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I played a 'board' game with the Mrs the other day. I would love to what the odds were of what happened when we were determining who went first.


The game consists of 106 tiles. 8 sets of numbers 1-13 and 2 'jokers'. These were all placed facedown on the table in random order, and we pick one each to see who has the higher number and thus gets to start.


This time however we both picked the same number (as each other, not literally the same number) on 3 successive occasions. Eventually at the 4th attempt, who should start was decided.


I think the odds on this happening must be huge, but I don't know how huge. Anyone know? Quids might if he reads this!


PS Each time we selected a tile of the same number they were returned to the table, so we were always selecting from a total of 106.

Ah, Rummikub!


(8/106) x (8/106) x (8/106)


There's an 8 in 106 chance of drawing the same numbered tile (you can sort of play around with the % a bit if 1 person draws first from a pool of 106 then the second person draws from the remaining pool of 105 but that's just getting OTT, we'll assume you're both reaching in at exactly the same time). To both draw a Joker would be a 2/106 chance. The Jokers complicate things a tiny bit further but not enough to really mess with that ballpark figure, technically it's an 8 in 104 chance of drawing the same numbered tile PLUS a 2 in 106 chance of drawing a Joker but for rough estimate and easier maths purposes...


Each draw is independent of the previous / next (again, you can complicate it if the tile is placed back and you can see it / remember it but we'll assume a blind draw).


Multiply by 100 to give percentage and it's 0.04% chance. Roughly 1 in 2500 chance.

Hi, I'm not sure about this. I think the above if for one person selecting the same number three times.


My calculations would be:

The chance of taking any one of the same number the first time would be 8/106 x 7/105 (ok, I'm allowing for the lower pool)

This can happen 13 different ways so the chance for the first draw is (8/106 x 7/105) x 13


For three times in a row this number is cubed, so the chance would be ((8/106 x 7/105) x 13)^3 = 0.027%



Rummikub is great

Rachel043 Wrote:

-------------------------------------------------------

> Yes but for each selection, they drew at the same

> time, more or less, so the second person couldn't

> select the first person's tile.

>

> After that they were replaced.


Yes so there's 8 5s in the pack, person A picks a 5 and now there's only 7 in the pack for Person B to match them.

thinking this out -

(1) you get a card that isn't a joker (pretty certain)

(2) your partner has a chance of matching that card (unlikely)


reset and repeat.


(104/106 x 7/105 )3


Ooh but you might match on the jokers too :)


so including jokers


(104/106 x 7/105) + (2/106 x 1/105) all powered to 3


Bet I'm well wrong LOL

seenbeen Wrote:

-------------------------------------------------------

> Alan Medic wrote

> 'PS Each time we selected a tile of the same

> number they were returned to the table, so we were

> always selecting from a total of 106.'


Is it 105 cards on the table when the partner picks a card ?


Both cards are replaced on the table after both have picked a card and they compare them to see if one wins ?.

JohnL Wrote:

-------------------------------------------------------

> seenbeen Wrote:

> --------------------------------------------------

> -----

> > Alan Medic wrote

> > 'PS Each time we selected a tile of the same

> > number they were returned to the table, so we

> were

> > always selecting from a total of 106.'

>

> Is it 105 cards on the table when the partner

> picks a card ?

>

> Both cards are replaced on the table after both

> have picked a card and they compare them to see if

> one wins ?.


Yes. Tiles though not cards.

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