This seems to fit best in the Nerdery, even though it doesn't really fit anywhere...

If I rolled a 26-sided die four times, what are the odds of me rolling the same number three out of the four times?

aggie_wes said:

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How many times is that number on the die?

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Once (numbers 1-26 are on the die)

0.00021883

(1/26)*(1/26)*(1/26)*(1/26)

Mathguy64 said:

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For any given number (say a "1") 3 of 4 times it's c(4,3)((1/26)^3)*((25/26)^1)

Since there are 26 possible numbers you can have this happen to multiply that by 26.

0.005690

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Username checks out! Thank you so much.

hypeiv said:

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Are you asking the probability of rolling the same number exactly 3 out of 4 times or "at least" 3 out of 4 times?

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Exactly 3 of 4.

Multiplying by 26 in the explanations above is how you account for this. If you were trying to find the probability of a specific number you wouldn't multiply by 26

My thinking would be as follows...

First dice roll is irrelevant, it merely sets the baseline original number

The second dice roll divides us into some different possible paths.

Second dice roll has a 1/26 chance of being the same as the first. In this case, the only possible number that can hit three times is the original number and you have two more throws to hit it again. You either hit it again on the third roll (1/26 * 1/26) or you hit it on the fourth after missing the third (1/26 * 25/26 * 1/26).

So the total probability of this path is (1/26 * 1/26) + (1/26 * 25/26 * 1/26) = 51/17576

The second dice roll also has a 25/26 chance of being different from the first roll. This sets up two different possible paths where either of the two numbers previously rolled could get rolled twice in a row. So the third roll has a 2/26 chance of hitting one of the two possible winning numbers, and then the last roll must be 1/26 to match the third roll.

The probability of this path is 25/26 * 2/26 * 1/26 = 50/17576

So by my count the probability of rolling a 26 sided dice and getting the same number to come up three times in less than four rolls is 101/17576 or 0.005746

I'm just glad that this wasn't a super obvious and easy answer that everyone would make fun of me for not knowing!

Azariah said:

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I always do this the hard way.

There are 26*26*26*26= 456976 possible rolls in this scenario.
Three 1's exist in 26+26+26+26 = 104 of them (1,1,1,1-26), (1,1,1-26,1), etc.
104/456976=0.00022758
There are 26 total numbers that can hit three out of four, so multiply by 26.

Equals 0.00591715 or 0.591715% chance.

I think the difference between mine and the total formula way is that I'm counting 1,1,1,1 as a valid three out of four. Since OP didn't specify 3/4 and NOT 4/4, I will stick with it.

Just double-checked, and yep, that's correct.

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You are quadruple counting your (1,1,1,1) combo because it appears in all 4 of your sets. There are 26+25+25+25= 101 sets of throws with at least three ones. After fixing that your odds will agree with mine i think.

This is a binomial random variable. Y'all are way overthinking this.

Mathguy64 said:

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This is a binomial random variable. Y'all are way overthinking this.

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OK, in that case..

Mathguy64 said:

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For any given number (say a "1") 3 of 4 times it's c(4,3)((1/26)^3)*((25/26)^1)

Since there are 26 possible numbers you can have this happen to multiply that by 26.

0.005690

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but if hitting 4 out of 4 also counts, then you need to add in:

c(4,4)((1/26)^4)) multiplied by 26 which would add 0.00005690 to your 0.005690 and then you would get 0.005746

It's still a binomial RV. If X = number of 1's seen in 4 rolls, you calculate P(X=3) or P(X>=3). Either way X is Binomial with n=4 and p=1/26.

And OP indicated it was exactly 3 out of 4. But yes for 3 or 4 out of 4 P(X>=3) = P(X=3) + P(X=4) since X is discrete and the intersection is empty.

Got another one for ya Ags...

What are the odds of pulling 6 hearts from a standard deck of cards BEFORE drawing
5 spades
4 diamonds or
4 clubs?
(If 5 spades, 4 diamonds, or 4 clubs are drawn the game is over)

Counterpoint said:

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Got another one for ya Ags...

