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The problem I am thinking of is to find the expected number of tosses needed to get either 3 heads or 3 tails. It needs not be consecutive, e.g., Head->Head->Tail->Head is a possibility.

For a fair or biased coin, this is bounded by 3 moves minimum and 5 moves maximum. For a fair coin, the expected number of tosses is 4.125.

The solution states "A bias towards either heads or tails is a bias in favor of MORE OF THE SAME. Given that the stopping rule is 3 of the same, the expected number of tosses to end the game must decrease." I verified this statement by deriving the equation for a biased coin and indeed saw that the expected number of tosses peaks for a fair coin and falls when biased in either direction.

But I would like to get a better intuition of this without the help of equations. I only understand the solution in part, in that yes, the bias will favor of more of the same for heads if biased towards heads, but it also comes at the cost of less of the same for tails. So the solution statement seems to imply that more of the same for heads outweighs the less of the same for tails, hence reducing the expected value, but how does one determine this intutively without equations?

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    $\begingroup$ If it always comes up heads, only $3$ tosses are required. As the probability of a tail increases, it sometimes requires a fourth toss, or even a fifth. $\endgroup$ – saulspatz Apr 4 '20 at 16:12
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Using equations informally, let us think about the chance of starting with three of the same. If the probability of heads is $p$, the chance of three in a row is $p^3+(1-p)^3$. For $p=\frac 12$, each term is $\frac 18$ and the total is $\frac 14$. If $p$ gets large, $p^3$ by itself can be greater than $\frac 14$. If $p=1, p^3=1$ and you are sure to get three heads. I think this gives a feel for the fact that having the probabilities imbalanced makes it more likely to get lots of the same.

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  • $\begingroup$ Ah I see. So basically we want to increase the weights (probabilities) on the 3 tosses, and the further we deviate away from $p=0.5$, the more we weight the 3 tosses, hence giving an expected value closer to 3? $\endgroup$ – roulette01 Apr 4 '20 at 18:19
  • $\begingroup$ Now that we're introducing informal equations. We can see that the derivative with respect to $p$ of $p^3 + (1-p)^3$ is $6-3p$. So the probability weights on the 3 tosses is minimized when $p=\frac{1}{2}$ $\endgroup$ – roulette01 Apr 4 '20 at 18:43
  • $\begingroup$ Yes, both of those are correct. $\endgroup$ – Ross Millikan Apr 4 '20 at 20:12

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