Intuition with probabilities of tossing coins Say I flipped $10$ unfair coins (with probabilities of tails $51\%$ and heads $49\%$). I want to know the probability of having at least $M$ tails given that there were at least $M$ identical results.
I need your help explaining the results.
After I done the calculations I used different $M$ values and watched the results, I have noticed that for numbers smaller than $5$ the probability is very high (with around $99\%$ when $M=1$ and $64\%$ when $M=5$)
and then when $M=6$ the probability drops to around $53\%$ and slowly going up until $M=10$ with around $59\%$
This result is very confusing for me since when $M$ is $1$ its almost $100\%$ then it decreases and then $M=6$ is the minimum and then slowly increasing again.
 A: With no condition, you expect the most likely number of tails to be $5$ and for it to decay in both directions, since $p$ is quite close to $1/2$. The condition does nothing for $M \leq 5$, but it changes things a bit for $M>5$. For example with $M=6$ you remove the case of $5$ tails; with $M=7$ you remove the case of $4$, $5$ or $6$ tails; etc. This makes the chances of having $M$ tails or more more likely because it removes the "middle" possibilities, i.e. the possibilities $x$ with $x$ and $10-x$ both less than $M$.
A: The intuition is that if $M$ is less than $6$ then you can have both $M$ heads and $M$ tails, and for $M$ very small you almost certainly do. 
For $M\geq 6$, you know that you either have $M$ tails or $M$ heads, but not both. Since tails are more likely than heads, the probability of having $M$ tails will be a bit more than $50\%$ for $M=6$. As you increase $M$, the number of extra coins of the more common type will increase, and the fact that tails are individually more likely will have a greater effect.
Suppose you instead plotted the probability of not having $M$ heads, given there are at least $M$ of the same type. This will look much more like you expect -- it just increases. The key point is that the two probabilities are equal for $M\geq 6$, but different for $M\leq 5$ (because you can have $M$ of both).
A: This question is quite interesting.
I do the calculations and get the same results.

For each $M$, the conditional probability is the sum of the numbers in green cells divided by the sum of the numbers in coloured cells.
It is not strange that for smaller $M$, the conditional probability is nearly $1$, as most coloured cells are coloured green. For $M\ge 6$, the number of green cells and the number of yellow cells are the same. The conditional probability is close to $0.5$. It is slightly larger than $0.5$ since the coin is biased.

Remark:
The following table shows the calculation when the coin is not biased.

