the chances of matching the nth card with its position with the stipulation that none of the n-1 cards have matched their position?"



(48/52)(4/51)
(48/52)(47/51)(4/50)
(48/52)(47/51)(46/50)(4/49)
(48/52)(47/51)(46/50)(45/49)(4/48)
(48/52)(47/51)(46/50)(45/49)(44/48)(4/47)

Is this the pattern you're referring to? I think I can find a formula to express this but I don't want to waste my time if this is not the question asked.
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Match52
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December 8th, 2020 at 7:01:30 PM permalink
Crystal Math - any chance you can provide a formula for 3, 4, and 5? I am stuck!

Thank you!!!
Match52
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December 8th, 2020 at 9:12:19 PM permalink
Hello and thank you so much for your time. I truly appreciate it.

So far I think we have
A= 1/13
2 = ((44/48)(4/51)+(4/48)(3/51)) (1-the probability of A)

I am not sure how to get to 3 - Can you assist?
Once I see the pattern I will spend the time running out all the numbers.

THANK YOU!

Quote: CrystalMath

I wrote a recursive program to calculate this. I'm going to stop here, because it really takes a lot of time doing this method. Anyhow, here are the results for the first 10 spots, which are very close to my simulation.

Card Number Probability
10.0769230769230769
20.0708898944193062
30.0653393665158372
40.0602320928371344
50.0555319563723086
60.0512058440398016
70.0472233915504426
80.0435567504101312
90.0401803752151958
100.0370708275356189

Match52
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December 11th, 2020 at 6:49:18 PM permalink
I will work on this. Thanks!
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