Next card is red General forum
20 replies. Last post: 20170310
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Carroll at 20170303
The game is played between two players with a deck of card in which you added a red joker.
One player turns up cards on after the other, the other player has to announce before you turn a card (and only once before the end of the deck):
“next card is red!”. He gets one point if he is right, otherwise his opponent gets one point.
What is the best strategy and what is the probability of winning for the player announcing “next card is red”?
Remark: it is believed the probability for best strategy in case of a normal deck is .5

Carroll at 20170303
Sorry I should have asked for either the expected value of best strategy (which is 0 for a standard deck) or the probability of winning a game.

ypercube ★ at 20170303
My wild guess is that the probability of winning is 27/53. With any strategy.

ypercube ★ at 20170303
It might be a more interesting problem if the announcement could be either “next card is red” or “next card is black” and the player was allowed to make the announcement when the deck has at least X cards left.
(If X=1, the best strategy, with 100%, would be to wait till the last card)

Carroll at 20170303
@Yper, in that case won’t you just wait until there are X cards left and call the most probable suit?

The_Burglar at 20170303
I would guess a target of the square root of (27/53) for the first move reducing to 0 for the last move in a straight line
make a guess when the target is below the real odds (average of 38th card)
best to first shrink the problem to less cards and then expand learnt results

Tommah ★ at 20170303
I agree with ypercube’s guess. I’m thinking that the odds of winning the game in a deck with R red cards and B black cards are R/(R+B). I’m busy now, but later on I will try to prove it using induction. A strategy will look like a set of constants alpha_R_B, each of which says that when the remaining deck is R red and B blue, you say “next card is red” with probability alpha_R_B. Except for the case R=1 and B=0, these alpha constants should drop out of the equation.

Martyn Hamer at 20170303
I’ve had a go at working out the probability of winning when you adopt the strategy of declaring immediately after drawing the first black card, as this seems a reasonable strategy and not too messy to work out. It seems to come out at 27/53, obviously this isn’t proof that a better strategy doesn’t exist but it does suggest that Ypercube is correct.

Carroll at 20170303
For Ypercube’s game “guess colour of next card before there will be X1 cards left”, let note a position with R red cards, B black cards (R,B,X).
I’ll tackle the X=26 game starting from a standard game of 52 cards, so initial position is (26,26,26).
If you guess now you proba is 26/52=0.5, not so good.
If you draw a card you come to two equivalent positions of (25,26,26) where p=26/51 a little better...
If ever you draw two red cards, something interesting happens:
at (24,26,26) if you call now ‘black!’ you have p=26/50 = .52 but if you draw a card, you go to (23,26,26) = 26/49 with proba 24/50 and to (24,25,26)=25/49 with proba 26/50 and summing these yields (26.24+25.26)/(49.50) = (26.49)/((49.50) = 26/50 so you don’t gain anything more.
So do you stop there with a difference of two or do you continue drawing cards?

Carroll at 20170303
For the initial game you are right, but the fact that probability is constant to R/(R+B) for me should induce the existence of a simple argument showing that this is obvious or maybe just true.
But I can’t find this simple argument, hence the post here on LG where smart people should be able to produce this simple argument...

The_Burglar at 20170303
from a 5 card spreadsheet with thresholds on each card drawn
the average for the 10 permutations will remain at 0.6 no matter what thresholds are used
design 1st section is 1 for red cards, 0 for not
2nd section is chance of red been draw , with thresholds to right
3rd section is reversed start with the 5th drawn
4th drawn if threshold met else 5th drawn etc
average on 1st drawn passed on values (which may = 2nd drawn etc)

Tommah ★ at 20170303
@Carroll: I cannot find a simple argument that is convincing enough to convince me. :) The main idea is that the strategy the player uses does not matter. If I start the game telling myself I am going to choose the 5th card, but then all of a sudden I change my mind and decide to choose the 10th card, my odds of winning are the same in both cases. When the first card is dealt, my strategy of choosing the 10th card is now choosing the 9th card. If I change now, it doesn’t affect my odds of winning. All that affects my odds is the sequence of cards dealt, which I can’t control.

Carroll at 20170303
Yes, not sure if recurrence is an obvious argument but if for any red,black less than r,b we have a probability of next card red red/red+black), then we get turning up one card in (r,b)=
r/(r+b)*(r1)/(r+b1)+b/(r+b)*r/(r+b1)
which simplifies to r(r1+b)/(r+b)(r+b1) or
r/(r+b) QED

Tommah ★ at 20170303
@Carroll: That recurrence was the induction that I had in mind. I realize now that my thing with the alpha constants is unnecessary. At the current turn, you either announce or you don’t announce. If you do, your probability of winning is R/(R+B). If you don’t, then your probability of winning is given by the calculation in your post, so it is again R/(R+B). The choice of move doesn’t matter.

ITR13 at 20170305
With four cards:1/4 chance first card is red 1*1/4 = 1/43/4 chance second card is black1/3 chance second card is red 3/4 * 1/3 = 1/42/3 chance second card is black1/2 chance third card is red 3/4 * 2/3 * 1/2 = 1/41/2 chance third card is black1 chance fourth card is black 3/4 * 2/3 * 1/2 = 1/4
Unless I misunderstand 
ITR13 at 20170305
Wops, my newlines got butchered, here’s a second try: (sorry, first time using this forum)
With four cards:
1/4 chance first card is red 1*1/4 = 1/4
3/4 chance second card is black
1/3 chance second card is red 3/4 * 1/3 = 1/4
2/3 chance second card is black
1/2 chance third card is red 3/4 * 2/3 * 1/2 = 1/4
1/2 chance third card is black
1 chance fourth card is black 3/4 * 2/3 * 1/2 = 1/4

mmKALLL ★ at 20170306
Hmm... Intuitively I would go through the deck, hoping that at some point more black cards are revealed than red ones. Of course that doesn’t make my chances any better. :)
An interesting problem I came up while thinking about this is that if the player turning the cards got to arrange the deck in whatever way they wanted beforehand, what would be the best mixed strategy for both players? I would imagine that random shuffle is the best, but consider a more realistic situation where one can expect some degree of gambler’s fallacy or patternsearching from the person calling red cards. What kinds of configurations would trick a player in realistic situations into lowering their chances of success?

Carroll at 20170306
Hum interesting question, do you have ways of attacking it, maybe on small decks?
I would arrange the deck so that there are more red cards first, for example 4 red cards and then one black (to trick him to call for a fifth red card) and then random order...
For Ypercube’s problem, I’ve made a small python script and the winner for (26,26,26) is... waiting for 26 cards with an exact probability of 82533729461/148835093922 or ~55.45% to win by calling the least seen card or any if 26/26.
This is 156631*526931/(2*3³*29*31*37*41*43*47)

Carroll at 20170310
For the initial “next card is red” game, I’ve been pointed to a reddit post that gives this elegant solution I was looking for, https://www.reddit.com/r/math/comments/22s0bk/red_or_black_xpost_rpoker/cgpu2nt
Look at a different game, where when you say ‘stop’, the BOTTOM card is revealed instead of the next card.
In this game, your winning probability is 1/2 since the bottom card never changes and is determined by the initial state of the deck.
On the other hand, at any point during (either) game, the (conditional) probability that the next card is red equals the probability that the bottom card is red, so this second game is equivalent to the first.