Odds of matching each of 52 cards
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!
Also, I assume you didn't mean to show two 8s in each row?
Quote: Match52A 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.
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
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.
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!
**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
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
Quote: Match52The 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.
**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
"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?
Quote: Match52You 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 |
|---|---|
| 1 | 0.0769153747058823 |
| 2 | 0.0708959829411765 |
| 3 | 0.0653428296078431 |
| 4 | 0.0602346490196078 |
| 5 | 0.0555321645098039 |
| 6 | 0.0512043825490196 |
| 7 | 0.0472218988235294 |
| 8 | 0.0435617060784314 |
| 9 | 0.0401810262745098 |
| 10 | 0.0370697531372549 |
| 11 | 0.0342054219607843 |
| 12 | 0.0315671260784314 |
| 13 | 0.0291355278431373 |
| 14 | 0.0274571731372549 |
| 15 | 0.0253012731372549 |
| 16 | 0.0233237278431373 |
| 17 | 0.0215010392156863 |
| 18 | 0.0198182801960784 |
| 19 | 0.018274458627451 |
| 20 | 0.0168560174509804 |
| 21 | 0.015547888627451 |
| 22 | 0.0143461131372549 |
| 23 | 0.0132320680392157 |
| 24 | 0.0122098090196078 |
| 25 | 0.0112663558823529 |
| 26 | 0.0103982415686275 |
| 27 | 0.00980000215686275 |
| 28 | 0.00903099823529412 |
| 29 | 0.00832462294117647 |
| 30 | 0.00767371509803922 |
| 31 | 0.00707545450980392 |
| 32 | 0.0065239737254902 |
| 33 | 0.00601478705882353 |
| 34 | 0.0055511431372549 |
| 35 | 0.00511578823529412 |
| 36 | 0.00472130901960784 |
| 37 | 0.00435657647058824 |
| 38 | 0.00402158039215686 |
| 39 | 0.00371179588235294 |
| 40 | 0.00349901235294118 |
| 41 | 0.00322404392156863 |
| 42 | 0.00297092843137255 |
| 43 | 0.00274013098039216 |
| 44 | 0.00252641921568627 |
| 45 | 0.00232914549019608 |
| 46 | 0.00214757019607843 |
| 47 | 0.00198080176470588 |
| 48 | 0.00182649901960784 |
| 49 | 0.00168572 |
| 50 | 0.00155507490196078 |
| 51 | 0.00143643392156863 |
| 52 | 0.00132484529411765 |
Quote: CrystalMathI 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 1 0.0769153747058823 2 0.0708959829411765 3 0.0653428296078431 4 0.0602346490196078 5 0.0555321645098039 6 0.0512043825490196 7 0.0472218988235294 8 0.0435617060784314 9 0.0401810262745098 10 0.0370697531372549 11 0.0342054219607843 12 0.0315671260784314 13 0.0291355278431373 14 0.0274571731372549 15 0.0253012731372549 16 0.0233237278431373 17 0.0215010392156863 18 0.0198182801960784 19 0.018274458627451 20 0.0168560174509804 21 0.015547888627451 22 0.0143461131372549 23 0.0132320680392157 24 0.0122098090196078 25 0.0112663558823529 26 0.0103982415686275 27 0.00980000215686275 28 0.00903099823529412 29 0.00832462294117647 30 0.00767371509803922 31 0.00707545450980392 32 0.0065239737254902 33 0.00601478705882353 34 0.0055511431372549 35 0.00511578823529412 36 0.00472130901960784 37 0.00435657647058824 38 0.00402158039215686 39 0.00371179588235294 40 0.00349901235294118 41 0.00322404392156863 42 0.00297092843137255 43 0.00274013098039216 44 0.00252641921568627 45 0.00232914549019608 46 0.00214757019607843 47 0.00198080176470588 48 0.00182649901960784 49 0.00168572 50 0.00155507490196078 51 0.00143643392156863 52 0.00132484529411765
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 ?
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?????
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 ?
Quote: ssho88gordonm888,
Prob D3 = 8664/132609 = 0.065339367 ?
yes
| Card Number | Probability |
|---|---|
| 1 | 0.0769230769230769 |
| 2 | 0.0708898944193062 |
| 3 | 0.0653393665158372 |
| 4 | 0.0602320928371344 |
| 5 | 0.0555319563723086 |
| 6 | 0.0512058440398016 |
| 7 | 0.0472233915504426 |
| 8 | 0.0435567504101312 |
| 9 | 0.0401803752151958 |
| 10 | 0.0370708275356189 |
Quote: Match52A 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.
Quote: USpapergamesCan 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?"
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.
Thank you!!!
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: CrystalMathI 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 1 0.0769230769230769 2 0.0708898944193062 3 0.0653393665158372 4 0.0602320928371344 5 0.0555319563723086 6 0.0512058440398016 7 0.0472233915504426 8 0.0435567504101312 9 0.0401803752151958 10 0.0370708275356189
