Match52
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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Match52
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?
USpapergames
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Match52
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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Match52
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
I heart Crystal Math.
gordonm888
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Match52
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
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
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
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
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

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