Probability of getting ten unique numbers

Math and probability wasn't ever my strong side, so I need to ask for a help in calculating simple (as I assume) value. Here is situation description.

Player A throws seven times with ten-side dice (numbers in range 1-10). Writes down results and shows them to player B. Player B throws three times with the same dice.

What is the probabillity that among all ten numbers (both players) there will be no repeats - i.e. all three player B's numbers will be exactly different than all player A's numbers?

This comes from a simple game, that I've been playing. Developer of this game suggests that above mentioned situation can appear fairly often (sometimes 2-3 times per particular game) completely random. While I'm pretty sure, that probability of such situation is so extremely low, that it can't happen that often, without any changes to game random generator, to make game itself much harder to complete. In other words -- I claim that game's random generator is not that random.

Thank you in advance for any help.

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It's about counting. How many possible throw sequences with 10 unique numbers (hint 10!). How many total possible throw sequences? –  Guy Sirton Jan 4 '13 at 20:55

As noted above, if you meant "no repeats among the 10 rolls", then we get $\frac{10!}{10^{10}} \approx 0.00036$.

However, if you're only requiring that none of A's numbers is rolled by B, then we can solve using inclusion-exclusion to get $\frac{1}{10^{10}}(10(9^7) + 270(8^7) + 720(7^7)) \approx 0.1207$

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I set up a Monte Carlo simulation mimicking this setup, and found that the probability of Player B's 3 rolls being different from any of Player A's 7 rolls is 12.1%.

Assumptions: - Equal probability of rolling 1-10 (10% each) - 1,000,000 trials

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Welcome to Math.SE! Thank you for sharing your experimental data, which is in perfect agreement with Michael Biro's answer. –  user53153 Jan 5 '13 at 8:09

This is just a bit of an explanation of the answers above:

This experiment is known as sampling with replacement, i.e. in a sense you 'return' numbers into the pool. The probability that the first number is unique is clearly 1. The second time it is $\frac{9}{10}$, then $\frac{8}{10}$, etc. If we denote the event that we got all unique numbers $A$, and each $k$'th number $A_k$, the result is the product of probabilities of independent events: $$P(A)=P(A_1)P(A_2)\ldots P(A_{10})=1 \cdot \frac{9}{10} \cdot \frac{8}{10} \cdots \frac{1}{10}=\frac{10!}{10^{10}}=0.00036288$$

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The probability of getting $6$ unique numbers in 6 consecutive throws of a regular dice would be $$\frac{6\times5\times4\times3\times2\times1}{6\times6\times6\times6\times6\times6}=\frac{720}{46,656}=.0154$$ or $1.5\%$ or approx $1$ in $65$ tries. Statistically it will happens once every $65$ tries.
But with $10$ numbers, it's about once every $2755$ tries a lot harder to get.