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Ten Out Of Tens


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At last week's friendly game it was our most right-brained member and least likely, statistically speaking, to instigate a prop bet who did just that. Rob got right to the point and didn't even give the opportunity for anyone to bow out. Though no one of course complained. "Vern, cut this shuffled deck. Ten dollars says there won't be a ten in the first ten cards." Ten quick cards flipped; $10 to Rob. The same result happened in turn to Coop, Huck, and myself. We played several rounds and although I managed to break even, Rob cleaned up. We all figured the sample space was just too small for this result to be truly representative but still I need to know, who had the advantage?

EDIT: additional challenge added

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There are (52 choose 10) possibilities for the first 10 cards in the deck. Each of these is equally likely. Of them, there are (48 choose 10) possible sets that don't have any tens. Thus, Rob's total probability of winning is (48 choose 10)/(52 choose 10), which comes out to about 41.3%.

Are you sure nothing funny was going on?

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Wouldn't the probility of not gtting a ten in the first 10 cards be something like 48/52 * 47/51 ... 38/42 ... or roughly 37%? (It has been quite a while since I took statistics) Then he plays the game 3 times seems like he should only win 1 of those on the average. Also is the deck being reshuffled between each play? If it isn't his odds get better since multiple 10's can be removed.

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Wouldn't the probility of not gtting a ten in the first 10 cards be something like 48/52 * 47/51 ... 38/42 ... or roughly 37%? (It has been quite a while since I took statistics) Then he plays the game 3 times seems like he should only win 1 of those on the average. Also is the deck being reshuffled between each play? If it isn't his odds get better since multiple 10's can be removed.

Your expression has 11 terms; you've gone one step too far. If you stop at 39/43, you get the same thing that I have.

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There are (52 choose 10) possibilities for the first 10 cards in the deck. Each of these is equally likely. Of them, there are (48 choose 10) possible sets that don't have any tens. Thus, Rob's total probability of winning is (48 choose 10)/(52 choose 10), which comes out to about 41.3%.

Are you sure nothing funny was going on?

Hmm. Not what I would have expected. I still think Rob has the advantage though. In a probabilistically (is that a real word?) perfect world, Rob wins about 41% of the time. His opponent wins 59% of the time. However, it sounds like he is rotating through four people who are splitting the 59%. So he wins 41% of the time while each of the other four win 15%? (of all of the games, not what they played). It makes sense to me right now. Not sure I'll be saying the same thing tomorrow morning. . .

:wacko:
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If one of the four tens is removed from the pack, then probability of no tens in ten deals is 51.2%. 63.7% is the result if two of four tens are up the sleeve

:mad:

hmm - a fair point. maybe Rob was more clever than we gave him credit for. Short sleeves for him on poker night from now on.

For Thalia's direction, let me ask:

What is the average number of cards drawn before a ten appears?

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yes indeed, mathboy. that's correct for the average number or cards flipped before a ten appears. and SeaCalMaster's answer to the OP is also right. and as SCM mentioned, Blavek's approach also works.

maybe i shouldn't bother pushing this but how can both be (mathboy's and SCM's)?

and Thalia, a hint - think symmetrically. maybe mathboy will reflect some more on his answer.

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I calculated the probability that the first 10 would appear for each draw 1-49 (a 10 has to appear before the 50th draw.

1 draw - 4/52

2nd draw - (1-4/52)*4/51 : The prob of not getting 10 on the first draw times the conditional probability of getting 10 on the 2nd draw given not getting it on the first.

3rd draw - keep following the logic of above.

etc.

Finally, multiply the draw number by its probability and sum the products for the answer - 10.6.

There is probably an easier, more elegant method, but I can't think of it.

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yes indeed, mathboy. that's correct for the average number or cards flipped before a ten appears. and SeaCalMaster's answer to the OP is also right. and as SCM mentioned, Blavek's approach also works.

maybe i shouldn't bother pushing this but how can both be (mathboy's and SCM's)?

and Thalia, a hint - think symmetrically. maybe mathboy will reflect some more on his answer.

I wondered this myself plainglazed. Without thinking too hard about it, I think it is like comparing a median of sorts and an average.

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I calculated the probability that the first 10 would appear for each draw 1-49 (a 10 has to appear before the 50th draw.

1 draw - 4/52

2nd draw - (1-4/52)*4/51 : The prob of not getting 10 on the first draw times the conditional probability of getting 10 on the 2nd draw given not getting it on the first.

3rd draw - keep following the logic of above.

etc.

Finally, multiply the draw number by its probability and sum the products for the answer - 10.6.

There is probably an easier, more elegant method, but I can't think of it.

Think about it this way:

Define a given (non-ten) card's position in relation to the tens. There are five places it could be; is any one of them more likely than any other?

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52 cards in deck, 4 tens.

Odds of drawing a 10 on Card 1: 48/52

Card 2: 47/51

Card 3: 46/50

...

Card 10: 39/43

Multiply all those together (some terms cancel, yay :)) to get: (42*41*40*39)/(52*51*50*49) = 41.3% tada!

AAAWWWOOOOO! (wolf howl)

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