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Paper Scissors Rock


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With (1/5) probability the second person matches the pick of the first. Draw.
With (4/5) probability the first two picks produce a win.

Say R and S were picked: the second contest has satchels containing P1 P2 S R.

With (1/6) probability P1 P2 are picked. Draw
With (5/6) probability the second two picks produce a win.

With (4/5)(5/6) = (2/3) probability there were two wins.
With (1/2)(2/3) = (1/3) probability they were won by the same person.

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I get 17/60.

If you win the first match (2/5 chance), you have 5/12 chance of winning the second match, equaling 1/6 chance of winning the first two (WWx).

If you tie the first match (1/5 chance), you have 1/3 chance of winning the second match. Of those 1/3 possibilities, you then have 1/2 chance to win the third match. The final probability of TWW is 1/30.

If you lose the first match (2/5 chance), you have 5/12 chance of winning the second match. Of those 5/12 possibilities, you then have 1/2 chance to win the third match. The final probability of LWW is 1/12.

1/6 + 1/30 +1/12 = 17/60

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Does the OP want to know what the chances are of you as a player winning two in a row are? Or what the chances are of either player winning two in a row?

If it's just the chance you will win two in a row, I'll stick with 17/60.

If it's the chance of either player to win two in a row, I'll double that to 17/30.

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Looks like dark magician got this one.

I forgot to exclude the LWT scenario from my LWW calculations. I mistakenly thought the chance of a win after LW was 1/2, but neglected to realize that is only the case 4/5 of the time.

LWW = 2/5 x 5/12 x (4/5 x 1/2) = 1/15

TWW = 1/5 x 1/3 x 1/2 = 1/30

WW? = 2/5 x 5/12 = 1/6

1/15 + 1/30 + 1/6 = 4/15 chance one specific player getting two wins in a row

(x2) = 8/15 chance either player gets two wins in a row

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The OP is not restrictive, 1/9 is for 2 contest wins in a row.

Possible "2 win in a row" posted above are all correct lateral thinking,
but the first thing that comes in mind is that- a contestant winnning
the 1st and 2nd pickings...or 2nd and 3rd pickings is what OP ask.
If the P S R P S R are on satchels no. 1 2 3 4 5 6 respectively
and contestants pick order are c1 c2 c1 c2 c1 c2,
win for c1 has 12 ways (13,16,21,24 . . . 65)for 1st pick;
win2 for c1 has 5 ways (excluding picked satchels) for 2nd pick;
there are 2 ways for pick 3 (ex. 1324w/l,1346t/t,1354w/l,1362w/l,1365w/l)
thus 12x5x2 = 120 for win-win-?,while for ?-win-win it is the same 120 ways.
All ways are equal to permut(6,6)=720 , Chance=240/720 or 1/3
but there is no pick 3 after win-win ...

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you have a 2/5 chance of winning the first match.

whether you won lost or drew, there are now 4 satchels left, and thus a 2/4 chance of winning.

whether you won lost or drew the second match, player 1 has either a gauranteed win or a gauranteed draw, with a 50% chance of either.

so i'd say 9/20 chance of winning 2 in a row for player 1.

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The OP does not say there is a third contest, but I can see how it's implied.

"In a row" is redundant if not.

No second contest neither,just best of 3 picks because there are six satchels.. sometimes out of the box can be away from the box.

but i figured a third is the right chance of winning 2 in a row for the lucky contestant..

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