What are the odds of pulling 6 hearts from a standard deck of cards BEFORE drawing
5 spades
4 diamonds or
4 clubs?
(If 5 spades, 4 diamonds, or 4 clubs are drawn the game is over)

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I'm not very good at writing polynomial notation so here is my hack at this:

The game will conclude after no fewer than 4 cards are drawn and no more than 16 cards drawn.

For Clubs to win in 4 cards there are 13*12*11*10 combinations or 17160 combinations.

Same applies for diamonds so there are a total of 34320 favorable combinations for Clubs or Diamonds and zero favorable combinations for Spades or Hearts.

For a 5 card draw, the winning combinations for Clubs are any 4 Clubs drawn plus 1 card of the 39 non-Clubs in the deck or (13*12*11*10*39). Same for Diamonds. But the winning combination for Spades is 5 Spades and no other cards so there are (13*12*11*10*9) combinations for Spades. There are still zero winning combinations for Hearts.

For a 6 card draw, the winning combinations for Clubs are any 4 Clubs drawn plus 2 cards from the remaining non-Clubs in the deck or (13*12*11*10*39*38). Same for Diamonds. But the winning combination for Spades is 5 Spades and 1 other card so there are (13*12*11*10*9*39) combinations for Spades. There are finally some (13*12*11*10*9*8) winning combinations for Hearts.

The pattern continues through all 16 card draws resulting in 77369947111822800000000 unfavorable outcomes for Hearts and 2948510699370930000000 favorable outcomes for Hearts.

Hearts wins 3.81% of the time.

lb3 said:

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Counterpoint said:

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Got another one for ya Ags...

What are the odds of pulling 6 hearts from a standard deck of cards BEFORE drawing
5 spades
4 diamonds or
4 clubs?
(If 5 spades, 4 diamonds, or 4 clubs are drawn the game is over)

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I'm not very good at writing polynomial notation so here is my hack at this:

The game will conclude after no fewer than 4 cards are drawn and no more than 16 cards drawn.

For Clubs to win in 4 cards there are 13*12*11*10 combinations or 17160 combinations.

Same applies for diamonds so there are a total of 34320 favorable combinations for Clubs or Diamonds and zero favorable combinations for Spades or Hearts.

For a 5 card draw, the winning combinations for Clubs are any 4 Clubs drawn plus 1 card of the 39 non-Clubs in the deck or (13*12*11*10*39). Same for Diamonds. But the winning combination for Spades is 5 Spades and no other cards so there are (13*12*11*10*9) combinations for Spades. There are still zero winning combinations for Hearts.

For a 6 card draw, the winning combinations for Clubs are any 4 Clubs drawn plus 2 cards from the remaining non-Clubs in the deck or (13*12*11*10*39*38). Same for Diamonds. But the winning combination for Spades is 5 Spades and 1 other card so there are (13*12*11*10*9*39) combinations for Spades. There are finally some (13*12*11*10*9*8) winning combinations for Hearts.

The pattern continues through all 16 card draws resulting in 77369947111822800000000 unfavorable outcomes for Hearts and 2948510699370930000000 favorable outcomes for Hearts.

Hearts wins 3.81% of the time.

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This is amazing, thank you. Crazy that just a few extra cards lower the odds for hearts that much!

Each roll of the die is unique and independent of any other roll.

The chances of getting any particular number for each roll are 1/26

The chances of getting any number EXCEPT any particular number are 25/26

Thus the chances of getting the same number 3 out of 4 rolls is:

(1/26)*(1/26)*(1/26)*(25/26)=0.0000547=0.00547%

92Ag95 said:

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Each roll of the die is unique and independent of any other roll.

The chances of getting any particular number for each roll are 1/26

The chances of getting any number EXCEPT any particular number are 25/26

Thus the chances of getting the same number 3 out of 4 rolls is:

(1/26)*(1/26)*(1/26)*(25/26)=0.0000547=0.00547%

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And if you add in the chances of getting all four the same:
+(1/26)*(1/26)*(1/26)*(1/26)
you get .0000569, the same answer as Mathguy64 above.