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Following Unfair coins at South Park Elementary

Wendy and Sally have another friend, called Timmy. Although Timmy has no coin, he wants to play this game in a fair condition with them. Wendy does the first coin flip then Sally does. Is it possible to play this game in a fair way for Timmy, even without him having a coin? If so, what should be the values of $p_1$ (Wendy's coin shows head probability) and $p_2$ (Sally's Head probability) that make this game a fair game for all of them?

Note: Timmy cannot use Sally's or Wendy's coin and one of them wins at the end!

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As Timmy has a "head" on his shoulders with probability of 1, he always wins at his turn. Then

Wendy = 1/3 and Sally = 1/2 makes the game fair.

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    $\begingroup$ JonTheMon: What I wanted to write overlapped so much with your solution that I did not want to post another solution. If you do not like my changes, then please accept my apologies and rollback. $\endgroup$ – Gamow Mar 16 '16 at 14:19
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    $\begingroup$ @Gamow Oh dear, the puns. I cringed at it, but it also amused me a lot, so it stands! $\endgroup$ – JonTheMon Mar 16 '16 at 14:26
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    $\begingroup$ Flip. Flip. "TIMMY!" $\endgroup$ – user662852 Mar 17 '16 at 11:21
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Unless I'm missing something

Because Tim has no coin, he can't possibly have an outcome of "heads", therefore his winning chance is always 0.

This means

Wendy and Sally must also have a winning chance of 0 so $p_1 = p_2 = 0$

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    $\begingroup$ Actually you found a solution where noone has chance to win or lose... Please assume that one of them wins every time they play this game so I cannot accept this answer :( $\endgroup$ – Oray Mar 16 '16 at 14:25
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Yes, we can have a game in which

$p_1 = \frac{1}{2}$ and $p_2 =\frac{1}{2}$

So,

We have two tosses, one each from Wendy and Sally.

And

After two turns, the number of heads(with proper ordering) is counted by Timmy.

Thus,

0 heads means Wendy wins, 1 heads (on Sally's coin only) means Sally wins and 2 heads means Timmy wins. Here, each case has probability $\frac{1}{4}.$

Assuming that the results from two coins tosses are independent.

In case of $\{Head, Tail\} $, i.e. Head on Wendy's coin and then Tail on Sally's coin, then we redo the whole two-step process!

In simple words,

$\{Tail, Tail\} =$ Wendy wins.
$\{Tail, Head\} =$ Sally wins.
$\{Head, Head\} =$ Timmy wins.
$\{Head, Tail\} =$ Again two coins are tossed in the same order( Wendy then Sally)

Thus, there's no need of biased coins.

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  • $\begingroup$ Why downvoted, please explain! $\endgroup$ – ABcDexter Mar 16 '16 at 14:44
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    $\begingroup$ I've retracted my downvote, since when it was given there was no mention of a loop. I do think that since there is a chance of an infinite loop, this doesn't satisfy "one of them wins at the end" $\endgroup$ – JonTheMon Mar 16 '16 at 15:16
  • $\begingroup$ Yes, you do have a valid argument. $\endgroup$ – ABcDexter Mar 16 '16 at 15:24
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    $\begingroup$ I think you have typo in your summary at the end $\endgroup$ – Kevin Mar 16 '16 at 17:07
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    $\begingroup$ @ABcDexter of course, that's why there is an order for flipping the coin. $\endgroup$ – Oray Mar 18 '16 at 9:25
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There are 3 kids so for this to be fair, each of them should have the change 1/3.
So Wendy goes first then p1 = 1/3.
Sally goes second.
so Her chance is 2/3 * p2 = 1/3. This results in p2 = 1/2.
And "Timmah" wins if neither Wendy or Sally's coins turn out to be Head.

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If Wendy wins with probability $\dfrac1{p_1}$, then Sally's coin must be $p_2=\dfrac1{p_1-1}$, as Xnor points out.

Any of these values will work, as they agree to split their winnings with Timmy in a $2:1$ ratio.

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